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Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 6 Mechanical Properties of Solids Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 6 Mechanical Properties of Solids

Question 1.
Explain the equilibrium state of solids at a given temperature.
Answer:

1. Solids are made up of atoms or a group of atoms placed in a definite geometric arrangement.
2. This arrangement is decided by nature so that the resultant force acting on each constituent due to others is zero. This is the equilibrium state of a solid at room temperature.
3. This equilibrium arrangement does not change with time but it can only change when an external stimulus, like compressive force is applied to a solid from all sides.
4. The constituents vibrate about their equilibrium positions even at very low temperature but cannot leave their fixed positions.
5. As a result, the solids possess a definite shape and size.

Question 2.
Explain the effects of applied force on a rigid body.
Answer:

1. When an external force is applied to a solid, the constituents are slightly displaced and restoring forces are developed in it.
2. These restoring forces try to bring the constituents back to their equilibrium positions so that the solid can regain its shape.
3. When the deforming forces are removed, the inter-atomic forces tend to restore the original positions of the molecules and thus the body regains its original shape and size.

Question 3.
Explain the concept of deforming force with the help of examples.
Answer:

1. When a force (within specific limit) is applied to a solid (which is not free to move), the size or shape or both change due to changes in the relative positions of molecules. Such a force is called deforming force.
2. The larger the deforming force on a body, the larger is its deformation.
3. Deformation could be in the form of change in length of a wire, change in volume of an object or change in shape of a body.
4. Examples:
   • When a deforming force such as stretching, is applied to a rubber band, it gets deformed (elongated) but when the force is removed, it regains its original length.
   • When a similar force is applied to a dough or a clay, it also gets deformed but it does not regain its original shape and size after removal of the deforming force.

Question 4.
Define plasticity.
Answer:
If a body does not regain its original shape and size and retains its altered shape or size upon removal of the deforming force, it is called a plastic body and the property is called plasticity.

Question 5.
When is a body said to be perfectly elastic? Give an example for perfectly elastic body and perfectly plastic body.
Answer:

1. If a body regains its original shape and size completely and instantaneously upon removal of the deforming force, then it is said to be perfectly elastic.

• • There is no solid which is perfectly elastic or perfectly plastic,
  • The best example of a near perfectly elastic body is quartz fibre and that of a plastic body is putty.

Question 6.
State SI unit and dimensions of strain.
Answer:
Strain is the ratio of two similar quantities. Hence strain is a dimensionless physical quantity and it has no units.

Question 7.
State and explain longitudinal stress (Tensile stress and Compressive stress).
Answer:
i. Stress produced by a deforming force acting along the length of a body or a rod is called longitudinal stress.

ii. Tensile stress: Consider a force F is applied along a length of a wire, or perpendicular to its cross-section A of a wire (or along its length).

[Note: Elongation of wire ∆l is exaggerated for explanation.]
This produces an elongation in the wire and the length of the wire increases, then

iii. Compressive stress: When a rod is pushed from two ends with equal and opposite forces.

[Note: Compression of wire ∆l is exaggerated for explanation.]
This restoring force per unit area is called compressive stress.
Compressive stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\)

Question 8.
State and explain longitudinal strain (Tensile strain or Linear strain).
Answer:
The strain produced by a tensile deforming force is called longitudinal strain (tensile strain or linear strain,).

where, L = Original length,
∆l = Change in length.

Question 9.
State and explain volume stress / hydraulic stress.
Answer:
When a deforming force acting on a body produces change in its volume, the stress is called volume stress.
Volume stress/hydraulic stress:

[Note: Change in size is exaggerated for explanation.]

1. Let \(\vec{F}\) be a force acting perpendicular to the entire surface of the body.
2. It acts normally and uniformly all over the surface area A of the body.
3. Such a stress which produces change in size but no change in shape is called volume stress.
   Volume stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\)
4. Volume stress produces change in size without change in shape of body, hence it is also called as hydraulic or hydrostatic volume stress.

Question 10.
Explain volume strain.
Answer:

1. A deforming force acting perpendicular to the entire surface of a body produces a volume strain.
2. 
   where ∆V = change in volume,
   V = original volume.

Question 11.
Define and explain shearing stress.
Answer:
The restoring force per unit area developed due to the applied tangential force is called shearing stress or tangential stress.
Shearing stress:

1. When a deforming force acting on a body produces change in the shape of a body shearing stress is produced.
2. Consider ABCD as a front face of a cube, a tangential force is applied to the cube so that the bottom of the cube is fixed and only the top surface is slightly displaced.
   

Question 12.
Explain shearing strain.
Answer:

1. The relative displacement of the bottom face and the top face of the cube is called shearing strain.
2. Shearing strain = \(\frac{\Delta l}{\mathrm{~L}}\) = tan θ
   where, ∆l = displaced length,
   L = Original length.
3. When ∆l is very small,
   tan θ ≈ θ and shearing strain = θ

Question 13.
State and explain Hooke’s law.
Answer:
Statement: Within elastic limit, stress is directly proportional to strain.
Explanation:

1. According to Hooke’s law,
   Stress-Strain
   
   This constant of proportionality is called modulus of elasticity.
2. Modulus of elasticity of a material is the slope of stress-strain curve in elastic deformation region and depends on the nature of the material.
3. The graph of strain (on X-axis) and stress (on Y-axis) within elastic limit is shown in the figure.
   

Question 14.
Define elastic limit.
Answer:
The maximum value of stress upto which stress is directly proportional to strain is called the elastic limit.

Question 15.
Define modulus of elasticity.
Answer:
The modulus of elasticity of a material is the ratio of stress to the corresponding strain.

Question 16.
State different types of modulus of elasticity.
Answer:

1. Young’s modulus (Y): It is the modulus of elasticity related to change in length of an object like a metal wire, rod, beam, etc., due to the applied deforming force.
2. Bulk modulus (K): It is the modulus of elasticity related to change in volume of an object due to applied deforming force.
3. Shear modulus or Modulus of rigidity (η): The modulus of elasticity related to change in shape of an object is called modulus of rigidity.

Question 17.
Explain the usefulness of Young’s modulus.
Answer:

1. Young’s modulus indicates the resistance of an elastic solid to elongation or compression.
2. Young’s modulus of a material is useful for characterization of an object subjected to compression or tension.

Question 18.
Within elastic limit, prove that Young’s modulus of material of wire is the stress required to double the length of wire.
Answer:

(i) Let, L = Initial length of wire
2 L = Final length of wire
∴ Increase in length = ∆l = 2L – L = L

(ii) Longitudinal strain of wire = \(\frac{\Delta l}{\mathrm{~L}}=\frac{\mathrm{L}}{\mathrm{L}}=1\)

(iii)

∴ Y = Longitudinal stress

(iv) Hence, Young’s modulus of material of wire is the stress required to double the length of wire.

Question 19.
What is bulk modulus? Derive an expression for bulk modulus.
Answer:
Definition:
Bulk modulus is defined as the ratio of volume stress to volume strain.
It is denoted by ‘K’.
Unit: N/m² or Pa in SI system
Dimensions: [L^-1M¹T^-2]
Expression for bulk modulus:

(i) If a sphere made from rubber is completely immersed in a liquid, it will be uniformly compressed from all sides.
Let, F = Compressive force,
dP = Change in pressure,
dV = Change in volume,
V = Original volume.

(ii) Volume stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\) = dP

(iii) The negative sign indicates that there is a decrease in volume.
The magnitude of the volume strain is \(\frac{\mathrm{dV}}{\mathrm{V}}\)

Question 20.
Explain the usefulness of bulk modulus.
Answer:

1. Bulk modulus indicates the resistance of gases, liquids or solids to change their volume.
2. Materials with small bulk modulus and large compressibility are easier to compress.

Question 21.
State few applications of bulk modulus.
Answer:

1. When a balloon is filled with air at high pressure, its walls experience a force from within. It tries to expand the balloon and change its size without changing shape. When the volume stress exceeds the limit of bulk elasticity, the balloon explodes.
2. A gas cylinder explodes when the pressure inside it exceeds the limit of bulk elasticity of its material.
3. A submarine when submerged under water is under volume stress. The depth it can reach within water depends upon its limit of bulk elasticity.

Question 22.
Define compressibility. State its unit and dimensions.
Answer:

1. The reciprocal of bulk modulus of elasticity is called compressibility of the material.
   
2. Compressibility is the fractional decrease in volume, (-dV/V) per unit increase in pressure.
   Compressibility = \(\frac{-\mathrm{dV}}{\mathrm{V} \mathrm{dP}}\)
3. Unit: m²/N or Pa^-1 in SI system.
4. Dimensions: [L¹M^-1T²]

Question 23.
What is modulus of rigidity? Derive an expression for it.
Answer:
Definition: It is defined (is the ratio of shear stress to shear strain within elastic limit.
It is denoted by ‘η’.
Unit: N/m² or Pa in SI system.
Dimension: [L^-1M¹T^-2]

Expression for modulus of rigidity:

Consider a solid cube as shown in the figure.

Let, F = Tangential force
A = Cross sectional area
∆l = Relative displaced length
l = Original length
θ = Shear strain

[Note: Displacement of upper surface is exaggerated for explanation.]
The forces applied on the block is subjected to a shear stress,
Shearing stress = F/A
The comer angle which changes by a small amount θ (expressed in radian) is given by,
Shearing strain = \(\frac{\Delta l}{l}\) ≈ θ
From definition,

Question 24.
What does modulus of rigidity indicate?
Answer:
Modulus of rigidity indicates resistance offered by solid to change in its shape.

Question 25.
Explain the change of diameter of a wire when it is stretched and compressed.
Answer:
i. When a wire is fixed at one end and a force is applied at its free end so that the wire gets stretched, length of the wire increases and at the same time, its diameter decreases, i.e., the wire becomes longer and thinner as shown in figure.

[Note: Linear expansion ∆l is exaggerated far explanation.]

ii. If equal and opposite forces are applied to an object along its length inwards, the object gets compressed. There is a decrease in dimensions along its length and at the same time there is an increase in its dimensions perpendicular to its length. When length of the wire decreases, its diameter increases.

[Note: Compression of wire ∆l is exaggerated for explanation.]

Question 26.
Derive expression for Poisson’s ratio.
Answer:
i. Let,
L = original length
l = increase/decrease in length
D = original diameter
d = change in diameter

ii. The ratio of change in dimensions to original dimensions in the direction of the applied force is called linear strain.
Linear strain = \(\frac{l}{\mathrm{~L}}\) …. (1)

iii. The ratio of change in dimensions to original dimensions in a direction perpendicular to the applied force is called lateral strain.
Lateral strain = \(\frac{\mathrm{d}}{\mathrm{D}}\) ….(2)

iv. Poisson’s ration,

Question 27.
A wire of length 20 m and area of cross section 1.25 × 10^-4 m² is subjected to a load of 2.5 kg. (1 kg wt = 9.8 N). The elongation produced in wire is 1 × 10^-4 m. Calculate Young’s modulus of the material.
Solution:
L = 20 m, A = 1.25 × 10^-4m²,
F = mg = 2.5 × 9.8 N, l = 10^-4 m
Formula Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
To find: Young’s modulus (Y)
Calculation: From formula,
Y = \(\frac{2.5 \times 9.8 \times 20}{1.25 \times 10^{-4} \times 10^{-4}}\) = 3.92 × 10¹⁰Nm^-2
Answer: The Young’s modulus of the material is 3.92 × 10¹⁰ Nm^-2.

Question 28.
A wire of diameter 0.5 mm and length 2 m is stretched by applying a force of 2 kg wt. Calculate the increase in length of the wire, (g = 9.8 m/s², Y = 9 × 10¹⁰ N/m²)
Solution:
Given:
L = 2 m, F = 2 kg wt, d = 0.5 mm = 5 × 10^-4 m,
∴ r = \(\frac{\mathrm{d}}{2}\) = 2.5 × 10^-4 m.
Y = 9 × 10¹⁰ N/m², g = 9.8 m/s²
To find: Increase in length (l)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{MgL}}{\pi \mathrm{r}^{2} l}\)
Calculation:
From formula,

Answer:
The increase in length of the wire is 2.218 × 10^-3 m.

Question 29.
A brass wire of length 4.5 m with cross-section area of 3 × 10^-5 m² and a copper wire of length 5.0 m with cross section area 4 × 10^-5 m² are stretched by the same load. The same elongation is produced in both the wires. Find the ratio of Young’s modulus of brass and copper.
Solution:
L_B = 4.5 m, A_B = 3 × 10^-5 m²
L_C = 5 m, A_C = 4 × 10^-5 m² l_B = _C, F_B = F_C
To find: Ratio of Young’s modulus \(\left(\frac{Y_{B}}{Y_{C}}\right)\)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Calculation: For brass,

= 1.2
Answer:
The ratio of Young’s modulus of brass and copper is 1.2 : 1.

Question 30.
The length of wire increases by 9 mm when weight of 2.5 kg is hung from the free end of wire. If all conditions are kept the same and the radius of wire is made thrice the original radius, find the increase in length.
Solution:
Given; l₁ = 9mm = 9 × 10^-3m,
M = 2.5 kg, r₂ = 3r₁,
Y₁ = Y₂ = Y (material is same)
To find: Increase in length (l₂)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Calculation: From formula,

Answer:
The longitudinal strain produced in 1st wire is

Question 31.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in the figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.(Y_s = 2.0 × 10¹¹ Nm^-2, Y_B = 0.91 × 10¹¹ Nm^-2) (NCERT)

Solution:
Given: D = 0.25 cm = 0.25 × 10^-2 m,
L_S = 1.5 m, L_B = 1 m,
Y_S = 2.0 × 10¹¹ Nm^-2,
Y_B = 0.91 × 10¹¹ Nm^-2
To find: Elongations of brass wire (l_B)
Elongations of steel wire (l_S)

Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)

Calculation: Since, A = \(\frac{\pi \mathrm{D}^{2}}{4}\)

Answer:
The elongation of the steel wire is 1.5 × 10^-4 m and that of brass wires is 1.32 × 10^-4 m.

Question 32.
One end of steel wire is fixed to a ceiling and load of 2.5 kg is attached to the free end of the wire. Another identical wire is attached to the bottom of load and another load of 2.0 kg, is attached to the lower end of this wire as shown in the figure. Compute the longitudinal strain produced in both the wires, if the cross-sectional area of wires is 10^-4m², (Y_steel = 20 × 10¹⁰N/m²)

Solution:
Given: M₁ = 2.5 kg, M₂ = 2kg, A = 10^-4m², Y_steel = 20 × 10¹⁰ N/m², L₁ = L₂ = L
To find: Longitudinal strain of 1^st wire (Strain₁).
Longitudinal strain of 2^nd wire (Strain₂)

Answer:
The longitudinal strain produced in 1^st wire is 1.225 × 10^-6 and in 2^nd wire is 2.205 × 10^-6

Question 33.
A steel wire having cross-sectional area 2 mm² is stretched by 10 N. Find the lateral strain produced in the wire. (Given: Y for steel = 2 × 10¹¹ N/m², Poisson’s ratio σ = 0.29)
Solution:
Given: A = 2 mm² = 2 × 10^-6 m²,
F = 10 N, Y_stee; = 2 × 10¹¹ N/m², σ = 0.29

To find: Lateral strain
Formulae:

Calculation: From formula (i),
longitudinal stress = \(\frac{10}{2 \times 10^{-6}}\)
= 5 × 10⁶ N/m²
From formula (ii),

From formula (iii),
lateral strain = σ × longitudinal strain
= 0.29 × 2.5 × 10^-5
∴ lateral strain = 7.25 × 10^-6
Answer:
Lateral strain produced in the wire is 7.25 × 10^-6.

Question 34.
A brass wire of radius 1 mm is loaded by a mass of 31.42 kg. What would be the decrease in its radius? (Y = 9 × 10¹⁰ N/m², Poisson’s ratio σ = 0.36)
Solution:
Given: R = 1 mm = 1 × 10^-3 m, M = 31.42 kg, Y = 9 × 10¹⁰ N/m², σ = 0.36
To find: Decrease in radius (r)

Formulae:

i. Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
ii. σ = \(\frac{\mathrm{Lr}}{l \mathrm{R}}\)

Calculation: From formula (i),

Answer:
The decrease in radius of the brass wire would be 3.92 × 10^-7 m

Question 35.
A metal cube of side 1 m is subjected to a force. The force acts normally on the whole surface of cube and its volume changes by 1.5 × 10^-5 m³. The bulk modulus of metal is 6.6 × 10¹⁰ N/m². Calculate the change in pressure.
Solution:
Given: L = 1 m, dV = 1.5 × 10^-5 m³,
K = 6.6 × 10¹⁰ N/m².
To find: Change in pressure (dP)
Formula: K = V\(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,

Answer:
The change in pressure is 9.9 × 10⁵ N/m².

Question 36.
Determine the volume contraction of solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ P.a. (Bulk modulus of copper = 140 × 10⁹ pa)
Solution:
Given: L = 10 cm =0.1 m, ∆P = 7 × 10⁶ Pa, K = 140 × 10⁹ pa
To find: Volume contraction (dV)
Formula: K = V × \(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,
dV = \(\mathrm{L}^{3} \times \frac{\mathrm{dP}}{\mathrm{K}}\) = (0.1)³ × \(\frac{7 \times 10^{6}}{140 \times 10^{9}}\)
∴ dV = 5 × 10^-8 m³
Answer:
The volume contraction of solid copper cube is 5 × 10^-8 m³

Question 37.
Calculate the modulus of rigidity of a metal, if a metal cube of side 40 cm is subjected to a shearing force of 2000 N. The upper surface is displaced through 0.5 cm with respect to the bottom. Calculate the modulus of rigidity of the metal.
Solution:
Given: L = 40 cm = 0.4 m,
F = 2000 N = 2 × 10³N
l = 0.5 cm = 0.005 m, A = L² = 0.16 m²
To find: Modulus of rigidity (η)

Formulae:

i. θ = \(\frac{l}{\mathrm{~L}}\)
ii. η = \(\frac{\mathrm{F}}{\mathrm{A} \theta}\)

Calculation:

From formula (i),
θ = \(\frac{0.005}{0.4}\) = 0.0125
From formula (ii),
η = \(\frac{2 \times 10^{3}}{0.16 \times 0.0125}\) = 1 × 10⁶ N/m²
Answer:
The modulus of rigidity of the metal cube is 1 × 10⁶ N/m².

Question 38.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N forces producing only elastic deformation. Calculate the resulting strain. (Rigidity modulus of copper = 42 × 10⁹ N m^-2)
Solution:
Given: A = 15.2 × 19.14 × 10^-6 m²,
F = 44500 N, η = 42 × 10⁹ N m^-2
To find: Strain (θ)
Formula: η = \(\frac{\mathrm{F}}{\mathrm{A} \theta}\)

Calculation:
From formula,

Answer:
The strain produced in the piece of copper is 3.64 × 10^-3

Question 39.
A copper metal cube has each side of length 1 m. The bottom edge of cube is fixed and tangential force 4.2 × 10⁸ N is applied to top surface. Calculate the lateral displacement of the top surface, if modulus of rigidity of copper is 14 × 10¹⁰ N/m².
Solution:
Given: L = h = 1 m, F = 4.2 × 10⁸ N, η = 1.4 × 10¹¹ N/m²
To find: Lateral displacement (x)
Formula: η = \(\frac{\mathrm{F}}{\mathrm{A} / \theta}\) = \(\frac{\mathrm{Fh}}{\mathrm{Ax}}\)
Calculation: From formula,
x = \(\frac{\mathrm{Fh}}{\mathrm{A\eta}}\)
∴ x = \(\frac{4.2 \times 10^{8} \times 1}{(1 \times 1) \times 1.4 \times 10^{11}}\) = 3 × 10^-3 m
∴ x = 3 mm
Answer:
The lateral displacement of top is 3 mm.

Question 40.
The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.0 15 mm. Find the strain and shearing force. (Modulus of rigidity: η = 4.5 × 10¹⁰ N/m²)
Solution:
Given: A = 0.5 m × 0.5 m = 0.25 m²,
h = 1 cm = 10^-2m,
x = 0.015 mm = 15 × 10^-6m
η = 4.5 × 10¹⁰ N/m²
To find: Strain (θ). Shearing force (F)

Formulae:

i. θ = \(\frac{\mathrm{x}}{\mathrm{h}}\)
ii. F = ηAθ

Calculations:

Using formula (i),
θ = \(\frac{15 \times 10^{-6}}{10^{-2}}\) = 1.5 × 10^-3
Using formula (ii),
F = 4.5 × 10¹⁰ × 0.25 × 1.5 × 10^-3
= 1.688 × 10⁷ N
Answer:
Shearing force is 1.688 × 10⁷ N and strain is 1.5 × 10^-3

Question 41.
Explain the behaviour of metal wire under increasing load.
Answer:
Consider a metal wire suspended vertically from a rigid support and stretched by applying load to its lower end. The load is gradually increased in small steps until the wire breaks. The elongation produced in the wire is measured during each step. Stress and strain are noted for each load and a graph is drawn by taking tensile strain along X-axis and tensile stress along Y-axis. It is a stress-strain curve as shown in the figure.

1. Proportional limit: The initial part of the graph is a straight line OA. This is the region in which Hooke’s law is obeyed and stress is directly proportional to stain. The straight line portion ends at A. The stress at this point is called proportional limit.
2. Yield point: If the load is further increased till point B is reached, stress and strain are no longer proportional and Hooke’s law is not valid. If the load is gradually removed starting at any point between O and B. The curve is retracted until the wire regains its original length. The change is reversible. The material of the wire shows elastic behaviour in the region OB. Point B is called the yield point. It is also known as the elastic limit.
3. Permanent Set: When the stress is increased beyond point B, the strain continues to increase. If the load is removed at any point beyond B, for example (at C), the material does not regain its original length. It follows the line CE. Length of the wire when there is no stress is greater than the original length. The deformation is irreversible and the material has acquired a permanent set.
4. Fracture point: Further increase in load causes a large increase in strain for relatively small increase in stress, until a point D is reached at which fracture takes place. The material shows plastic flow or plastic deformation from point B to point D. The material does not regain its original state when the stress is removed. The deformation is called plastic deformation.

Question 42.
Stress-strain curve for two materials A and B are shown in the figure. The graphs are drawn to the same scale.

1. Which material has greater Young’s modulus?
2. Which of the two is the stronger material? (NCERT)

Answer:

1. For a given strain, stress for A is more than that of B. Hence, Young’s modulus (= stress/strain) is greater for A than that of B.
2. Material A is stronger than B because A can bear greater stress before the breaking of the wire.

Question 43.
Figure shows the stress-strain curve for a given material. What are

1. Young’s modulus and
2. approximate yield strength for this material? (NCERT)

Answer:

1. The graph, implies stress of 150 × 10⁶ N m^-2 corresponds to a strain of 0.002. Therefore, Young’s modulus is,
   
2. The yield strength for the material is less than 300 × 10⁶ N m^-2, i.e.. 3 × 10⁸ N m^-2 and greater than 2.5 × 10⁸ N m^-2.

Question 44.
Explain the following terms:

1. Ductile
2. Malleable

Answer:

1. Metals such as copper, aluminium, wrought iron, etc. have large plastic range of extension. They lengthen considerably and undergo plastic deformation till they break. They are called ductile.
2. Metals such as gold, silver which can be hammered into thin sheets are called malleable.

Question 45.
Define strain energy.
Answer:
The elastic potential energy gained by a wire during elongation b a stretching force is called as strain energy.

Question 46.
A steel wire is acted upon by a load of 10 N. Calculate the extension produced in the wire, if the strain energy stored in the wire is 1.1 × 10^-3 J.
Solution:
Given: F = 10 N, Strain energy =1.1 × 10^-3 J,
To find: Extension (l)
Formula: W = \(\frac{1}{2}\) × F × l
Calculation:
From formula,
l = \(\frac{2 \mathrm{~W}}{\mathrm{~F}}\) = \(\frac{2 \times 1.1 \times 10^{-3}}{10}\) = 2.2 × 10^-4 m.
Answer:
The extension produced in the wire is 2.2 × 10^-4 m.

Question 47.
Calculate the strain energy per unit volume in a brass wire of length 3 m and area of cross-section 0.6 mm² when it is stretched by 3 mm and a force of 6 kg-wt is applied to its free end.
Solution:
Given: L = 3 m, F = 6 kg wt = 6 × 9.8N = 58.8N, A = 0.6 mm² = 0.6 × 10^-6 m², l = 3 mm = 3 × 10^-3 m
To find: Strain energy per unit volume (u)

Formulae:

i. Stress = \(\frac{F}{A}\)
ii. Strain = \(\frac{l}{L}\)
iii. u = \(\frac{1}{2}\) × Stress × Strain

Calculation:

From formula (i),
Stress = \(\frac{F}{A}\) = \(\frac{58.8}{0.6 \times 10^{-6}}\) = 98 × 10⁶ N/m²
From formula (ii),
Strain = \(\frac{l}{L}\) = \(\frac{3 \times 10^{-3}}{3}\) = 10^-3
From formula (iii),
u = \(\frac{1}{2}\) × Stress × Strain
= \(\frac{1}{2}\) × 98 × 10⁶ × 10^-3
u = 49 × 10³ J/m³
Answer:
Strain energy per unit volume in the brass wire is 49 × 10³ J/m³

Question 48.
A steel wire of diameter 1 × 10^-3 m is stretched by a force of 20 N. Calculate the strain energy per unit volume. (Y_steel = 2 × 10¹¹ N/m²)
Solution:
Given: d = 1 × 10^-3 m, r = 5 × 10^-4 m, y_steel = 2 × 10¹¹ N/m²
To find: Strain energy per unit volume
Formula: Strain energy per unit volume
= \(\frac{1}{2}\) × \(\frac{\text { (stress) }^{2}}{\mathrm{Y}}\)
Calculation:
From formula.
Strain energy per unit volume

Answer:
The strain energy per unit volume of the steel wire is 1621 J.

Question 49.
A uniform steel wire of length 3 m and area of cross section 2 mm² is extended through 3 mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. (Y_steel = 20 × 10¹⁰ N/m²)
Solution:
Given: L = 3 m, A = 2 mm² = 2 × 10^-6 m²,
l = 3 mm = 3 × 10^-3 m,
Y_steel = 20 × 10¹⁰ N/m²
To find: Energy stored (U)
Formula: W = \(\frac{1}{2}\) × F × l
Calculation: Since, Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)

Answer:
The energy stored in the steel wire is 0.6 J.

Question 50.
Explain the following terms:
i. Hardness
ii. Strength
iii. Toughness
Answer:

i. Hardness:

1. Hardness is the property of a material which enables it to resist plastic deformation.
2. Hard materials have little ductility and they are brittle to some extent.
3. The term hardness also refers to stiffness or resistance to bending, scratching, abrasion or cutting.
4. It is the property of a material which gives it the ability to resist permanent deformation when a load is applied to it.
5. The greater the hardness, greater is the resistance to deformation.
6. Hardness of material is different from its strength and toughness.
7. The most well known example for hard material is diamond, while metal with very low hardness is aluminium.

ii. Strength:

1. If a force is applied to a body, it produces deformation in it.
2. Higher is the force required for deformation, the stronger is the material, i.e., the material has more strength.
3. Steel has high strength whereas plasticine clay is not strong because it gets easily deformed even by a small force.

iii. Toughness:

1. Toughness is the ability of a material to resist fracturing when a force is applied to it.
2. Plasticine clay is relatively tough as it can be stretched and deformed due to applied force without breaking.

Question 51.
Explain the concept of frictional force.
Answer:

1. Whenever the surface of one body slides over another, each body exerts a certain amount of force on other body.
2. These forces are tangential to the surfaces. The force on each body is opposite to the direction of motion between two bodies.
3. It prevents or opposes the relative motion between two bodies.
4. It is common experience that an object placed on any surface does not move easily when a small force is applied to it.
5. This is because of certain force of opposition acting between the surface of the object and the surface on which it is placed.
6. To initiate any motion between pair of surfaces, we need a certain minimum force. Also, after the motion begins, it is constantly opposed by some natural force.
7. This mechanical force between two solid surfaces in contact with each other is called as frictional force.

Question 52.
Explain few examples of friction.
Answer:

1. A rolling ball comes to rest after covering a finite distance on playground because of friction.
2. Our foot ware is provided with designs at the bottom of its sole so as to produce force of opposition to avoid slipping. It is difficult to walk without such opposing force. When we try to walk fast on polished flooring at home with soap water spread on it. There is a possibility of slipping due to lack of force of friction.
3. Relative motion between solids and fluids (i.e. liquids and gases) is also opposed naturally by friction, eg.: a boat on the surface of water experiences opposition to its motion.

Question 53.
Explain the origin of friction.
Answer:

1. If smooth surfaces are observed under powerful microscope, many irregularities and projections are observed.
2. Friction arises due to interlocking of these irregularities between two surfaces in contact.
3. The surfaces can be made extremely smooth by polishing to avoid irregularities but then case also, friction does not decrease but may increase.
4. Hence the interlocking of irregularities is not the real cause of friction.
5. According to modem theory, cause of friction is the force of attraction between molecules of two surfaces in actual contact in addition to the. force due to the interlocking between the two surfaces.
6. When one body is in contact with another body, the real microscopic area in contact is very small due to irregularities in contact.
   
7. Due to small area, pressures at points of contact is very high. Hence there is strong force of attraction between the surfaces in contact.
8. When the surfaces in contact become more and more smooth, the actual area of contact goes on increasing.
9. Due to this, the force of attraction between the molecules increases and hence the friction also increases.

Question 54.
State the following terms:

1. Cohesive force
2. Adhesive force

Answer:

1. When two Surfaces are of the same material, the force of attraction between them is called cohesive force.
2. When two surfaces are of different materials, the force of attraction between them is called adhesive force.

Question 55.
Explain the concept of static friction.
Answer:

1. Consider a wooden block placed on a horizontal surface as shown in the figure and small horizontal force F is applied to it.
   
2. The block does not move with this force as it cannot overcome the frictional force between the block and horizontal surface.
3. In this case the force of static friction is equal to F and balances it.
4. The frictional force which balances applied force when the body is static is called force of static friction. In other words, static friction prevents sliding motion.
5. If we keep increasing F, a stage will come
   • For F < F_max, the force of static friction is equal to F.
   • when for F = F_max, the object will start moving.
   • For F ≥ F_max, the kinetic friction comes into play.
6. Static friction opposes impending motion i.e. the motion that would take place in absence of frictional force under the applied force.

Question 56.
Define limiting force of friction.
Answer:
Just before the body starts sliding over another body, the value of frictional force is maximum, it is called as limiting force of friction.

Question 57.
Explain the concept of kinetic friction.
Answer:

1. Once the sliding of block on the surface starts the force of friction decreases.
2. The force required to keep the body sliding steadily is thus less than the force required to just start its sliding.
3. The force of friction that comes into play when a body is in steady state of motion over another surface is called kinetic fòrce of friction.
4. Friction between two surfaces in contact when one body is actually sliding over the other body, is called kinetic friction or dynamic friction.

Question 58.
Why ball bearings are used in machines?
Answer:

1. For same pair of surfaces, the force of static friction is greater than the force of kinetic friction while the force of kinetic friction is greater than force of rolling friction.
2. As ball bearing undergo rolling friction and rolling friction is the minimum, ball bearings are used to reduce friction in parts of machines to increase its efficiency.

Question 59.
The coefficient of static friction between a block of mass 0.25 kg and a horizontal surface is 0.4. Find the horizontal force applied to it.
Solution:
Given: μ_s = 0.4, m = 0.25 kg, g = 9.8 m/s²
To find: Force (F_L)
Formula: F_L = μ_sN = μ_s(mg)
Calculation: From formula,
F_L = 0.4 × 0.25 × 9.8 = 0.98 N
Answer:
The horizontal force applied to it is 0.98 N.

Question 60.
Calculate the force required to move a block of mass 20 kg resting on a horizontal surface, if μ_s = 0.3 and g = 9.8 m/s².
Solution:
Given: m = 20 kg, μ_s = 0.3, g = 9.8 m/s²
To find. Force required (F)
Formula: F_S = μ_sN = μ_smg
Calculation: From formula,
F_S = 3 × 20 × 9.8 = 58.8N
Answer:
The force required to move the block is 58.8 N.

Question 61.
A block of mass 40 kg resting on a horizontal surface just begins to slide when a horizontal force of 120 N is applied to it. Once the motion starts, It can be maintained by a force of 80 N. Determine the coefficients of static friction and kinetic friction (g = 9.8 m/s²)
Solution:
Given: F_L= 120N, F_k = 80N, m = 40 kg, g = 9.8 m/s²

To find:

i. Coefficient of static friction (μ_s)
ii. Coefficient of kinetic friction (μ_k)

Formulae:

i. μ_s = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}\)
ii. μ_k = \(\frac{\mathrm{F}_{\mathrm{k}}}{\mathrm{N}}\)

Calculation:

From formula (i) we get.
N = mg = 40 × 9.8 = 392 N
∴ μ_s = \(\frac{F_{L}}{N}\) = \(\frac{120}{392}\) = 0.306
From formula (ii) we get,
∴ μ_k = \(\frac{F_{k}}{N}\) = \(\frac{80}{392}\) = 0.204
Answer:

1. The coefficient of static friction is 0.306.
2. The coefficient of kinetic friction is 0.204.

Question 62.
A 20 kg metal block is placed on a horizontal surface. The block just begins to slide when horizontal force of 100 N is applied to it. Calculate the coefficient of static friction. If coefficient of kinetic friction is 0.4, then find minimum force to maintain its uniform motion.
Solution:
Given: m = 20 kg, F_L = 100 N, μ_k = 0.4

To find:

i. Coefficient of static friction (μ_s)
ii. Minimum force required (F_k)

Formulae:

i. μ_s = \(\frac{F_{L}}{N}\) = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{mg}}\)
ii. μ_s = \(\frac{F_{k}}{N}\)

Calculation:

From formula (i),
μ_s = \(\frac{100}{20 \times 9.8}\) = 0.5102
From formula (ii),
F_k = μ_kN = μ_k × mg = 0.4 × 20 × 9.8
∴ F_k = 78.4 N
Answer:

1. The coefficient of static friction is 0.5102.
2. The minimum force required is 78.4 N.

Question 63.
The Mariana trench is located in the Pacific Ocean and at one place it is nearly 11 km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches the bottom? (K = 1.6 × 10¹¹ N/m²)
Answer:
Given: V = 0.32 m³, K = 1.6 × 10¹¹ N/m²,
P_W= 1.1 × 10⁸ Pa, P_atm = 1.01 × 10⁵ Pa
∴ dP = P_W – P_atm = 1.1 × 10⁸ – 1.01 × 10⁵ ≈ 1.1 × 10⁸ Pa
As, bulk modulus is given as,
K = V × \(\frac{d P}{d V}\)
∴ dV = \(\frac{\mathrm{V} \times \mathrm{dP}}{\mathrm{K}}\) = \(\frac{\left(1.1 \times 10^{8}\right) \times 0.32}{1.6 \times 10^{11}}\)
∴ dV = 2.2 × 10^-4 m³
The change in the volume of the ball when it reaches the bottom will be 2.2 × 10^-4 m³.

Question 64.
Two spheres A and B having same volume are dropped from same height in ocean. Sphere A is made up of brass and sphere B is made up of steel. Will there be same change in volume of spheres at a certain depth inside water? What will be the ratio of change in volumes of the two spheres at this depth?
Answer:

1. Brass and copper have different elastic moduli. Hence, there won’t be same change in the volume of spheres at a certain depth inside water.
2. As two spheres are dropped from same height and are at same depth in water pressure exerted on them remains same.
   ∴ dP_A = dP_B = dP
   V_A = V_B = V
3. If dV_A and dV_B be the change in volume of two spheres A and B then,
   dV_A = \(\frac{\mathrm{V} \mathrm{dP}}{\mathrm{K}_{\mathrm{A}}}\) and dV_B = \(\frac{\mathrm{V} \mathrm{dP}}{\mathrm{K}_{\mathrm{B}}}\)
4. Ratio of change in volume,
   
   where, K_A and K_B are bulk modulus of material of spheres A and B respectively.

Question 65.
What is the basis of deciding the thickness of metallic ropes used in crane to lift the heavy weight?
Answer:

1. The thickness of the metallic ropes used in cranes is decided on the basis of the elastic limit of the material of the rope and the factor of safety.
2. To lift a load upto 10⁴ kg, the rope is made for a factor of safety of 10.
3. It should not break even when a load of (original load × factor of safety = 10⁴ × 10) 10⁵ kgf i.e., 10⁵ × 9.8 N is applied on it.
4. If ‘r’ is the radius of the rope, then maximum stress = \(\frac{10^{5}}{\pi r^{2}}\).
5. The maximum stress (ultimate stress) should not exceed the breaking stress (5 to 20 × 10⁸ N/m²) as well as elastic limit of steel Y_s(30 × 10⁷ N/m²).
   
6. In order to have flexibility, the rope is made up of large number of thin wires twisted together.

Question 66.
Multiple choice questions

Which one of the following substances possesses the highest elasticity?
(A) Rubber
(B) Glass
(C) Steel
(D) Copper
Answer:
(C) Steel

Question 1.
S.I unit of stress is
(A) Newton/ metre
(B) Newton/ metre²
(C) Newton²/metre
(D) Newton/metre³
Answer:
(B) Newton/ metre²

Question 2.
A wire is stretched to double of its length. The strain is
(A) 2
(B) 1
(C) zero
(D) 0.5
Answer:
(B) 1

Question 3.
A and B are two steel wires and the radius of A is twice that of B. If they are stretched by the same load, then the stress on B is ____.
(A) four times that of A
(B) two times that of A
(C) three times that of A
(D) same as that of A
Answer:
(A) four times that of A

Question 4.
Two wires of the same material have radii r_A and r_B respectively. The radius of wire A is twice the radius of wire B. If they are stretched by same load then stress on wire B is ___
(A) equal to that of A
(B) half that of A.
(C) two times that of A.
(D) four times that of A.
Answer:
(D) four times that of A.

Question 5.
The length of a wire increases by 1% by a load of 2 kg wt. The linear strain produced in the wire will be
(A) 0.02
(B) 0.001
(C) 0.01
(D) 0.002
Answer:
(C) 0.01

Question 6.
A wire of length ‘L’, radius ‘r’ when stretched with a force ‘F’ changes in length by l’. What will be the change in length of a wire of same material having length ‘2L’ radius ‘2r and stretched by a force ‘2F’?
(A) l/2
(B) l
(C) 2l
(D) 4l
Answer:
(B) l

Question 7.
A force of 100 N produces a change of 0.1% in a length of wire of area of cross section 1 mm². Young’s modulus of the wire is ____
(A) 10⁵ N/m²
(B) 10⁹ N/m²
(C) 10¹¹ N/m²
(D) 10¹² N/m²
Answer:
(C) 10¹¹ N/m²

Question 8.
A copper wire (Y = 1 × 10¹¹ N/m²) of length 6 m and a steel wire (Y = 2 × 10¹¹ N/m²) of length 4 m each of cross-section 10^-5 m² are fastened end to end and stretched by a tension of 100 N. The elongation produced in the copper wire is
(A) 0.2 mm
(B) 0.4 mm
(C) 0.6 mm
(D) 0.8 mm
Answer:
(C) 0.6 mm

Question 9.
When a force is applied to the free end of a metal wire, metal wire undergoes
(A) longitudinal and lateral extension.
(B) longitudinal extension and lateral contraction.
(C) longitudinal contraction and lateral extension.
(D) longitudinal and lateral contraction.
Answer:
(B) longitudinal extension and lateral contraction.

Question 10.
A force of 1 N doubles the length of a cord having cross-sectional area 1 mm². The Young’s modulus of the material of cord is
(A) 1 N m^-2
(B) 0.5 × 10⁶ N m^-2
(C) 10⁶ N m^-2
(D) 2 × 10⁶ N m^-2
Answer:
(C) 10⁶ N m^-2

Question 11.
In equilibrium the tensile stress to which a wire of radius r is subjected by attaching a mass ‘m’ is ____.
(A) \(\frac{\mathrm{mg}}{\pi \mathrm{r}}\)
(B) \(\frac{\mathrm{mg}}{2 \pi \mathrm{r}}\)
(C) \(\frac{\mathrm{mg}}{\pi \mathrm{r}^{2}}\)
(D) \(\frac{\mathrm{mg}}{2 \pi \mathrm{r}^{2}}\)
Answer:
(C) \(\frac{\mathrm{mg}}{\pi \mathrm{r}^{2}}\)

Question 12.
When load is applied to a wire, the extension is 3 mm, the extension in the wire of same length but half the radius by the same load is
(A) 0.75 mm
(B) 6 mm
(C) 1.5 mm
(D) 12.0 mm
Answer:
(D) 12.0 mm

Question 13.
The S.I. unit of compressibility is ____.
(A) \(\frac{\mathrm{m}^{2}}{\mathrm{~N}}\)
(B) Nm²
(C) \(\frac{\mathrm{N}}{\mathrm{m}^{2}}\)
(D) \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
Answer:
(A) \(\frac{\mathrm{m}^{2}}{\mathrm{~N}}\)

Question 14.
When the pressure applied to one litre of a liquid is increased by 2 × 10⁶ N/m². Its volume decreases by 1 cm³. The bulk modulus of the liquid is
(A) 2 × 10⁹ N/m²
(B) 2 × 10³ dyne/cm²
(C) 2 × 10³ N/m²
(D) 0.2 × 10⁹ N/m²
Answer:
(A) 2 × 10⁹ N/m²

Question 15.
A cube of aluminium of each side 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be
(A) 0.02
(B) 0.1
(C) 0.005
(D) 0.002
Answer:
(D) 0.002

Question 16.
A wire has Poisson’s ratio of 0.5. It is stretched by an external force to produce a longitudinal strain of 2 × 10^-3. If the original diameter was 2 mm, the final diameter after stretching is
(A) 2.002 mm
(B) 1.998 mm
(C) 0.98 mm
(D) 1.999 mm
Answer:
(B) 1.998 mm

Question 17.
The compressibility of a substance is the reciprocal of ___.
(A) Young’s modulus
(B) Bulk modulus
(C) Modulus of rigidity
(D) Poisson’s ratio
Answer:
(B) Bulk modulus

Question 18.
For which of the following is the modulus of rigidity highest?
(A) Aluminium
(B) Quartz
(C) Rubber
(D) Water
Answer:
(B) Quartz

Question 19.
An elongation of 0.2% in a wire of cross-section 10^-4 m² causes a tension of 1000 N. Then its Young’s modulus is
(A) 6 × 10⁸ N/m²
(B) 5 × 10⁹ N/m²
(C) 10⁸ N/m²
(D) 10⁷ N/m²
Answer:
(B) 5 × 10⁹ N/m²

Question 20.
Within the elastic limit, the slope of graph between stress against strain gives ____
(A) compressibility
(B) Poisson’s ratio
(C) modulus of elasticity
(D) extension
Answer:
(C) modulus of elasticity

Question 21.
Solids which break or rupture before the elastic limits are called
(A) brittle
(B) ductile
(C) malleable
(D) elastic
Answer:
(A) brittle

Question 22.
Substances which break just after their elastic limit is reached are ___.
(A) ductile
(B) brittle
(C) malleable
(D) plastic
Answer:
(B) brittle

Question 23.
Which of the following substances is ductile?
(A) glass
(B) high carbon steel
(C) Steel
(D) copper
Answer:
(D) copper

Question 24.
The Young’s modulus of a material is 10¹¹ Nm^-2 and its Poisson’s ratio is 0.2. The modulus of rigidity of the material is
(A) 0.42 × 10¹¹ N/m²
(B) 0.42 × 10¹⁴ N/m²
(C) 0.42 × 10¹⁶ N/m²
(D) 0.42 × 10¹⁸ N/m²
Answer:
(A) 0.42 × 10¹¹ N/m²

Question 25.
Two pieces of wire A and B of the same material have their lengths in the ratio 1:2, and their diameters are in the ratio 2:1. If they are stretched by the same force, their elongations will be in the ratio
(A) 2 : 1
(B) 1 : 4
(C) 1 : 8
(D) 8 : 1
Answer:
(C) 1 : 8

Question 26.
Young’s modulus of material of wire is ‘Y’ and strain energy per unit volume is ‘E’, then the strain is

Answer:
(C) \(\sqrt{\frac{2 \mathrm{E}}{\mathrm{Y}}}\)

Question 27.
Strain energy per unit volume is given by

Answer:
(A) \(\frac{1}{2} \times \frac{(\text { stress })^{2}}{\mathrm{Y}}\)

Question 28.
The strain energy per unit volume of the wire under increasing load is
(A) \(\frac{1}{2}\) × (stress)² × strain
(B) \(\frac{1}{2}\) × stress × (strain)²
(C) 0.5 × stress × strain
(D) 0.5 × (strain)² × \(\frac{1}{Y}\)
Answer:
(C) 0.5 × stress × strain

Question 29.
The energy stored per unit volume of a strained wire is
(A) \(\frac{1}{2}\) × (load) × (extension)
(B) \(\frac{1}{2}\) × \(\frac{Y}{(\text { strain })^{2}}\)
(C) \(\frac{1}{2}\) × Y × (strain)²
(D) Stress × strain
Answer:
(C) \(\frac{1}{2}\) × Y × (strain)²

Question 30.
If work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material, but with double the radius and half the length by 1 mm in joule is
(A) 1/4
(B) 4
(C) 8
(D) 16
Answer:
(D) 16

Question 31.
When the load on a wire is slowly increased from 3 to 5 kg wt, the elongation increases from 0.61 to 1.02 mm. The work done during the extension of wire is
(A) 0.16 J
(B) 0.016 J
(C) 1.6 J
(D) 16 J
Answer:
(B) 0.016 J

Question 32.
A body lies on a table. Its weight is balanced by the ___.
(A) frictional force
(B) normal force
(C) force causing motion on the body
(D) surface of the table
Answer:
(B) normal force

Question 33.
The friction that exists between the surface of two bodies in contact when one body is sliding over the other, is called ___.
(A) rolling friction
(B) friction
(C) kinetic friction
(D) static friction.
Answer:
(C) kinetic friction

Question 34.
Limiting force of static friction does NOT depend on
(A) actual area of contact.
(B) geometrical area of contact.
(C) interlocking between surfaces in contact.
(D) intermolecular forces of attraction between molecules of the two surfaces.
Answer:
(C) interlocking between surfaces in contact.

Question 35.
In case of coefficient of static friction (µ_s), kinetic friction (µ_k) and rolling friction (µ_r), which of the following relation is true
(A) µ_s > µ_k > µ_r
(B) µ_s > µ_r > µ_k
(C) µ_r > µ_s > µ_k
(D) µ_r > µ_k > µ_s
Answer:
(A) µ_s > µ_k > µ_r

Question 36.
A body of weight 50 N is placed on a smooth surface. If the force required to move the body on the surface is 30 N, the coefficient of friction is
(A) 0.6
(B) 0.4
(C) 0.3
(D) 0.8
Answer:
(A) 0.6

Question 37.
A wooden block of 50 kg is at rest on the table. A force of 70 N is required to just slide the block. The coefficient of static friction is
(A) \(\frac{6}{7}\)
(B) \(\frac{5}{7}\)
(C) \(\frac{7}{35}\)
(D) \(\frac{1}{7}\)
Answer:
(D) \(\frac{1}{7}\)

Question 38.
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:
(A) \(\frac{1}{2}\)Mgl
(B) \(\frac{1}{2}\)MgL
(C) Mgl
(D) MgL
Answer:
(A) \(\frac{1}{2}\)Mgl
Hint: elastic potential energy: \(\frac{1}{2}\) × F × L = \(\frac{1}{2}\)Mgl

Question 67.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1 π ms^-2, What will be the tensile stress that would be developed
(A) 6.2 × 10⁶ N m^-2
(B) 5.2 × 10⁶ N m^-2
(C) 3.1 × 10⁶ N m^-2
(D) 4.8 × 10⁶ N m^-2
Answer:
(C) 3.1 × 10⁶ N m^-2

Question 68.
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to:
(A) 10⁶ N m^-2
(B) 10⁴ N m^-2
(C) 10⁸ Nm^-2
(D) 10³ N m^-2
Answer:
(A) 10⁶ N m^-2
Hint: d = 6 mm = 0.006 m, l = 42 cm = 0.42 m, ∆l = 20 cm = 0.2 m, m = 0.02 kg, v = 20 m/s
Energy stored in the catapult,

Question 69.
The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively),
(A) Plastic and ductile
(B) Ductile and brittle
(C) Brittle and ductile
(D) Brittle and plastic
Answer:
(C) Brittle and ductile
Hint: Ductile materials have a fracture strength lower than the ultimate Tensile strength (i.e., the points are far apart.) whereas in brittle materials, the fracture strength is equivalent to ultimate tensile strength (i.e., the points are close.)
∴ Material X is brittle and Y is ductile in nature.

Question 70.
The compressibility of water is ‘o’ per unit atmospheric pressure. The decrease in its volume ‘V’ due to atmospheric pressure ‘P’ will be
(A) \(\frac{\sigma \mathrm{V}}{\mathrm{P}}\)
(B) \(\frac{\sigma P}{\mathrm{~V}}\)
(C) σPV
(D) \(\frac{\sigma}{\mathrm{PV}}\)
Answer:
(C) σPV

Question 71.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount?
(A) 9 F
(B) 6 F
(C) 4 F
(D) F
Answer:
(A) 9 F
Hint: As material is same, Young’s modulus of two wires is same.
Also, volume of both wires is same.
V₁ = V₂

Question 72.
A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, \(\left(\frac{\mathrm{dr}}{\mathrm{r}}\right)\),is:
where negative sign indicates volume decreases with increase in pressure.
(A) \(\frac{\mathrm{mg}}{\mathbf{3} \mathrm{Ka}}\)
(B) \(\frac{\mathrm{mg}}{\mathrm{Ka}}\)
(C) \(\frac{\mathrm{Ka}}{\mathrm{mg}}\)
(D) \(\frac{\mathrm{Ka}}{3 \mathrm{mg}}\)
Answer:
(A) \(\frac{\mathrm{mg}}{\mathbf{3} \mathrm{Ka}}\)
Hint:
Bulk modulus is given as,
K = \(\left(\frac{-\mathrm{dP}}{\mathrm{dV} / \mathrm{V}}\right)\)
where negative sign indicates volume decreases with increase in pressure.
∴ Fractional decrease in volume will be,
\(\frac{\mathrm{dV}}{\mathrm{V}}\) = \(\frac{\mathrm{dP}}{\mathrm{K}}\)
If area of cross-section of cylinder is a, then,

Question 73.
Two metal wires ‘P’ and ‘Q’ of same length and material are stretched by same load. Their masses are in the ratio m₁ : m₂. The ratio of elongations of wire ‘P’ to that of’Q’ is

Answer:
(C) m₂ : m₁
Hint:

Question 74.
The increase in energy of a metal bar of length ‘L’ and cross-sectional area ‘A’ when compressed with a load ‘M’ along its length is (Y = Young’s modulus of the material of metal bar)
(A) \(\frac{\mathrm{FL}}{2 \mathrm{AY}}\)
(B) \(\frac{\mathbf{F}^{2} \mathbf{L}}{\mathbf{2 A Y}}\)
(C) \(\frac{\mathrm{FL}}{\mathrm{AY}}\)
(D) \(\frac{F^{2} L^{2}}{2 A Y}\)
Answer:
(B) \(\frac{\mathbf{F}^{2} \mathbf{L}}{\mathbf{2 A Y}}\)
Hint:
U = \(\frac{1}{2}\) × F × l
= \(\frac{1}{2}\) × F × \(\frac{\mathrm{FL}}{\mathrm{AY}}\) = \(\frac{\mathrm{F}^{2} \mathrm{~L}}{2 \mathrm{AY}}\)

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 10 Human Health and Diseases Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 10 Human Health and Diseases

Multiple Choice Questions

Question 1.
Infectious stage of Plasmodium is …………………
(a) trophozoite
(b) sporozoite
(c) cryptozoite
(d) metacercaria
Answer:
(b) sporozoite

Question 2.
After birth, antibodies are transferred from mother to infant through …………………
(a) colostrum
(b) placenta
(c) blood
(d) tissue fluid
Answer:
(a) colostrum

Question 3.
Which cells give rise to T-lymphocytes?
(a) Thymocytes
(b) Bone marrow cells
(c) Erythrocytes
(d) Leucocytes
Answer:
(a) Thymocytes

Question 4.
Where is antigen D is present?
(a) On Rhesus factor
(b) On the surface of RBCs
(c) On A-antigen
(d) On AB-antigen
Answer:
(b) On the surface of RBCs

Question 5.
Erythroblastosis foetalis is caused when mother is …………………
(a) Rh +ve
(b) with antibody ‘a’
(c) Rh -ve
(d) with antibody ‘b’
Answer:
(c) Rh -ve

Question 6.
Which of the following is NOT a parasitic vector insect?
(a) Mosquito
(b) Housefly
(c) Honey bee
(d) Head louse
Answer:
(c) Honey bee

Question 7.
Which is the proper sequence in the developmental stages of Plasmodium?
(a) Merozoites → Sporozoite → Trophozoites → Schizonts
(b) Trophozoites → Merozoites → Sporozoite → Schizonts
(c) Sporozoite → Merozoites → Trophozoites → Schizonts
(d) Schizonts → Merozoites → Sporozoite → Trophozoites.
Answer:
(c) Sporozoite → Merozoites → Trophozoites → Schizonts

Question 8.
There is no vaccination on this disease till today.
(a) Typhoid
(b) Tuberculosis
(c) Polio
(d) AIDS
Answer:
(d) AIDS

Question 9.
Charas, hashish, ganja are obtained from …………………
(a) Papaver somnijerum
(b) Erythroxylum coca
(c) Atropa belladorta
(d) Cannabis sativa
Answer:
(d) Cannabis sativa

Question 10.
………………. Plant is used to obtain cocaine alkaloid.
(a) Marijuana
(b) Papaver somntferum
(c) Cannabis sativa
(d) Coca
Answer:
(d) Coca

Question 11.
………………… fish is released in the waterbody to prevent the spread of Malaria and Filaria.
(a) Pomfret
(b) Tilapia
(c) Gambusia
(d) Gold fish
Answer:
(c) Gambusia

Question 12.
The carcinogen that can cause vaginal cancer is …………………
(a) Vinyl chloride
(b) Diethylstilboestrol
(c) Mustard gas
(d) Cadmium oxide
Answer:
(b) Diethylstilboestrol

Question 13.
Prostate cancer can be caused due to exposure to …………………
(a) cadmium oxide
(b) mustard gas
(c) asbestos
(d) Nickel and chromium compounds
Answer:
(a) cadmium oxide

Question 14.
Choose the correct definition of health …………………
(a) Health is not contracting any disorder or disease.
(b) State of complete physical, mental and social well-being.
(c) Health is complete absence of any disease.
(d) Health is feeling good all the time.
Answer:
(b) State of complete physical, mental and social well-being

Question 15.
The interval between infection and appearance of disease symptoms is called …………………
(a) inoculation
(b) penetration
(c) infection period
(d) incubation period
Answer:
(d) incubation period

Question 16.
What is injected in vaccination ?
(a) Half killed pathogen
(b) Dead pathogens
(c) Live pathogens
(d) Readymade antibodies
Answer:
(a) Half killed pathogen

Question 17.
Who among the following is considered as father of immunology ?
(a) Ferdinand Kohn
(b) Robert Koch
(c) Louis Pasteur
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 18.
Who coined the term antibiotics ?
(a) Charles Darwin
(b) Louis Pasteur
(c) Alexander Fleming
(d) Selman Waksman
Answer:
(d) Selman Waksman

Question 19.
Who coined the term antibody?
(a) Selman Waksman
(b) Alexander Fleming
(c) Paul Ehrlich
(d) Edward Jenner
Answer:
(c) Paul Ehrlich

Question 20.
Widal test is used for the diagnosis of …………………
(a) Malaria
(b) Typhoid
(c) Diabetes mellitus
(d) HIV/AIDS
Answer:
(b) Typhoid

Question 21.
Which bacterial genus out of the following is the common pathogen causing pneumonia ?
(a) Streptococcus sps
(b) Lactobacillus sps
(c) Pseudomonas sps
(d) Salmonella sps
Answer:
(a) Streptococcus sps

Question 22.
Given below are some statements. Which among them are symptoms of pneumonia ?
(i) Greenish, yellow sputum coughed out.
(ii) Hepatomegaly and hypoglycemia.
(iii) High fever with shaking chills.
(iv) Thickening of skin and underlying tissues.
(v) Stabbing chest pain with shortness of breath.
(vi) Mood swings and joint pains along with nausea and vomiting.
(a) (i), (ii), (iii), (iv)
(b) (i), (iii), (v), (vi)
(c) (i), (ii), (iv), (vi)
(d) (ii), (iii), (iv), (v)
Answer:
(b) (i), (hi), (v), (vi)

Question 23.
Which of the following is not the common way to prevent common cold ?
(a) Using hand sanitizers
(b) Blowing nose in open
(c) Staying away from people suffering from cold
(d) Sipping warm water
Answer:
(b) Blowing nose in open

Question 24.
Common cold is not cured by antibiotics because it is …………………
(a) caused by a virus
(b) caused by a Gram-positive bacterium
(c) caused by a Gram-negative bacterium
(d) not an infectious disease
Answer:
(a) caused by a virus

Question 25.
Motile zygote of Plasmodium occurs in …………………
(a) gut of female Anopheles
(b) salivary glands of Anopheles
(c) Human RBCs
(d) Human liver
Answer:
(a) gut of female Anopheles

Question 26.
Haemozoin is …………………
(a) a precursor of haemoglobin
(b) a toxin from Streptococcus
(c) a toxin from Plasmodium
(d) a toxin from Hemophilus
Answer:
(c) a toxin from Plasmodium

Question 27.
Vaccination against malaria is not possible because …………………
(a) they produce antibodies and antitoxins
(b) they do not produce antibodies and antitoxins
(c) antibodies resistant to vaccines are produced
(d) none of these
Answer:
(b) they do not produce antibodies and antitoxins

Question 28.
The active form of Entamoeba histolytica feeds upon …………………
(a) blood only
(b) erythrocytes, mucosa and submucosa of colon
(c) mucosa and submucosa of colon only
(d) food in intestine
Answer:
(b) erythrocytes, mucosa and submucosa of colon

Question 29.
Eating unwashed and raw green leafy vegetables grown along the railway tracks in Mumbai may cause …………………
(a) malaria
(b) influenza
(c) amoebic colitis
(d) ringworm
Answer:
(c) amoebic colitis

Question 30.
Which method of water purification can terminate amoebae ?
(a) Chlorination
(b) Sedimentation
(c) Filtration
(d) Boiling
Answer:
(d) Boiling

Question 31.
Which of the following measures should be taken to control amoebic dysentery ?
(A) Using insecticidal sprays to kill flies.
(B) Not allowing stagnant water to be accumulated over a long time.
(C) To avoid eating uncovered food.
(D) Drinking only boiled water.
(E) Eating plenty of fruits.
(a) (A) (C) (D)
(b) (A) (B) (D)
(c) (B) (C) (E)
(d) (A) (B) (E)
Answer:
(a) (A) (C) (D)

Question 32.
Which of the following should be avoided for endemic spread of amoebiasis ?
(a) Cleaning bathroom taps and toilet seats with disinfectants.
(b) Washing hands and using hand sanitizers.
(c) Proper sewage disposal and treatment.
(d) Eating uncovered roadside food.
Answer:
(d) Eating uncovered roadside food.

Question 33.
Name the disease in which the genital organs are grossly affected due to infective helminth.
(a) Ascariasis
(b) Ring worm
(c) Scabies
(d) Filariasis
Answer:
(d) Filariasis

Question 34.
Find the odd organism:
(a) Wuchereria bancrofti
(b) Brugia malayi
(c) Brugia timori
(d) Ascaris lumbricoides
Answer:
(d) Ascaris lumhricoides

Question 35.
When is hydrocele formed in a man ?
(a) When testis are not functioning properly.
(b) When scrotum is infected with filarial worms.
(c) When testis are injured due to accident.
(d) When there is water accumulation in testis.
Answer:
(b) When scrotum is infected with filarial worms

Question 36.
Which medicine is used for eradicating microfilariae from endemic areas ?
(a) Diethyle carbamacine
(b) Mebendazole
(c) Albendazole
(d) Rimfampcin
Answer:
(a) Diethyle carbamacine

Question 37.
Which of the following fungi are causative organisms of ringworm ?
(a) Microsporum
(b) Candida
(c) Thrush
(d) Tinea pedis
Answer:
(a) Microsporum

Question 38.
On which material present on the outer skin surfaces of the human body does the fungus causing infections feed on ?
(a) Melanin
(b) Keratin
(c) Lignin
(d) Suberin
Answer:
(b) Keratin

Question 39.
Which of the following statements is correct ?
(a) Fungus grows well on dry skin.
(b) Fungus cannot survive on the outside of the hair shafts.
(c) Fungus thrives well on the warm and moist skin.
(d) Nails can never show fungal infections.
Answer:
(c) Fungus thrives well on the warm and moist skin.

Question 40.
Which of the following pair is viral diseases ?
(a) Common cold, AIDS
(b) Dysentery, Common cold
(c) Typhoid, Tuberculosis
(d) Ringworm, AIDS
Answer:
(a) Common cold, AIDS

Question 41.
Which one of the following glands is often referred in relation with AIDS ?
(a) Thymus
(b) Adrenal
(c) Thyroid
(d) Pancreas
Answer:
(a) Thymus

Question 42.
The first patient of AIDS was detected in India in …………………
(a) 1980
(b) 1986
(c) 1990
(d) 1996
Answer:
(b) 1986

Question 43.
Why is it said that for AIDS prevention is the only cure ?
(a) AIDS does not have any cure, once it is contracted.
(b) By prevention AIDS cannot be cured.
(c) AIDS can be cured by proper medication and vaccination.
(d) Only prevention helps as there is no cure for AIDS.
Answer:
(d) Only prevention helps as there is no cure for AIDS.

Question 44.
After a person is detected to be having AIDS by ELISA test, which is the next confirmatory test ?
(a) Western blot
(b) Southern blot
(c) PCR
(d) Northern blot
Answer:
(a) Western blot

Question 45.
What is full form of ELISA ?
(a) Enzyme Linked Inductive Assay
(b) Enzyme Linked Iron Sorbent Assay
(c) Enzyme Linked Immunosorbent Assay
(d) None of the above
Answer:
(c) Enzyme Linked Immunosorbent Assay

Question 46.
The possible ways of transmission of AIDS are …………………
(A) Intimate sexual contact
(B) Hugging and kissing
(C) Blood transfusion without properly checking it
(D) Eating from the same plate
(E) Sharing bed linen
(F) Transplacental infection from infected mother
(G) Sharing same tattoo gun and syringes
(a) (A), (C), (F), (G)
(b) (B), (D), (E), (G)
(c) (C), (D), (E), (F)
(d) (A), (B), (C), (D)
Answer:
(a) (A), (C), (F), (G)

Question 47.
Which one of the following statements is correct ?
(a) Benign tumours show the property of metastasis.
(b) Heroin accelerates body functions.
(c) Malignant tumours may exhibit metastasis.
(d) Patients who have undergone surgery are given cannabinoids to relieve pain.
Answer:
(c) Malignant tumours may exhibit metastasis.

Question 48.
Heroin or smack is chemically …………………
(a) diclofenac
(b) diacetyl morphine
(c) benzodiazepine
(d) amphetamines
Answer:
(b) diacetyl morphine

Question 49.
Ecstasy is a drug that is used in most of the Rev parties which is chemically a derivative of …………………
(a) Barbiturates
(b) Amphetamines
(c) Catecholamine
(d) Morphine
Answer:
(b) Amphetamines

Question 50.
From which plant is charas obtained?
(a) Cannabis sativa
(b) Erythroxylum coca
(c) Papaver somniferum
(d) Atropa belladonna
Answer:
(a) Cannabis sativa

Question 51.
Opium is obtained from the latex of the unripe fruits of …………………
(a) Cannabis sativa
(b) Thea siensis
(c) Papaver somniferum
(d) Erythroxylon coca
Answer:
(c) Papaver somniferum

Question 52.
Use of Cannabis products results in …………………
(a) depressed brain activity and feeling of calmness
(b) suppressed brain function and relief of pain
(c) stimulation of nervous system, increased alertness and activity
(d) alteration in perception, thoughts and feelings
Answer:
(d) alteration in perception, thoughts and feelings

Question 53.
Marijuana, ganja and LSD are …………………
(a) narcotics
(b) stimulants
(c) hallucinogens
(d) all of these
Answer:
(c) hallucinogens

Question 54.
What is the source of LSD ?
(a) Poppy seeds
(b) Datura plant
(c) Sugar
(d) Claviceps purpurea
Answer:
(d) Claviceps purpurea

Question 55.
Narcotics are …………………
(a) amphetamines and caffeine
(b) morphine and heroine
(c) LSD and cocaine
(d) barbiturates and benzodiazepine
Answer:
(b) morphine and heroine

Question 56.
Which is an incorrectly matched pair ?
(a) LSD – Ergot fungus
(b) Heroin – Opium
(c) Amphetamines – Depressant
(d) Benzodiazepine – Calmpose tablets
Answer:
(c) Amphetamines – Depressant

Question 57.
Choose the incorrect statement
(a) The excessive use of anabolic steroids cause severe acne.
(b) In both the sexes there is increased aggressiveness and mood swings, due to steroids.
(c) In females, anabolic steroids cause breast enlargement.
(d) In males, anabolic steroids cause enlargement of prostate gland.
Answer:
(c) In females, anabolic steroids cause breast enlargement.

Question 58.
Who are the first ones to note the danger signs of drug or alcohol abuse in the adolescents ?
(a) Alert parents and teachers
(b) Neighbours
(c) Relatives
(d) Doctors
Answer:
(a) Alert parents and teachers

Question 59.
Who can give professional help for the deaddiction?
(a) Highly qualified psychiatrist
(b) Parents
(c) Teachers
(d) Friends
Answer:
(a) Highly qualified psychiatrist

Match the columns

Question 1.

Column I	Column II
(a) Metchnikoff	(i) ABO Blood group system
(b) Fleming	(ii) Concept of immunity
(c) Edward Jenner	(iii) Phagocytic cells
(d) Karl Lands teiner	(iv) Lysozyme

Answer:

Column I	Column II
(a) Metchnikoff	(iii) Phagocytic cells
(b) Fleming	(iv) Lysozyme
(c) Edward Jenner	(ii) Concept of immunity
(d) Karl Lands teiner	(i) ABO Blood group system

Question 2.

Disease	Vector species
(a) Dengue	(i) Anopheles
(b) Malaria	(ii) Housefly
(c) Filaria	(iii) Culex
(d) Typhoid	(iv) Aedes

Answer:

Disease	Vector species
(a) Dengue	(iv) Aedes
(b) Malaria	(i) Anopheles
(c) Filaria	(iii) Culex
(d) Typhoid	(ii) Housefly

Classify the following to form Column B as per the category given in Column A

Question 1.
Benzyl penicillin, Chloromycetin, Mebendazole, Levamisole, Pyrimethamine Ampicillin, Ty21a vaccine, Sulfadoxine.

Column A (Disease)	Column B (Treatment)
(1) Pneumonia	————–
(2) Malaria	————–
(3) Ascariasis	————–
(4) Typhoid	————–

Answer:

Column A (Disease)	Column B (Treatment)
(1) Pneumonia	Benzyl penicillin, Ampicillin
(2) Malaria	Pyrimethamine, Sulfadoxine
(3) Ascariasis	Mebendazole, Levamisole
(4) Typhoid	Chloromycetin, Ty21a vaccine

Question 2.
Lung cancer, Pituitary, Spleen, Skin cancer, Cancer of adipose tissue, lymph nodes, Adrenal, Bone tumour.

Column (A Type of cancer)	Column B (Organs affected)
(1) Carcinoma	————–
(2) Sarcoma	————–
(3) Lymphoma	————–
(4) Adenocarcinoma	————–

Answer:

Column (A Type of cancer)	Column B (Organs affected)
(1) Carcinoma	Lung cancer, Skin cancer
(2) Sarcoma	Cancer of adipose tissue, Bone tumour
(3) Lymphoma	Spleen, Lymph nodes
(4) Adenocarcinoma	Pituitary, Adrenal

Very Short Answer Questions

Question 1.
By which process T-cells and B-cells are produced?
Answer:
T-cells and B-cells are produced from the stem cells called haemocytoblasts, in bone marrow of adults and in liver of the foetus, by the process of haematopoiesis and in the bone marrow in adult.

Question 2.
What is a hinge?
Answer:
Hinge is the region of Y-shaped structure, holding arms and stem of antibody where four polypeptide chains of antibody are held together by disulfide bonds (-s-s-) to form a ‘Y’-shaped structure.

Question 3.
What is epitope?
Answer:
Epitope is antigenic determinant, which is present on antigens.

Question 4.
What is paratope?
Answer:
Paratope is antigen binding site that is present on the antibodies.

Question 5.
Which antigen is present in Rh +ve person?
Answer:
Antigen D is present in Rh +ve person.

Question 6.
Give the role of flushing action of lachrymal secretions.
Answer:
The conjunctiva is freed from foreign particles by the flushing action of lachrymal secretions.

Question 7.
What happens when lachrymal secretion is absent in eyes?
Answer:
Eyes become susceptible to infection when lachrymal secretion is absent.

Question 8.
In tears which antibacterial substance is present?
Answer:
Lysozyme is the antibacterial substance present in the tears.

Question 9.
Which kind of immunity is provided by vaccination?
Answer:
Artificial acquired active and passive immunity is provided by vaccination.

Question 10.
Who was Edward Jenner?
Answer:
Edward Jenner was the British scientist who developed cowpox vaccine for the protection against small pox virus.

Question 11.
Mrunmayi is called universal blood acceptor. What is her blood group?
Answer:
Blood group of Mrunmayi is AB.

Question 12.
What are antigens?
Answer:
Different foreign substances that invade the body and are capable of stimulating an immune response are called antigens.

Question 13.
In which animal Rh factor was discovered at first?
Answer:
Rh factor was first discovered in Rhesus monkey for the first time.

Question 14.
What is elephantiasis?
Answer:
Elephantiasis is one of the symptoms of lymphatic filariasis, in which there is thickening of skin and underlying tissues due to presence of malarial parasite.

Question 15.
What is dermatophytosis ?
Answer:
Dermatophytosis is a clinical condition in which fungal infection of skin occur in humans, pets and cattle which is commonly called as ringworm.

Question 16.
What is sporozoite?
Answer:
Sporozoite is a developmental stage of Plasmodium produced by rupture of oocyst. Sporozoite can enter the bloodstream in human body and then infect hepatocytes or liver cells, where they multiply into merozoites.

Question 17.
Why does male mosquito not spread Malaria?
Answer:
Male mosquito feed only on plant sap and not blood of human beings; therefore it does not spread Malaria.

Question 18.
Where does Plasmodium reproduce asexually?
Answer:
Plasmodium reproduce asexually in the liver cells and red blood cells of infected human being.

Question 19.
Where does Plasmodium reproduce sexually?
Answer:
Plasmodium undergoes sexual reproduction by the process of fertilization and development in the intestine of mosquito.

Question 20.
Which fish can be used for mosquito control?
Answer:
Gambusia fish can be used for mosquito control.

Question 21.
What is arthralgia?
Answer:
Arthralgia means joint pains.

Question 22.
What do you mean by hepatomegaly?
Answer:
Hepatomegaly means enlargement of liver.

Question 23.
Which organism yields LSD?
Answer:
Ergot fungus, Claviceps purpurea yield LSD.

Question 24.
Enlist various types of barriers which prevent entry of foreign agents into the body.
Answer:

1. Epithelial surface
2. Antimicrobial substances in blood and tissues
3. Cellular factors in innate immunity
4. Fever
5. Acute phase proteins (APPs).

Question 25.
Which is the gastro-intestinal disease by which 15% Indian population is affected?
Answer:
Amoebiasis or amoebic dysentery is the gastro-intestinal disease by which 15% Indian population is affected.

Question 26.
What are the anti-helminthic drugs which are used in treatment of Ascariasis?
Answer:
Anti-helminthic drugs like Piperazine, Mebendazole, Levamisole, Pyrantel are used against Ascaris lumbricoidies.

Question 27.
Which are the diseases that can be avoided by eradication of mosquitoes in your area?
Answer:
Malaria, dengue, chikungunya and filariasis or elephantiasis can be avoided by eradication of mosquitoes.

Define the following

Question 1.
Serology
Answer:
A branch of immunology which deals with the study of antigen-antibody interactions is called serology.

Question 2.
Hygiene
Answer:
Hygiene is the science of health, which aims at preserving, maintaining and improving the health of an individual or the community as a whole.

Question 3.
Disease
Answer:
Disease is a condition of disrupted or deranged functioning of one or more organs or systems of the body caused due to infection, detective diet or heredity.

Question 4.
Immune system
Answer:
The system which protects us from various infectious agents is called immune system.

Question 5.
Resistance
Answer:
Resistance is an ability to ward off damage or disease through our defence mechanism.

Question 6.
Immunity
Answer:
The immunity is defined as the general ability of a body to recognize and neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.

Question 7.
Antibody
Answer:
The protective chemicals produced by immune cells in response to antigens is called antibodies.

Question 8.
Opsonisation
Answer:
The process of coating of bacteria to facilitate their subsequent phagocytosis by macrophages is called opsonisation.

Question 9.
Pathogen
Answer:
Pathogen are living agents such as viruses, ricketssia, bacteria, fungi, protozoans, helminth and certain insect larvae which are capable of causing diseases.

Question 10.
Parasite
Answer:
An organism that lives in or on another organism called host and takes its nourishment from it (host) is called parasite.

Question 11.
Pathogenicity
Answer:
The ability of an organism to enter a host and cause a disease is called pathogenicity.

Question 12.
Infectious disease
Answer:
The disease which is transmitted from infected person to another healthy person either directly or indirectly is called infectious disease or communicable disease.

Question 13.
Non-infectious disease
Answer:
The disease that cannot be transmitted from one infected person to another healthy person, either directly or indirectly is called non infectious or non-communicable disease.

Question 14.
Innate immunity
Answer:
Innate immunity is defined as the resistance to infections that an individual possesses due to his or her genetic make-up and thus it is inborn defence mechanism present naturally in the body.

Question 15.
Acquired immunity
Answer:
The resistance developed during lifetime is called acquired immunity.

Question 16.
APP proteins
Answer:
APP proteins or acute phase proteins are certain collection of plasma proteins which are suddenly increased by the infection caused after injury.

Question 17.
Incubation period
Answer:
Incubation period is the time interval from the invasion of a pathogen to the development of clinical manifestations or symptoms.

Name the following

Question 1.
Name the cell that produces lymphokines.
Answer:
Helper T-cells

Question 2.
Blood group systems in human beings.
Answer:
ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay blood group.

Question 3.
Types of sarcoma.
Answer:

1. Osteosarcoma (bone)
2. Myosarcoma (muscle)
3. Chondrosarcoma (cartilage)
4. Liposarcoma (adipose tissue)

Question 4.
Therapies used for treatment of cancer.
Answer:

1. Chemotherapy
2. Radiotherapy
3. Surgery
4. Immunotherapy
5. Supportive therapy

Question 5.
Name the term for the transmission of HIV from pregnant mother to foetus.
Answer:
Transplacental.

Question 6.
Factors that maintain good health.
Answer:

1. Balanced diet
2. Personal hygiene
3. Regular exercise
4. Right attitude of mind
5. Good habits.

Question 7.
Two examples of ascaricides.
Answer:
Mebendazole and Albendazole, etc.

Question 8.
Two vaccines for typhoid.
Answer:

1. Oral Ty21a
2. Injectable Typhoid polysaccharide vaccine or Typhium vi/ Typherix.

Question 9.
Parasites causing lymphatic filariasis.
Answer:

1. Wuchereria bancrojti
2. Brugiamalayi
3. Brugia timori.

Question 10.
Parasites causing Subcutaneous Filariasis.
Answer:

1. Loa loa
2. Mansonella spp.

Question 11.
Name the scientists who discovered AB blood group?
Answer:
Decastallo and Sturti

Distinguish between the following

Question 1.
Inborn immunity and acquired immunity.
Answer:

Inborn Immunity	Acquired Immunity
1. Inborn immunity or innate immunity is also called natural immunity.	1. Acquired immunity is also called adaptive immunity.
2. Innate immunity is present right from the birth.	2. Acquired immunity is not present at birth, but is acquired during lifetime of the individual.
3. Inborn immunity does not depend upon the previous exposure to a pathogen or foreign substance.	3. Acquired immunity always depends upon the previous exposure to a pathogen or foreign substance.
4. It is non-specific immunity as it can offer resistance to any pathogen.	4. It is specific immunity as it can offer resistance only to a particular pathogen.
5. Innate immunity consists of various types of barriers for defence against the pathogens.	5. Acquired immunity consists of various types of cells which are able to produce antibodies.
6. Inborn immunity shows immediate effect in the body.	6. Acquired immunity requires several days to become activated.
7. Inborn immunity is seen in all animals.	7. Acquired immunity is seen only in vertebrates.
8. Inborn immunity is genetic in nature and is heritable.	8. Acquired immunity is non-genetic in nature and is non-heritable.

Question 2.
Communicable and non-communicable diseases.
Answer:

Communicable diseases	Non-communicable diseases
1. Diseases transmitted from infected person to healthy person are called communicable or infectious diseases.	1. Diseases that are not passed from one person to other are non-communicable or non-infectious diseases.
2. Communicable diseases spread through pathogens.	2. Non-communicable diseases do not spread through pathogens.
3. Communicable diseases are not inherited from parental generation to offspring.	3. Non-communicable diseases like cancer can be from parental generation to offspring.
4. Vectors play the major role in spreading disease from one person to another.	4. Caused due to allergy, illness, malnutrition or abnormalities in cell proliferation, changes in lifestyle, environment play a significant role.
5. Treated by conventional methods using antibiotics and other drugs.	5. Treated conservatively for a long time or surgically.
6. Diseases are acute which develop suddenly due to infections. E.g. Pneumonia, Tuberculosis, AIDS, Typhoid, Cholera, Malaria.	6. Diseases are chronic which develop and persist for a long time. E.g. Cancer, Rickets, Allergies, Kwashiorkor, Diabetes, Heart disease, etc.

Question 3.
Ascariasis and Filariasis.
Answer:

Ascariasis	Filariasis
1. Only one species Ascaris lumbricoid.es cause ascariasis.	1. There are many species of nematode that can cause filariasis.
2. Ascaris causes the infection of alimentary canal.	2. Wuchereria bancrofti causes the infection of lymphatic system.
3. Ascaris does not cause swellings of upper and lower limbs.	3. Filariasis cause swellings of extremities.
4. Ascaris is caused due to faeco-oral transmission.	4. Filariasis is caused due to vector transmission (Culex mosquito).
5. Medicines for treatment of Ascariasis are Piperazine, Mebendazole, Levamisole, Pyrantel.	5. Medicines for treatment of filariasis are diethyl- carbamazine citrate.

Short answer questions

Question 1.
Despite constant exposure to variety of pathogens, why do most of us remain healthy?
Answer:

1. All human beings are exposed to various foreign bodies, including infectious agents like bacteria, viruses, etc. which are called pathogens.
2. But human body can resist almost all types of these pathogens.
3. For this purpose, there is immune system which protects us from various infectious agents.
4. There is resistance and prevention of the damage or disease, through our defence mechanisms.
5. Thus, despite constant exposure to variety of pathogens, most of us remain healthy.

Question 2.
What are the unique features of acquired immunity?
Answer:
Following are the unique features of acquired immunity:

1. Specificity : Production of specific antibody or T-lymphocyte against a particular antigen/ pathogen is called specificity.
2. Diversity : Ability to recognize vast variety of diverse pathogens or foreign molecules by immunity is called diversity.
3. Discrimination between self and non¬self : Acquired immunity can differentiate between own body cells (self) and foreign (non-self) molecules.
4. Memory : The first immune response upon encounter of a specific foreign agent and its elimination is retained as a memory. This results in quicker and stronger immune response when the same pathogen is encountered again.

Question 3.
Describe the polypeptide chains seen in the structure of an antibody.
Answer:

1. There are four polypeptide chains which make the antibody.
2. There are two heavy or H-chains and two light or L-chains.
3. Each chain has two distinct regions, the variable region and the constant region.
4. Variable regions carry the antigen binding site or paratope.
5. This part of antibody recognizes and bindsto the specific antigen to form an antigen- antibody complex.

Question 4.
Which are the antimicrobial substances in blood and tissues?
Answer:

1. There are more than 30 serum proteins, circulating in the blood in an inactive state, which forms the complement system.
2. ‘Complement cascade’ is activated by presence of pathogens and thus they eliminate pathogens.
3. Cells which are affected by viruses secrete interferons which are a class of cytokines. These soluble proteins attack the pathogens.
4. Some leucocytes stimulate other cells to protect themselves from viral infection.

Question 5.
Is developing fever a bad sign or a good action? Explain.
Answer:

1. Fever is the innate immunity mechanism. When there is fever, the body rises.
2. This is in response to infection in a natural way.
3. Developing fever is a natural defence mechanism.
4. Fever helps to accelerate the physiological processes by which the invading pathogens are destroyed.
5. Fever stimulates the production of interferon and helps in recovery from viral infections.
6. Taking all these points into consideration, getting fever is a good action that innate immunity is working properly, however, entry of pathogen in the body causing illness is a bad sign.

Question 6.
With the help of a chart explain the compatibility of human blood groups.
Answer:
Different types of blood groups:

1. Person with blood group A can donate blood to persons having blood group A or AB and receive blood from A or O.
2. Person with blood group B can donate blood to persons having blood group B or AB and receive blood from B or O.
3. Person with blood group O can donate blood to all the persons having blood group either A, B, AB or O, because blood group O is universal donor. But can receive blood only from person having blood group O.
4. Person with blood group AB can donate blood to only AB but can receive blood from all the persons having blood group either, A, B, O or AB because AB is universal recipient.

Question 7.
The blood group of Krutika is O Rh +ve. What would be the possible blood groups of her parents?
Answer:

1. Krutika has O blood group, therefore her genotype is I°I°.
2. Her parents can be of following combinations.
3. They may be having blood group A or B with heterozygous genotype respectively, i.e. IAI° and IBI°. If both of them are heterozygous, having either A or B blood group, Krutika can be of O type.
4. Other possibility is both the parents have to be O.
5. For being Rh positive, her at least one parent should be Rh positive. Rh negative is a recessive phenotype and hence needs double dose of these genes.

Question 8.
Can a person with blood group O Rh+ve donate blood to a patient with blood group O Rh-ve? Why?
Answer:
No. blood group O may be common to both donor and recipient but their Rh factor is different. A person with RH+ve blood cannot donate to a patient with Rh-ve blood group. In Rh+ve blood there is antigen D. This antigen D when enters the body of recipient, there are production of anti-RH antibodies in his or her body. These anitibodies with cause agglutination of the Rh +ve blood which will be given to the patient. This agglutination will cause clots inside the vital organ of the recipient and the death may follow.

Question 9.
Why do we suffer from common cold repetitively in our life, but other viral diseases like Influenza or Small pox only once?
Answer:
Influenza infection causes production of antibodies in our body, once the virus attacks us. Therefore, second encounter with the virus may not cause effect. But in case of common cold, large number of different virus families are responsible for developing infection of common cold. Many a times different allergens are also inducing agents for common cold. Thus we may suffer from common cold again and again.

Question 10.
What are the symptoms of pneumonia?
Answer:

1. Main symptoms of infectious pneumonia are cough producing greenish or yellow sputum or phlegm and a high fever with chills.
2. Shortness of breath, stabbing chest pain, coughing up blood, headaches, sweaty and clammy skin, loss of appetite, fatigue, blueness of the skin, nausea, vomiting, mood swings and joint pains or muscle aches are some other symptoms.

Question 11.
What is the diagnosis and treatment of filariasis? How can we control this disease ?
Answer:
I. Diagnosis and Treatment : For the patient, diethyl-carbamazine citrate is the drug used for twice a day for three weeks. Thereafter for five days every six months the same treatment is repeated. This becomes effective against filarial worms.

II. Prevention and Control:

1. Mosquito eradication should be done for controlling filariasis.
2. In the areas with mosquitoes, avoid mosquito bite by using mosquito nets and insect repellents.

Question 12.
What are the signs and symptoms of filariasis?
Answer:
Signs and symptoms of filariasis:

1. As the lymphatic drainage does not take place, there is oedema with thickening of skin and underlying tissue.
2. Extremities like legs, arms, breasts, scrotum, etc. are affected by nematode causing lymphatic filariasis, i. e. Wuchereria bancrofti.
3. Lymph vessels and lymph nodes are enlarged and swollen.
4. Elephantiasis is seen in which limbs are swollen like legs of elephant.
5. Lymphoedema, i.e. accumulation of lymph fluid is seen in tissue causing swelling.
6. Hydrocele condition develops in which testis are enlarged due to accumulation of lymphatic fluid in testis.

Question 13.
What are the various ways in which mosquitoes can be eradicated from any area?
Answer:
Eradication of mosquitoes:

1. Removal of all stagnant water pools around the houses.
2. If such water bodies are there, they should be sprayed with insecticides.
3. But better option which is eco-friendly is releasing mosquito eating fish like Gambusia or Tilapia.
4. Use of mosquito repellent plants like Citronella. Use of coils and repellent creams.
5. Fumigation of the area to kill the mosquito.
6. Aedes sps. breed in man-made containers, especially plastic and cement tanks. Care should therefore be taken to dispose such containers properly. Water should not be accumulated in them.

Question 14.
What precautions will you take if you are travelling in an area which has lot of mosquitoes?
Answer:

1. Avoiding areas where mosquitoes are in more concentration.
2. Carrying mosquito repellent creams or coils.
3. Use of mosquito nets and other fumigation devices.
4. Wearing full clothing in light colours.
5. Staying indoors when mosquitoes are swarming, especially in the evening.
6. Taking anti-malarial pills as a precautionary measure.

Question 15.
Deaddiction may be difficult but not impossible. Collect information about NGOs, working in the field of deaddiction.
Answer:
There are many NGOs that work in the field of deaddiction. In different cities, there are different organizations. Even the central Government has started helpline number 1800-11-0031 for those drug and alcohol addicts who need help to come out of these addictions. Muktangan in Pune is one such reputed organization which does the great work in the field of deaddiction.

Question 16.
What are the most common warning signs of drug and alcohol abuse among youth ?
Answer:
The most warning signs of addictions are as follows:

1. Drop in academic performance, absenteeism from school or college.
2. No interest in personal hygiene and hobbies.
3. Withdrawal from the society, increased tendency of isolation and depression.
4. Aggressive and rebellious behaviour resulting into strained relationships with family and friends.
5. Fatigue and change in sleeping and eating habits.
6. Fluctuations in weight, appetite, deteriorating health, etc.

Question 17.
What are the preventive measures for malaria?
Answer:
Preventive measures of malaria :

1. Transmission of malarial parasite can be reduced by preventing mosquito bites. Therefore, mosquitoes should be controlled or totally eradicated.
2. This can be done by using of mosquito nets and insect repellents.
3. Mosquito control measures such as spraying insecticides inside houses and draining stagnant water where mosquitoes lay their eggs.
4. The mosquito larvae can be eradicated by releasing Gambusia fish which can feed upon these larvae.
5. Vaccine against malaria is also under preparation.

Question 18.
How does Entamoeba histolytica causes amoebiosis ?
Answer:

1. Amoebiosis is spread through ingestion of the cyst form of Entamoeba histolytica. This is a commensal organism.
2. Cyst is a semi-dormant and hardy structure found in faeces of infected person.
3. Non-encysted amoebae are called trophozoites. The trophozoites die quickly after leaving the body but may also be present in faeces.
4. Trophozoites are rarely the source of new infections.
5. The infection may remain asymptomatic for many days as Amoeba can remain latent in the gastrointestinal tract.

Question 19.
Describe the signs and symptoms of amoebiasis.
Answer:
Amoebiasis shows following common symptoms:

1. Diarrhoea, flatulence, stool with mucus and abdominal pains (cramps) are common.
2. Stool sticky with mucus and blood.
3. Amoebae form cysts in the liver, in such case there is hepatomegaly, i.e. enlargement of liver.
4. Liver shows amoebic liver abscess accompanied with fever and pain in right side of the abdomen.

Question 20.
How can amoebiasis be prevented?
Answer:
Prevention of amoebiasis is to be done at two levels, viz. at home and at endemic level.
1. Prevention of the spread of amoebiasis at the home level:

• Washing hands with soap and water after using the toilet or changing a baby’s diaper and before handling and eating food.
• Cleaning bathrooms and toilets properly with germicides.
• Avoiding raw vegetables when in endemic areas where they are grown in soil fertilized by human faeces.
• Boiling and purifying the drinking water.

2. Prevention of the spread of amoebiasis at endemic level:

• Avoiding consumption of street foods especially in public places.
• Following good sanitary practice, as well as using proper sewage disposal or treatment.
• E. histolytica cysts are usually resistant to chlorination; therefore sedimentation and filtration of water supplies are necessary to reduce the incidence of infection.
• Avoiding shared towels or face washers.

Question 21.
Describe the symptoms of ascariasis.
Answer:

1. After infection by Ascaris lumbricoides, there is appearance of eggs in stools in 60 – 70 days.
2. In larval ascariasis, symptoms are seen in 4-16 days after infection.
3. The final symptoms are gastrointestinal discomfort, colic and vomiting, fever and appearance of live worms in faeces.
4. Some patients may have pulmonary symptoms. Inflammation of alveolar walls is seen. This is known as pneumonitis.
5. Some may show neurological disorders during migration of the larvae.
6. Loss of appetite which reflects in weight loss.
7. A bolus of worms may obstruct the intestine.
8. Larvae that migrate may also cause eosinophilia, i.e. increase in number of eosinophils.

Question 22.
What are the preventive measures against ascariasis?
Answer:

1. Prevention of ascariasis can be done by adopting the following measures :
2. Use of proper toilet facilities.
3. Safe disposal of excreta.
4. Protection of food from dirt and soil.
5. Washing of vegetables before cooking and avoiding eating raw, unwashed vegetables and fruits.
6. Hand washing and use of safe food. Observing personal hygiene.
7. Use of pharmaceutical drugs such as Mebendazole and Albendazole can kill Ascaris.

Question 23.
Discuss the clinical manifestation of AIDS.
Answer:
There are four stages of clinical manifestations or symptoms of AIDS.

1. Stage I : This is initial infection with the virus and formation of antibodies, usually 2-8 weeks after initial infection.
2. Stage II : In this stage the person is asymptomatic carrier. Incubation takes place with a period ranging for 6 months to 10 years.
3. Stage III : This is called AIDS related complex (ARC). In this stage, one or more of the following clinical signs are seen. E.g. Recurrent fever for longer than one month, fatigue, unexplained diarrhoea, night sweats, shortness of breath, loss of more than 10 per cent body weight, etc.
4. Stage IV : This is the end stage in which patient shows full blown AIDS. Thus it is called the end stage of HIV infection. Life threatening opportunistic infections (like pneumonia, tuberculosis, Kaposi sarcoma, etc.) are easily caught during this period.

Question 24.
Through which modes HIV infection does not spread?
Answer:

1. HIV does not spread through casual contact such as hugging, etc.
2. Insect bite such as mosquito bites does not transmit HIV.
3. Participation in sports is not the mode by which HIV transmits.
4. Contact between articles used by AIDS patient, a hand shake with him or her does not transmit HIV
5. HIV infections do not occur through swimming pool or by sharing clothes, utensils, etc.

Question 25.
Write about laboratory diagnosis and treatment of AIDS.
Answer:
I. Laboratory diagnosis :

1. There are two tests for diagnosis of AIDS.
2. First test is ELISA (Enzyme-Linked Immunosorbent Assay) which is used to detect the HIV antibodies.
3. The second confirmatory test is Western Blot, which is used to weed out any false positive results. It is a highly specific test.
4. It is based on detecting specific antibody to viral core protein and envelope glycoprotein.

II. Treatment of AIDS:

1. AIDS cannot be cured.
2. Antiretroviral drugs are used to reduce the viral load and prolong the life of HIV patient. E.g. Antiretroviral therapy (ART) uses drugs such as TDF (tenofovir), EFV (Efavirenz), Lamivudine (3TC), etc.

Question 26.
Describe the structure of HIV with a suitable diagram.
Answer:

1. Human Immunodeficiency Virus or HIV is spherical and 100 to 140 nm in diameter.
2. It has centrally located two ss RNA molecules along with reverse transcriptase enzymes.
3. There are coverings of two layers of proteins. The outer layer formed by matrix protein (pi7) while in inner layer is of capsid protein (p24).
4. An additional layer of lipids is seen over the matrix protein layers. This layer is impregnated with glycoprotein GP120 and GP41.
   

Question 27.
What are the modes of transmission of HIV or AIDS?
Answer:
The transmission of HIV occurs through following routes:

1. Sexual relations, mainly unsafe sexual contact including oral, vaginal and anal sex.
2. Through blood and blood products either by blood transfusions or sharing needles and syringes.
3. Transplacental From pregnant mother to her foetus through placenta. Nursing mother can also transmit HIV to her baby through lactation.
4. Spreading the virus is very rare in case of accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs.
5. HIV is seen in urine, tears, saliva, breast milk and vaginal secretions but unless these body fluids enter the injuries and wounds, transmission is not easy.

Question 28.
What are the measures of prevention and control of AIDS ?
Answer:

1. Preventive measures : AIDS has no cure, hence prevention is the best choice. The
   following steps help in preventing this dreadful disease.
2. High risk group people should be educated about HIV transmission. They should never donate blood.
3. Use of disposable needles and syringes should be done with proper disposal.
4. Risky sexual habits should be avoided.
5. Tooth brushes, razors, other articles that can become contaminated with blood should not be shared.
6. Blood should be screened before receiving it.
7. Routine screening of blood and semen donors, organ donors (kidney, liver, lung, cornea), and patients undergoing haemodialysis must be done.
8. Pregnant women or those women who are contemplating pregnancy should be regularly screened.

Question 29.
Explain the ill-effects of opioids and cannabinoids on health.
OR
What are harmful effects of drug abuse?
Answer:

1. Opioids bind to specific opioid receptors which are present in central nervous system and gastrointestinal tract. Some opioids e.g. heroin act like depressants and slow down all the body functions.
2. Cannabinoids have the capacity to interact with receptors present in the brain. Inhalation or ingestion of cannabinoids such as marijuana, hashish, charas and ganja have adverse effect on cardiovascular system.
3. Cannabinoid like LSD causes hallucinations.
4. All these substances are addictive and hence cause adverse effects on the body and health.

Question 30.
Give the adverse effects of opioids, cannabinoids and morphine on human health.
Answer:
1. Opioids:

• Opioids bind to specific opioid receptors present in the central nervous system and in gastrointestinal tract.
• They are depressants and slow down the body functions.

2. Cannabinoids:

• Cannabinoids interact with receptors in the brain.
• They affect cardiovascular system of the body.

3. Morphine:

• Morphine is an effective sedative and pain killer when used for medicinal purpose.
• When abused it affects physical, physiological and psychological functions.

Give reasons

Question 1.
Vaccines are safe.
Answer:
During their manufacture, vaccines are rigorously tested. Many rounds of study, examination and research are carried out before they are used for general public. Extensive research and evidences are gathered to check their safety. Sometimes, some vaccines produce side effects but these are rare and mild. Hence vaccines are considered to be safe.

Question 2.
Innate immunity is also known as non-specific immunity.
Answer:
Innate immunity is non-specific because it does not depend on previous exposure to foreign substances. It is inborn capacity of the body to resist the pathogen that causes the disease. It is natural immunity and hence it remains non-specific, trying to protect the body in case of any invasion of foreign body.

Question 3.
Vaccination is important for preventing pneumonia.
Answer:
Vaccinations for pneumonia are available against Haemophilus influenzae and Streptococcus pneumoniae. If given earlier in life, they reduce the chances of catching pneumococcal infections. The deaths can be prevented which are common due to lung infections. Since it is a common and chronic j disease for all age groups, for the prevention one must take vaccination.

Question 4.
Common cold is the most frequent ! infectious disease in humans.
Answer:
Common cold is caused by virus which is abundantly present in congested city environments. The upper respiratory tract is infected due to these rhinoviruses and corona viruses. Average adult contracts such infections 2 to 4 times in a year while children capture it 6 to 12 times in a year. Therefore, it is said to be the most frequent infectious disease of human beings.

Question 5.
Typhoid is food and water-borne disease.
Answer:
Typhoid is caused due to Salmonella typhi which isa Gram-negative bacterium, transmitted from a patient or carrier to another healthy person through contaminated food or water. Flying insects, mostly houseflies transmit the bacteria from faeces to the food. Poor hygienic habits and improper public sanitation system spreads typhoid. Therefore, it is said to be food and water-borne disease.

Question 6.
Malignant tumour is more dangerous than benign tumour.
Answer:
Malignant tumour cells can show metastasis and hence can spread far and wide in the body, affecting other healthy cells and tissues. They are difficult to cure by any therapy as one does not know exact location of the cancerous cells. They also produce variety of symptoms depending on their i location. On the contrary, benign tumours can be treated surgically. Since they are covered by a cyst like membrane, the cancerous cells do not spread from them. Thus malignant tumour can be lethal as against the benign tumour.

Question 7.
Prevention is better than cure for AIDS.
Answer:
Till this time, there is no preventive vaccination for AIDS. There is also no cure for AIDS. The medicines are also costly and may not give complete cure. The only way to remain away from AIDS is the complete awareness about it. Thus it should be prevented by not allowing HIV to enter our body. Once HIV finds the entrance, the cure is impossible. Therefore, it is said that prevention is better them cure for AIDS.

Write short notes

Question 1.
Cellular factors in innate immunity.
Answer:

1. Phagocytic cells ingest and destroy the pathogens.
2. This is natural defence against the invasion of pathogenic microorganisms and other foreign particles in blood and tissues.
3. Phagocytic cells are of two types, viz. microphages and macrophages. They can remove foreign particles that enter the body.
4. Natural killer (NK) cells is a class of lymphocytes which carry out important and non-specific defence against viral infections and tumours.

Question 2.
Acute phase proteins (APPs).
Answer:

1. Acute phase proteins are involved in innate immune mechanism.
2. When there is an infection or injury, it leads to a sudden increase in concentration of certain plasma proteins, which are called acute phase proteins or APPs.
3. These include C Reactive Protein (CRP), Mannose binding protein, Alpha-1-acid glycoprotein, Serum Amyloid P etc.
4. APPs enhance host resistance, prevent tissue injury and promote repair of inflammatory lesions.

Question 3.
Rh factor.
Answer:
Rh factor:

1. Rh factor is the term adapted from Rhesus monkey.
2. In rhesus monkey, there is antigen D on the surface of their RBCs.
3. Landsteiner and Wiener discovered this antigen and termed it as Rh factor.
4. Persons having Rh factor or D antigen are called Rh positive while those lacking D antigen or Rh factor are called Rh negative.

Question 4.
Erythroblastosis foetalis.
Answer:

1. Erythroblastosis foetalis is condition in which there is destruction of the erythrocytes of the foetus. It is the haemolytic disease of the newborn (HDN).
2. This is caused in foetus, if mother is Rh -ve and father is Rh +ve. Rh +ve is the dominant allele, the foetus becomes Rh +ve, when its father is RH +ve.
3. Rh +ve blood groups have D antigen which induces a strong immunogenic response when introduced into Rh -ve individuals.
4. During foetal life, there is connection between mother and foetus through placenta, therefore Rh +ve antigen D from the foetus enters maternal circulation.
5. This triggers formation of anti-Rh antibodies in mother. Subsequently Rh+ve foetus receives anti-Rh antibodies produced by mother.
6. This causes agglutination reaction resulting into haemolysis in foetus. In order to prevent HDN, Rh -ve mother is injected with the anti-Rh antibody during all her pregnancies if her husband is Rh +ve.

Question 5.
Common cold.
Answer:

1. The common cold (nasopharyngitis or rhinopharyngitis) is a viral infectious disease of the upper respiratory system. The causative organisms are rhinoviruses and coronaviruses.
2. Symptoms include cough, sore throat, runny nose and fever.
3. There is no known treatment, however, symptoms usually resolve spontaneously in 7 to 10 days.
4. The best prevention for the common cold is to stay away from infected people and places where infected individuals have been.
5. Hand washing with plain soap and water is recommended. Also alcohol-based hand sanitizers provide very little protection.

Question 6.
Life cycle of Plasmodium.
Answer:

1. Anopheles Female mosquito which is a carrier carries sporozoites. When it bites the human, these sporozoites enter human circulation.
2. Sporozoites undergo asexual reproduction through fission or schizogony in the liver cells or erythrocytes of the human.
3. It forms merozoites. The cells formed within erythrocytes function as gametocytes. They undergo gamogony.
4. Upon biting such person, the gametocytes enter into female Anopheles, fertilization occurs in its gut.
5. Diploid zygote transforms into oocyst. Oocyst forms large number of haploid sporozoites through meiosis (sporogony).
6. Sporozoites migrate to salivary glands and are ready to infect new human host.
7. Again Sporozoite → Merozoite → Trophozoite → Schizont sequence is carried on for plasmodial stages in human body.
8. The sexual life cycle of Plasmodium occurs in mosquito body which acts as a vector. While its asexual phase takes place in human body.

Question 7.
Pneumonia.
Answer:

1. Pneumonia is an inflammatory condition of alveoli in the lungs causing formation of fluid in the lungs. This condition is called consolidation and exudation.
2. Causes of pneumonia are infection due to bacteria, viruses, fungi or parasites, chemical burns or physical injury to the lungs.
3. Influenza virus, adenovirus, para influenza and Respiratory Syncytial Virus (RSV) are some viruses that can cause pneumonia. Bacteria like Streptococcus pneumoniae and fungal pathogens e.g. Pneumocystis jirovecii and Pneumocystis carinii can also spread infection of pneumonia. Chemical burns of physical injury to lungs also cause similar infection.
4. Main symptoms of infectious pneumonia are cough producing greenish or yellow sputum or phlegm and a high fever with chills.
5. Shortness of breath or dyspnea, stabbing chest pain, coughing up blood, headaches, sweaty and clammy skin, loss of appetite, fatigue, blueness of the skin, nausea, vomiting, mood swings and joint pains or muscle aches are some other symptoms.
6. Preventive vaccination against pneumonia is available. Medicines such as Benzyl penicillin, Ampicillin and Chloramphenicol are effective to prevent pneumonia.

Question 8.
Ringworm.
Answer:

1. Ringworm or Dermatophytosis is a clinical condition caused by Trichophyton and Microsporum fungal infection of the skin. This infection is seen in humans and pets.
2. Dermatophytes are the fungi that feed on keratin. Keratin is the material found in the outer layer of skin, hair and nails.
3. These fungi attack various parts of the body. Infections on the body forms enlarged raised red rings. These patches have intense itching. Infection on the skin of the feet may cause athlete’s foot and jock itch.
4. When the nails are infected it causes onychomycosis. During this the nails thicken, discolour and finally crumble and fall off.
5. Prevention of ringworm infection is to be done by avoiding sharing of clothing, sports equipment, towels or sheets. Clothes should be washed in hot water with fungicidal soap after suspected exposure to ringworm. One should not walk barefoot but use appropriate footwear.
6. Diagnosis of ringworm is done by physical examination and treatment is done with uses drugs like nystatin, fluconazole, itraconazole, etc.

Question 9.
Dengue.
Answer:

1. Dengue is a viral disease causing high fever. It is a painful, debilitating vector-borne disease.
2. There are four closely related dengue viruses that cause infection.
3. Vector of Dengue virus is female Aedes mosquito. The mosquito takes up the dengue virus when it sucks blood of a person suffering from dengue.
4. The spread of dengue is not directly from one person to another person.

Question 10.
Performance enhancers.
Answer:

1. Performance enhancers are certain drugs used by sportspersons to enhance their performance during competitions.
2. Narcotic analgesics, anabolic steroids, diuretics and certain hormones are misused by such sportspersons to increase muscle strength and bulk. It also promotes aggressiveness and improve overall performance.
3. Use of anabolic steroids cause side effects.
4. Females show masculinization, increased aggressiveness, mood swings, depression, abnormal menstrual cycles, excessive hair growth on the face and body, enlargement of clitoris, deepening of voice.
5. Males show acne, increased aggressiveness, mood swings, depression, and reduction of size of the testicles, decreased sperm production, kidney and liver dysfunction, breast enlargement, premature baldness, enlargement of the prostate gland.
6. These effects may be permanent with prolonged use. Using such drugs is illegal and punishable.

Chart Based Questions

Question 1.
Complete the chart of ABO blood group system and answer the questions given below:

Questions:
(i) Which blood group from the above table is called universal acceptor?
(ii) Which blood group from the above table is called universal donor?
Answer:

(i) Blood group AB is called universal acceptor.
(ii) Blood group O is called universal donor.

Question 2.
Complete the table

Plasmodium species	Incubation period	Pattern of fever
————	————–	High fever after 48 hours.
————–	28 days	————–
————–	17 days	————-
————	————-	High fever at irregular intervals between 22 to 48 hours.

Answer:

Plasmodium species	Incubation period	Pattern of fever
P. vivax	14 days	High fever after 48 hours.
P. malariae	28 days	High fever after 72 hours interval
P. ovale	17 days	High fever after 48 hours interval
P. falciparum	12 days	High fever at irregular intervals between 22 to 48 hours.

Question 3.
Complete the following table

Answer:

Question 4.
Complete the following table:

Carcinogen	Organ affected
N-nitrosodimethlene	————–
Aflatoxin	———-
————–	Vagina
————–	Urinary bladder
————–	Prostate
————–	Skin and lungs

Answer:

Carcinogen	Organ affected
N-nitrosodimethlene	Lungs
Aflatoxin	Liver
Diethylstilboestrol	Vagina
2-naphthylamine and 4-aminobiphenyl	Urinary bladder
Cadmium oxide	Prostate
Soot, coal tar (2-4 benzopyrene)	Skin and lungs

Diagram Based Questions

Question 1.
Label the given diagram
Img 7
Answer:

1. Antigen binding site.
2. Variable region of heavy chain
3. Varible region of light chain
4. Constant region of light chain
5. Constant region of heavy chain
6. Disulphide bond
7. Hinge
8. Light chain
9. Heavy chain

Question 2.
Observe the given diagram and answer the following questions:

(1) What is I and II in the above diagram?
Answer:
I is a virus which is trying to cause infection, II are two antigen molecules which are trying to attack the virus.

(2) What structures are responsible for antigen and antibody complex? Identify them in the above diagram.
Answer:
(a) is epitope which is antigen determinant and
(b) is a paratope which is part of the antibody. Epitope and paratope are specific to each other and hence they form a complex.

(3) What is the study of antigen-antibody interactions called?
Answer:
The study of antigen-antibody interactions is called serology.

Question 3.
Fill in the blanks after observing the diagram.

On the right side of the diagram, the stages of plasmodium are passed in the body of ……………….. Whereas on the left side of the diagram, those take place in the body of ………………, ………………. is the stage that is dormant in the liver of human host. From this ………………… and then ………………… is the stage in the erythrocytes, which rupture and gives rise to …………….. Microgamete and macrogamete fuse with each other to form …………………… which later gives rise to ookinete which forms ………………… This enters the salivary glands of mosquito …………………. phase of Plasmodium occurs in mosquito body, whereas …………………… phase is in human body.
Answer:
On the right side of the diagram, the stages of plasmodium are passed in the body of human. Whereas on the left side of the diagram, those take place in the body of mosquito. Hypnozoite is the stage that is dormant in the liver of human host. From this schizont and then merozoites Trophozoite is the stage in the erythrocytes, which rupture and gives rise to gamerocyte. Microgamete and macrogamete fuse with each other to form zygote which later gives rise to ookinete which forms sporozoites. This enters the salivary glands of mosquito. Sexual phase of Plasmodium occurs in mosquito body, whereas asexual phase is in human body.

Question 4.
Observe the given diagram and answer the following questions

(1) Enlist the stages of Entamoeba histolytica you see in the above diagram.
Answer:
Trophozoite, pre-cystic form, cyst, binucleate cyst, quadrinucleate cyst, Metacycstic amoeba, amoebulae are the different stage of Entamoeba histolytica that are seen in the above diagram.

(2) Where are these stages passed?
Answer:
These stages are passed in the lumen of intestine of the host human being.

(3) How does Entamoeba come out of the body of the host?
Answer:
Encysted Entamoeba pass out with the faecal matter of the host.

Question 5.
Observe the given diagram and answer the following questions:

(1) In life cycle of Ascaris at what stage do they enter the human body?
Answer:
When there is development of infective larva inside the egg of Ascaris, it enters the human body.

(2) How do they enter the human body and through which organ do they enter?
Answer:
Ascaris eggs are deposited in the faeces. They mix in the soil, if faeces is exposed in open. From there, it can enter into nearby water body or it may contaminate vegetables or other food stuffs. Such unhygienic food or unclean hands pass these eggs in the body of human through the mouth.

(3) What are the vital organs affected by the Ascaris during its development within the body of host human ?
Answer:
Ascaris can affect trachea, lungs, heart, brain and eyes too.

(4) In which organ do they copulate and produce fertilized eggs?
Answer:
Adult male and female copulate in the intestine of the host human.

Question 6.
Observe the given diagram and answer the following questions:

(1) In which host stage II is completed?
Answer:
Stage II is completed in human.

(2) What happens in step 1 and step 2?
Answer:
Humans are infected at step 1 when mosquito bites human and larvae enter blood stream. In step 2 adult Wuchereria worms are formed in lymphatics.

(3) Describe the events in step 3.
Answer:
Mosquito carries the blood as it bites the human in step 4 and ingests microfilariae in human blood. Later the microfilariae start growing in the midgut of mosquito.

(4) In which host stage I is completed?
Answer:
Stage I is completed in mosquito.

Question 7.
Sketch and label the diagram of pathogen that causes typhoid.
Answer:
Pathogen of typhoid is Salmonella typhi.

Question 8.
Sketch and label disease causing agents of pneumonia.
Answer:

Question 9.
Sketch and label benign and malignant tumours.
Answer:

Question 10.
Sketch and label structure of HIV.
Answer:

Long answer questions

Question 1.
Describe different ways in which epithelial surface offers the innate immunity.
Answer:

1. Skin and mucous covering, when intact protects the body against the invasion by pathogens. The healthy skin has bactericidal activity due to the salts present in drying sweat.
2. Sebaceous glands in the skin produce secretions and long chain of fatty acids. These are bactericidal and fungicidal.
3. Respiratory tract is provided with mucosa which prevents entry of microorganisms to a large extent.
4. The inhaled particles are arrested through hair in the nasal passage. The particles that pass beyond nasal passage are caught by mucus lining the epithelium. They are swept back to pharynx. Then they are either swallowed or coughed out.
5. The cough reflex is an important defence mechanism of respiratory tract.
6. There is saliva in the mouth which has inhibitory effect on microorganisms. Gastric secretions has acidity and hence microorganisms are destroyed in stomach.
7. The flushing action of urine eliminates bacteria from the urethra. Semen too has antibacterial substances, e.g. Spermine and zinc.

Question 2.
Explain ABO blood group system in human being with a suitable chart.
Answer:

1. In ABO system, the blood groups are determined by the antigen present on the surface of red blood cells.
2. The blood groups are of four types, viz. A. B, AB and O.
3. In person with blood group A there is antigen ‘A’ on the surface of their red blood cells (RBCs) and antibodies ‘b’ in their plasma.
4. In person with blood group B there is antigen ‘B’ on the surface of their red blood cells (RBCs) and antibodies ‘a’ in their plasma.
5. In person with blood group AB there are both antigens ‘A’ and ‘B’ on the surface of their RBCs and no antibodies in their plasma.
6. In person with blood group ‘O’ there are no antigens ‘A’ and ‘B’ on the surface of their RBCs but have both ‘a’ and ‘b’ antibodies in their plasma.
7. During blood transfusion compatibility of blood has to be taken into consideration.
8. Person with ‘O’ blood group is called universal donor while the person with ‘AB’ blood group is called universal recipient. Individuals with blood group O can donate blood to anyone, while those individuals with blood group AB can receive blood from any person.
   

Question 3.
What are the main causes of cancer?
Answer:
Causes of Cancer : Following carcinogenic factors are responsible for causing cancer.

1. Chemicals : Many induce development of cancer. E.g. nicotine, caffeine, polycyclic hydrocarbons and products of combustion of coal and oil. Sex hormone and steroids, if given or secreted in excess, can cause cancer. E.g. Breast cancer.
2. Radiation : Radiations such as X-rays, gamma-rays, cosmic rays, ultra-violet rays are carcinogenic.
3. Viruses : Virus possessing oncogenes (v-onc genes) are carcinogenic. E.g. EBV (Epstein-barr virus), HPV (Human papiloma virus) are oncogenic viruses.
4. Oncogenes : Cellular oncogenes (c-onc genes) or proto-oncogenes can cause cancer. They are present in normal cells but if activated they lead to oncogenic transformation of cells.
5. Addiction : Addictive substances like cigarette smoke, tobacco lead to cancer of mouth, lips and lungs. Alcohol can cause cancer of oesophagus, stomach, intestine and liver. Drugs like marijuana or anaerobic steroids can also cause cancer.

Question 4.
What are the different ways of treating cancer?
Answer:
Cancer treatment consists of combination of a number of therapies which are follows:
(1) Chemotherapy : Chemotherapy means giving certain anticancer drugs. These drugs check cell division by inhibiting DNA synthesis. But these are more toxic to cancerous cell than to normal cells. Chemotherapy shows side effects such as hair loss or anaemia.

(2) Radiotherapy : In addition to chemotherapy, radiations are given. The cancer cells are bombarded with the radiations from radioactive materials such as cobalt, iridium and iodine. The X-rays, gamma rays and charge particles are used to destroy the cancerous tissue or cells. They cause minimum damage to the surrounding normal tissue or cells.

(3) Surgery : Entire cancerous tissue or cells are removed surgically. E.g. breast tumour or uterine tumour. After removing the cancerous tissue, additionally other treatments are also given.

(4) Immunotherapy : For tackling with tumour, patients are given biological response modifiers such as a-interferon which activates their immune system to destroy the tumour.

(5) Supportive therapy : With supportive therapy, patient’s quality of life is increased. To treat symptoms of cancer and side effects of cancer treatments, this therapy is used. This therapy varies depending upon condition of individual patient.

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 9 Control and Co-ordination Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 9 Control and Co-ordination

Multiple choice questions

Question 1.
The supporting cells that produce myelin sheath in the CNS are ……………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(a) Oligodendrocytes

Question 2.
Human brain develops to its full size at an age of year/s.
(a) 1
(b) 6
(c) 12
(d) 18
Answer:
(b) 6

Question 3.
Telencephalon is the other name of ……………….
(a) pons varolii
(b) medulla oblongata
(c) cerebrum
(d) cerebellum
Answer:
(c) cerebrum

Question 4.
Olfactory tracts merge in olfactory area of lobe.
(a) Frontal
(b) Parietal
(c) Occipital
(d) Temporal
Answer:
(d) Temporal

Question 5.
………………. is the largest commissure of the human brain.
(a) Corpora striata
(b) Corpora quadrigemina
(c) Habencular commissure
(d) Corpus callosum
Answer:
(d) Corpus callosum

Question 6.
Grey matter of the brain shows large collection of ……………….
(a) dendrons
(b) cytons
(c) axons
(d) synapsis
Answer:
(b) cytons

Question 7.
Masses of grey matter in white matter of the cerebrum are called ……………….
(a) corpora striata
(b) corpus callosum
(c) paracoel
(d) basal ganglia
Answer:
(d) basal ganglia

Question 8.
Parietal and temporal lobes of cerebrum are separated by sulcus.
(a) lateral
(b) parieto occipital
(c) central
(d) median longitudinal
Answer:
(a) lateral

Question 9.
……………… area is motor speech area
(a) Acoustic
(b) Wernike’s
(c) Somato sensory
(d) Broca’s
Answer:
(d) Broca’s

Question 10.
Maxillary nerve is a branch of nerve.
(a) Occulomotor
(b) Trochlear
(c) Trigeminal
(d) Facial
Answer:
(c) Trigeminal

Question 11.
Spinal accessory is the cranial nerve.
(a) IV
(b) VI
(c) IX
(d) XI
Answer:
(d) XI

Question 12.
Rotation of eye ball is controlled by ……………….
(a) Optic nerve
(b) Pathetic nerve
(c) Auditory nerve
(d) Hypoglossal nerve
Answer:
(b) Pathetic nerve

Question 13.
The spinal nerves emerge out of vertebral column through ……………….
(a) intervertebral foramina
(b) neural canal
(c) central canal
(d) foramen magnum
Answer:
(a) intervertebral foramina

Question 14.
The neuro transmitter is removed by an enzyme called ……………….
(a) noradrenaline
(b) acetylcholine
(c) hyaluronidase
(d) cholinesterase
Answer:
(d) cholinesterase

Question 15.
The reflex action originates in ……………….
(a) sensory neuron
(b) motor neuron
(c) receptor organ
(d) effector organ
Answer:
(c) receptor organ

Question 16.
Cytons of neurons are located in dorsal root ganglion.
(a) afferent
(b) efferent
(c) adjustor
(d) association
Answer:
(a) afferent

Question 17.
Wall of carotid arteries contain ……………….
(a) thermoreceptors
(b) mechanoreceptors
(c) baroreceptors
(d) statoacoustic receptors
Answer:
(c) baroreceptors

Question 18.
The electronegativity inside the membrane is due to ……………….
(a) less anions than cations
(b) less cations than anions
(c) bicarbonates
(d) carbonates
Answer:
(b) less cations than anions

Question 19.
The neuro transmitters stimulate ……………….
(a) presynaptic membrane
(b) cyton
(c) axon terminals
(d) postsynaptic membrane
Answer:
(d) postsynaptic membrane

Question 20.
………………. is an extero-receptor.
(a) Thermoreceptor
(b) Baroreceptor
(c) Proprioreceptor
(d) Enteroreceptor
Answer:
(a) Thermoreceptor

Question 21.
The are described as windows for brain.
(a) sensory neurons
(b) motor neurons
(c) effectors
(d) sense organs
Answer:
(d) sense organs

Question 22.
Otolith organ is formed of ……………….
(a) cochlea and vestibule
(b) sacculus and utriculus
(c) semicircular canals
(d) ear ossicles
Answer:
(b) sacculus and utriculus

Question 23.
Olfactory bulbs are extensions of brain’s ……………….
(a) cerebrum
(b) limbic system
(c) RAS
(d) pons varolii
Answer:
(b) limbic system

Question 24.
Gustatory senses are noted by ……………….
(a) retina
(b) skin
(c) nose
(d) tongue
Answer:
(d) tongue

Question 25.
………………. is attached to the eardrum.
(a) Malleus
(b) Incus
(c) Stapes
(d) Cochlea
Answer:
(a) Malleus

Question 26.
Eustachian tube is present in ……………….
(a) external ear
(b) internal ear
(c) heart
(d) middle ear
Answer:
(d) middle ear

Question 27.
The internal ear is a fluid filled structure called ……………….
(a) cochlea
(b) vestibule
(c) labyrinth
(d) otolith
Answer:
(c) labyrinth

Question 28.
The space within cochlea is known as ……………….
(a) scala vestibule
(b) scala tympani
(c) aqueous chamber
(d) scala media
Answer:
(d) scala media

Question 29.
Thermoregulatory centre in the body is ……………….
(a) hypothalamus
(b) cerebellum
(c) spinal cord
(d) pituitary
Answer:
(a) hypothalamus

Question 30.
Which of the following is a sensory nerve?
(a) Vagus
(b) Auditory
(c) Facial
(d) Lumbar
Answer:
(b) Auditory

Question 31.
Chemical transmission in synapse occurs due to ……………….
(a) cholesterol
(b) ADH
(c) acetylcholine
(d) cholinesterase
Answer:
(c) acetylcholine

Question 32.
Voluntary muscular coordination is under the control of ……………….
(a) medulla
(b) pons
(c) hypothalamus
(d) cerebrum
Answer:
(d) cerebrum

Question 33.
All involuntary vital activities are under the control of ……………….
(a) medulla oblongata
(b) cerebellum
(c) cerebral hemispheres
(d) pons Varolii
Answer:
(a) medulla oblongata

Question 34.
Cerebellum is controlling centre for ……………….
(a) muscular strength
(b) memory
(c) equilibrium
(d) muscular coordination
Answer:
(c) equilibrium

Question 35.
Which receptors are present in the retina?
(a) Chemoreceptors
(b) Thermoreceptors
(c) Photoreceptors
(d) Baroreceptors
Answer:
(c) Photoreceptors

Question 36.
Breathing is controlled by ……………….
(a) trachea
(b) medulla oblongata
(c) lungs
(d) hypothalamus
Answer:
(b) medulla oblongata

Question 37.
Corpus callosum is a nerve fibre bridge which connects ……………….
(a) two cerebral hemispheres
(b) cerebrum and cerebellum
(c) cerebellum and medulla
(d) midbrain and hindbrain
Answer:
(a) two cerebral hemispheres

Question 38.
Centre for thirst and hunger are located in ……………….
(a) cerebrum
(b) cerebellum
(c) hypothalamus
(d) medulla
Answer:
(c) hypothalamus

Question 39.
Gyri in the brain are present in ……………….
(a) cerebral cortex
(b) olfactory lobes
(c) medulla oblongata
(d) hypothalamus
Answer:
(a) cerebral cortex

Question 40.
Which of the following is a structure of mesencephalon?
(a) Inferior colliculi
(b) Thalamus
(c) Cerebellum
(d) Pons varolii
Answer:
(a) Inferior colliculi

Question 41.
Third ventricle lies in ……………….
(a) midbrain
(b) forebrain
(c) cerebellum
(d) medulla oblongata
Answer:
(b) forebrain

Question 42.
Medulla oblongata encloses ……………….
(a) third ventricle
(b) fourth ventricle
(c) first ventricle
(d) second ventricle
Answer:
(b) fourth ventricle

Question 43.
Loss of memory may result from injury to the ……………….
(a) corpora quadrigemina
(b) pons varolii
(c) cerebellum
(d) cerebrum
Answer:
(d) cerebrum

Question 44.
Terminal non-nervous part of spinal cord is ……………….
(a) funiculus
(b) filum terminale
(c) cauda equina
(d) conus terminalis
Answer:
(b) filum terminale

Question 45.
Which part of the pituitary is neurohaemal organ?
(a) Pars distalis
(b) Infundibulum
(c) Pars nervosa
(d) Pars intermedia
Answer:
(c) Pars nervosa

Question 46.
Development of secondary sexual characteristics in female is under the control of ……………….
(a) growth hormone
(b) TSH
(c) estrogen
(d) progesterone
Answer:
(c) estrogen

Question 47.
Hypersecretion of STH in children causes ……………….
(a) cretinism
(b) gigantism
(c) dwarfism
(d) myxoedema
Answer:
(b) gigantism

Question 48.
Milk secretion in lactating woman is controlled by ……………….
(a) LH
(b) prolactin
(c) relaxin
(d) oestrogen
Answer:
(b) prolactin

Question 49.
ADH is secreted by
(a) adrenal gland
(b) thyroid
(c) hypothalamus
(d) pancreas
Answer:
(c) hypothalamus

Question 50.
BMR is increased by the administration of ……………….
(a) insulin
(b) GH
(c) thyroxine
(d) testosterone
Answer:
(c) thyroxine

Question 51.
The largest endocrine gland in the body is ……………….
(a) pituitary
(b) adrenal
(c) liver
(d) thyroid
Answer:
(d) thyroid

Question 52.
Diabetes insipidus is caused by the deficiency of ……………….
(a) calcitonin
(b) oxytocin
(c) atrial natriuretic factor
(d) vasopressin
Answer:
(d) vasopressin

Question 53.
Simple goitre is caused by the deficiency of ……………….
(a) TSH
(b) thyrocalcitonin
(c) thyroxine
(d) iodine
Answer:
(d) iodine

Question 54.
Exopthalmic goitre is also known as ……………….
(a) Grave’s disease
(b) Gull’s disease
(c) Simple goitre
(d) Cushing’s disease
Answer:
(a) Grave’s disease

Question 55.
Cushing’s syndrome is developed due to ……………….
(a) hyposecretion of ACTH
(b) hypersecretion of corticoids
(c) hyposecretion of thyroxine
(d) hypersecretion of thyroxine
Answer:
(b) hypersecretion of corticoids

Question 56.
Pituitary gland is under the control of ……………….
(a) thyroid
(b) adrenal
(c) pineal
(d) hypothalamus
Answer:
(d) hypothalamus

Question 57.
FSH is secreted by ……………….
(a) pituitary gland
(b) thyroid gland
(c) ovary
(d) adrenal gland
Answer:
(a) pituitary gland

Question 58.
ICSH stimulates ……………….
(a) ovary
(b) Leydig cells
(c) seminiferous tubules
(d) kidney
Answer:
(b) Leydig cells

Question 59.
Which of the following secrete LH ?
(a) Pituitary
(b) Thyroid
(c) Ovary
(d) Adrenal
Answer:
(a) pituitary

Question 60.
TSH regulates secretion.
(a) thyroxine
(b) MSH
(c) androgens
(d) insulin
Answer:
(a) thyroxine

Question 61.
Deficiency of thyroxine in adults cause ……………….
(a) cretinism
(b) myxoedema
(c) diabetes
(d) Cushing’s disease
Answer:
(b) myxoedema

Question 62.
Osmotic pressure and blood pressure are maintained by ……………….
(a) glucocorticoids
(b) aldosterone
(c) TRF
(d) MSH
Answer:
(b) aldosterone

Question 63.
Hormone secreted by corpus luteum is ……………….
(a) aldosterone
(b) progesterone
(c) testosterone
(d) cortisol
Answer:
(b) progesterone

Question 64.
………………. is also called hypophyseal stalk.
(a) Infundibulum
(b) Median eminence
(c) Pars intermedia
(d) Sphenoid bone
Answer:
(a) infundibulum

Question 65.
………………. is like a collar around hypophyseal stalk.
(a) Pars distalis
(b) Pars nervosa
(c) Pars intermedia
(d) Pars tuberalis
Answer:
(d) Pars tuberalis

Question 66.
Herring bodies are the parts of ……………….
(a) hypothalamo-hypophyseal tracts
(b) pituicytes
(c) hypothalamo-hypophyseal portal system
(d) pituitary cleft
Answer:
(a) hypothalamo-hypophyseal tracts

Question 67.
Corticotropin is the other name of ……………….
(a) ACTH
(b) STH
(c) Aldosterone
(d) ADH
Answer:
(a) ACTH

Question 68.
Adrenal failure leads to ……………….
(a) Acromegaly
(b) Simmond’s disease
(c) Midget
(d) Addison’s disease
Answer:
(d) Addison’s disease

Question 69.
Prolactin inhibiting factor is secreted by ……………….
(a) Hypophysis
(b) Hypothalamus
(c) Thyroid
(d) Mammary glands
Answer:
(b) Hypothalamus

Question 70.
Which one of the following is not applicable to prolactin ?
(a) Mammotropin
(b) Lactogenic hormone
(c) Somatotropin
(d) Luteotropin
Answer:
(c) Somatotropin

Question 71.
………………. is a gonadotropic hormone.
(a) STH
(b) LTH
(c) ACTH
(d) FSH
Answer:
(d) FSH

Question 72.
Rhythmic integrated contractions of jejunum are controlled by ……………….
(a) coherin
(b) insulin
(c) glucagon
(d) ADH
Answer:
(a) coherin

Question 73.
Thyroid gland is derived from of embryo.
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) ecto-endoderm
Answer:
(c) endoderm

Question 74.
Deficiency of thyroxine in infants causes ……………….
(a) Cretinism
(b) Grave’s disease
(c) Myxoedema
(d) Exophthalmos
Answer:
(a) Cretinism .

Question 75.
………………. is a hypercalcemic hormone.
(a) PTH
(b) TCT
(c) Thyroxine
(d) ACTH
Answer:
(a) PTH

Question 76.
………………. is a middle layer of adrenal cortex.
(a) Zona fasciculata
(b) Zona pellicida
(c) Zona glomerulosa
(d) Zona reticularis
Answer:
(a) Zona fasciculata

Question 77.
Decrease in the blood calcium level is ……………….
(a) hyperglycemia
(b) hypercalcemia
(c) hypoglycemia
(d) hypocalcemia
Answer:
(d) hypocalcemia

Question 78.
………………. stimulates RBC production.
(a) Aldosterone
(b) Cortisol
(c) Epinephrine
(d) Parathormone
Answer:
(b) Cortisol

Question 79.
Chemicals which are released at the synaptic junction are called ……………….
(a) hormones
(b) neurotransmitters
(c) cerebrospinal fluid
(d) lymph
Answer:
(b) neurotransmitters

Question 80.
Potential difference across resting membrane is negatively charged. This is due to differential distribution of the following ions.
(a) Na⁺ and K⁺ ions
(b) Ca⁺⁺ and Cl^– ions
(c) Ca⁺⁺ and Mg⁺⁺ ions
(d) Mg⁺⁺ and Cl^– ions
Answer:
(a) Na⁺ and K⁺ ions

Question 81.
Which of the following is not involved in Knee-jerk reflex?
(a) Muscle spindle
(b) Motor neuron
(c) Brain
(d) Inter neurons
Answer:
(c) Brain

Question 82.
An area in the brain which is associated with strong emotions is ……………….
(a) Cerebral cortex
(b) Cerebellum
(c) Limbic system
(d) Medulla
Answer:
(c) Limbic system

Question 83.
Which is the vitamin present in Rhodopsin?
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D
Answer:
(a) Vitamin A

Question 84.
Wax gland present in the ear canal is modified ……………….
(a) Sweat gland
(b) Vestibular gland
(c) Cowper’s gland
(d) Sebaceous gland
Answer:
(d) Sebaceous gland

Question 85.
The part of internal ear responsible for hearing is ……………….
(a) cochlea
(b) semicircular canal
(c) utriculus
(d) sacculus
Answer:
(a) cochlea

Question 86.
The organ of corti is a structure present in ……………….
(a) external ear
(b) middle ear
(c) semi circular canal
(d) cochlea
Answer:
(d) cochlea

Question 87.
Select the right match of endocrine gland and their hormones among the options given below.
A. Pineal i. Epinephrine
B. Thyroid ii. Melatonin
C. Ovary iii. Estrogen
D. Adrenal medulla iv. Tetraiodothyronine
(a) A-iv, B-ii, C-iii, D-i
(b) A-ii, B-iv, C-i, D-iii
(c) A-iv, B-ii, C-i, D-iii
(d) A-ii, B-iv, C-iii, D-i
Answer:
(d) A-ii, B-iv, C-iii, D-i

Question 88.
Listed below are the hormones of anterior pituitary origin. Tick the wrong entry.
(a) Growth hormone
(b) FSH
(c) Oxytocin
(d) ACTH
Answer:
(c) Oxytocin

Question 89.
Mary is about to face an interview. But during the first five minutes before the interview she experiences sweating, increased rate of heartbeat, respiration, etc. Which hormone is responsible for her restlessness?
(a) Estrogen and progesterone
(b) Oxytocin and vasopressin
(c) Adrenaline and noradrenaline
(d) Insulin and glucagon
Answer:
(c) Adrenaline and noradrenaline

Question 90.
The steroid responsible for balance of water and electrolytes in our body is ……………….
(a) Insulin
(b) Melatonin
(c) Testosterone
(d) Aldosterone
Answer:
(d) Aldosterone

Question 91.
Thymosin is responsible for ……………….
(a) raising the blood sugar level
(b) raising the blood calcium level
(c) increased production of T lymphocytes
(d) decrease in blood RBC
Answer:
(c) increased production of T lymphocytes

Question 92.
In the mechanism of action of a protein hormone, one of the second messengers is ……………….
(a) Cyclic AMP
(b) Insulin
(c) T₃
(d) Gastrin
Answer:
(a) Cyclic AMP

Question 93.
Leydig cells produce a group of hormones called ……………….
(a) androgens
(b) estrogen
(c) aldosterone
(d) gonadotropins
Answer:
(a) androgens

Question 94.
Corpus luteum secretes a hormone called ……………….
(a) prolactin
(b) progesterone
(c) aldosterone
(d) testosterone
Answer:
(b) progesterone

Question 95.
Cortisol is secreted from ……………….
(a) pancreas
(b) thyroid
(c) adrenal
(d) thymus
Answer:
(c) adrenal

Question 96.
A hormone responsible for normal sleep – wake cycle is ……………….
(a) epinephrine
(b) gastrin
(c) melatonin
(d) insulin
Answer:
(c) melatonin

Question 97.
Match the pairs and choose the correct answer among the following options.
A. Epinephrine
i. Increase in muscle growth
B. Testosterone
ii. Decrease in blood pressure
C. Glucagon
iii. Decrease in liver glycogen content
D. Atrial natriuretic factor
iv. Increase in heartbeat
(a) A-ii, B-i, C-iii, D-iv
(b) A-iv, B-i, C-iii, D-ii
(c) A-i, B-ii, C-iii, D-iv
(d) A-i, B-iv, C-ii, D-iii
Answer:
(b) A-iv, B-i, C-iii, D-ii

Question 98.
Blood calcium level is a resultant of how much dietary calcium is absorbed, how much calcium is lost in the urine, how much bone dissolves releasing calcium into the blood and how much calcium from blood enters tissues. Several factors play an important role in these processes. Mark the one which has no role.
(a) Vitamin D
(b) Parathyroid hormone
(c) Thyrocalcitonin
(d) Thymosin
Answer:
(d) Thymosin

Question 99.
All the following tissues in mammals except one consists of a central ‘medullary’ region surrounded by a cortical region. Mark the wrong entry.
(a) Ovary
(b) Adrenal
(c) Liver
(d) Kidney
Answer:
(c) Liver

Match the columns

Question 1.
Match the hormones with their source

Column A	Column B
(1) Glucagon	(i) Neurohypophysis
(2) Adrenaline	(ii) Islets of Langerhans
(3) Somato tropins	(iii) Adenohypophysis
(4) ADH	(iv) Medulla

Answer:

Column A	Column B
(1) Glucagon	(ii) Islets of Langerhans
(2) Adrenaline	(iv) Medulla
(3) Somato tropins	(iii) Adenohypophysis
(4) ADH	(i) Neurohypophysis

Question 2.
Match the layer of adrenal with its hormone.

Column A (Layers of adrenal cortex)	Column B (Hormones)
(1) Zona glomerulosa	(A) Cortisols
(2) Zona fasciculata	(B) Androgens
(3) Zona reticularis	(C) Aldosterone

Answer:

Column A (Layers of adrenal cortex)	Column B (Hormones)
(1) Zona glomerulosa	(C) Aldosterone
(2) Zona fasciculata	(A) Cortisols
(3) Zona reticularis	(B) Androgens

Question 3.
Match the disorder with the gland associated with it.

Column A (Disorder)	Column B (Associated Gland)
(1) Addison’s disease	(A) Hypothalamus
(2) Grave’s disease	(B) Pituitary
(3) Diabetes insipidus	(C) Thyroid
(4) Acromegaly	(D) Adrenal

Answer:

Column A (Disorder)	Column B (Associated Gland)
(1) Addison’s disease	(D) Adrenal
(2) Grave’s disease	(C) Thyroid
(3) Diabetes insipidus	(A) Hypothalamus
(4) Acromegaly	(B) Pituitary

Classify the following to form Column B as per the category given in Column A

Question 1.

Column A	Column B
(1) Forebrain	————–
(2) Midbrain	————–
(3) Hindbrain	————–

Answer:

Column A	Column B
(1) Forebrain	Olfactory lobes, Corpus callosum
(2) Midbrain	Superior colliculi, Iter
(3) Hindbrain	Pons varolii, Vermis

Question 2.
Types of nerves
Occulomotor, Facial, Optic, Vagus, Abducens, Vestibulocochlear

Column A	Column B
(1) Sensory	————–
(2) Motor	————–
(3) Mixed	————–

Answer:

Column A	Column B
(1) Sensory	Optic, Vestibulocochlear
(2) Motor	Occulomotor, Abducens
(3) Mixed	Facial, Vagus

Question 3.
Hormones
Estrogen, Glucagon, Epinephrine, Relaxin, Somatostatin, Nor-Adrenalin

Column A	Column B
(1) Ovary	————
(2) Pancreas	————
(3) Adrenal Medulla	————

Answer:

Column A	Column B
(1) Ovary	Estrogen, Relaxin
(2) Pancreas	Glucagon, Somatostatin
(3) Adrenal Medulla	Epinephrine, Nor-Adrenaline

Question 4.
Disorders
Dwarfism, Myxoedema, Addison’s disease, Cushing’s disease, Gigantism, Goitre

Column A	Column B
(1) Pituitary	————
(2) Thyroid	————
(3) Adrenal Cortex	————

Answer:

Column A	Column B
(1) Pituitary	Dwarfism, Gigantism
(2) Thyroid	Myxoedema, Goitre
(3) Adrenal Cortex	Addison’s disease, Cushing’s disease

Very short answer questions

Question 1.
What is the need for the control and coordination in multicellular animals?
Answer:
Multicellular animals need control and coordination to maintain constancy of internal environment, i.e. homeostasis.

Question 2.
How do plants carry out control and coordination?
Answer:
Plants carry out control and coordination by sending chemical signals and bringing about various types of movements.

Question 3.
What is the type of nervous system of Earthworm?
Answer:
Earthworm is an annelid having ventral, ganglionated nervous system. It consists mainly of nerve ring, nerve cord and peripheral segmentally arranged nerves.

Question 4.
What kind of nervous system is seen in Hydra, earthworm and cockroach?
Answer:
In Hydra, the nervous system is in the form of nerve net, while in earthworm and cockroach the nervous system is ganglionated.

Question 5.
What is a gland?
Answer:
An organized collection of secretory epithelial cells capable of producing of some secretion is called a gland.

Question 6.
What do you mean by discrete endocrine glands ?
Answer:
The glands which are exclusively endocrine in function are discrete endocrine glands.

Question 7.
What is genu and splenium?
Answer:
Genu is anterior and splenium is posterior fold of corpus callosum.

Question 8.
Which is the largest basal nucleus of brain ?
Answer:
Corpus striatum is the largest basal nucleus of the brain.

Question 9.
What is EEG?
Answer:
EEG is electro encephalography. It is used to detect the electrical changes taking place in the brain.

Question 10.
Mention the major sulci present in the cerebral hemispheres.
Answer:
The sulci present in the cerebral hemisphere are – central sulcus, parieto-occipital sulcus and lateral sulcus.

Question 11.
What is the difference in ‘tract’ and ‘nerve’?
Answer:
A bundle of axons within CNS is called a ‘tract’ while the one outside the CNS is called ‘nerve’.

Question 12.
What is a choroid plexus? State its locations.
Answer:
Network of blood capillaries associated with the brain is called choroid plexus, which are anterior choroid plexus present on the roof of epithalamus and posterior choroid plexus located on the roof of medulla oblongata.

Question 13.
What is a synapse ?
Answer:
The interconnection between two neurons or neuron with motor organ is called synapse.

Question 14.
What is a polarised membrane?
Answer:
The cell membrane of a neuron at resting stage is called polarised membrane. In such membrane, the outer side of the cell membrane is more electropositive due to more Na⁺ ions.

Question 15.
What is summation effect?
Answer:
Many weak stimuli given in a quick succession may produce a response due to summation or additive effect of stimuli, which is known as summation effect.

Question 16.
What is synaptic delay?
Answer:
The time required for a nerve impulse to cross a synapse during transmission of a nerve impulse is called synaptic delay, which is about 0.3-0.5 milliseconds.

Question 17.
What is refractory period?
Answer:
The time interval between two consecutive nerve impulses is called refractory period, during which a nerve cannot be stimulated. Nerve is stimulated only after completion of this period.

Question 18.
What is a synaptic cleft?
Answer:
Small intercellular space between two successive neurons which is about 20¬30 nm in width, is called synaptic cleft.

Question 19.
What is synaptic transmission?
Answer:
The process by which the impulse from the pre-synaptic neuron is conducted to the post-synaptic neuron or cell is called synaptic transmission.

Question 20.
What is a pre-synaptic neuron?
Answer:
The neuron carrying an impulse to the synapse is the pre-synaptic neuron.

Question 21.
What is transmission terminal and generator region?
Answer:
The pre-synaptic membrane of a neuron is transmission terminal while the post- synaptic membrane is called generation region.

Question 22.
What is synaptic fatigue?
Answer:
The synaptic fatigue is the time lag or halting of the transmission of nerve impulse, temporarily at synapse due to exhaustion of its neurotransmitter.

Question 23.
What is the width of synaptic cleft?
Answer:
The width of synaptic cleft is about 20-30 nm.

Question 24.
How much is the resting potential of axon?
Answer:
The resting potential of axon is -70 mV.

Question 25.
How does Na⁺ – K⁺ pump work?
Answer:
During repolarisation of certain part of a nerve, Na⁺ – K⁺ Pump, pumps out 3 Na⁺ ions for every 2 K⁺ ions they pump into the cell.

Question 26.
What is blood-brain barrier?
Answer:
The barrier that keeps a check on passage of ions and large molecules from the blood to the brain tissue is called blood-brain barrier. Endothelial cells lining the blood capillaries help in this process along with the astrocytes.

Question 27.
Enlist meninges of human brain.
Answer:
Dura mater, arachnoid mater or membrane and pia mater are the meninges of human brain.

Question 28.
What is the source of CSF?
Answer:
CSF is secreted by the choroid plexuses of brain and ependymal cells lining the ventricles of brain and central canal of spinal cord.

Question 29.
What is the function of tympanic membrane?
Answer:
Tympanic membrane vibrates on receiving the sound waves and then transfers the vibrations to malleus, the first of the three ear ossicles.

Question 30.
Mention the role of semicircular canals in ear.
Answer:
Semicircular canals help in balancing the equilibrium of the body.

Question 31.
What is the cause for diabetes insipidus?
Answer:
Diabetes insipidus is caused by the deficiency of ADH.

Question 32.
Give role of Parathormone.
Answer:
Parathormone regulates the calcium balance in the body. It increases blood calcium level by increasing reabsorption of calcium from bones.

Question 33.
What is meningitis?
Answer:
Infection of meninges is called meningitis.

Question 34.
Where is general motor area located? What is the function of this area?
Answer:
General motor area is located on precentral gyrus of frontal lobe. It is concerned with all motor activities.

Question 35.
What is the role of Wernicke’s area?
Answer:
Wernicke’s area is the sensory speech area responsible for understanding and formulating written and spoken language.

Question 36.
What is the location of Wernicke’s area?
Answer:
Wernicke’s area is located in the area of contact between temporal, parietal and occipital lobes of cerebrum.

Question 37.
What is the role of Broca’s area?
Answer:
The role of Broca’s area which is the motor speech area that translates thoughts into speech and controls movement of tongue, lips and vocal cords.

Question 38.
What is the location of Broca’s area?
Answer:
Broca’s area is present in the frontal lobe of cerebrum.

Question 39.
Which areas are present in the post central gyrus?
Answer:
General sensory areas concerned with sensation of temperature, touch, pressure, pain and speech are present in the post central gyrus.

Question 40.
Which areas are located on temporal lobe?
Answer:
The areas concerned with sense of taste, sense of hearing and sense of smell are located on temporal lobe.

Question 41.
What is the main function of occipital lobe?
Answer:
Occipital lobe carries sensory visual area which is concerned with the sense of sight. Association visual area is concerned with perception, analysis and storing of information obtained by sight.

Question 42.
What is neuroendocrine system?
Answer:
Neuroendocrine system is an integrated and coordinated system consisting of nervous and endocrine system which brings about the coordination of the body.

Question 43.
What is the quantity of cerebrospinal fluid in adult human being?
Answer:
Cerebrospinal fluid is about 120 ml in adult human being.

Question 44.
How many neurons are present in the brain?
Answer:
There are about 30,000 million neurons in the brain.

Question 45.
Which part of the brain forms 80-85% of the brain?
Answer:
Cerebrum forms 80-85% of the brain.

Question 46.
Which sense is poorly developed in human beings and what is the reason for this?
Answer:
Sense of smell is poorly developed in human beings and it is due to less developed olfactory lobes.

Question 47.
Where are lateral ventricles situated and what is its roof called?
Answer:
Lateral ventricles are the cavities filled with cerebrospinal fluid, present inside the cerebrum and its roof is called pallium.

Question 48.
What is foramen of Monro?
Answer:
Foramen of Monro is a narrow opening that connects lateral ventricle present in cerebrum with third ventricle present in diencephalon.

Question 49.
What is the roof and floor of diencephalon called respectively?
Answer:
Roof of diencephalon is epithalamus and floor of diencephalon is hypothalamus.

Question 50.
Which hormones are produced from pineal gland? Name their functions.
Answer:
Serotonin and melatonin are hormones of pineal gland which are concerned with metabolic activities and regulation of biological rhythm respectively.

Question 51.
What is Habenular commissure?
Answer:
Habenular commissure connects the lateral walls of diencephalon or thalami with each other.

Question 52.
What structure is present on the anterior side of the hypothalamus?
Answer:
Optic chiasma or crossing of two optic nerves is seen on the anterior side of the hypothalamus.

Question 53.
Name the parts of corpora quadrigemina and give their functions.
Answer:
The upper two lobes of corpora quadrigemina are called superior colliculi which receive impulse from optic nerves while the lower two lobes are called inferior colliculi which receive auditory stimuli. Both colliculi control and coordinate head movements.

Question 54.
What is arbour vitae?
Answer:
Arbor vitae is white, branching tree-like processes of white matter that are sent into grey cortex of cerebellum.

Question 55.
Which cranial nerve is called a dentist nerve?
Answer:
Trigeminal, the Vth cranial nerve is called a dentist nerve.

Question 56.
What is central canal?
Answer:
Central canal is the narrow central cavity present inside the spinal cord.

Question 57.
Which cranial nerves originate from the midbrain?
Answer:
Cranial nerves (III) Occulomotor, (IV) Pathetic and (VI) Abducens originate from the midbrain.

Question 58.
What is the other name for the auditory cranial nerve?
Answer:
Vestibulocochlear.

Question 59.
What is the difference between unconditioned and conditioned reflexes?
Answer:
Unconditioned reflexes are inborn or hereditary and permanent while the conditioned reflexes are temporary, learnt or acquired during lifetime.

Question 60.
Give examples of unconditional reflexes.
Answer:
Blinking of eyes, withdrawing of hand upon pricking, suckling to breast by infant, swallowing, knee jerk, sneezing and coughing are some of the unconditional reflexes.

Question 61.
What do you mean by accommodation power of the lens?
Answer:
The ability of the lens by which the light ray from far and near objects is focused on the retina is called accommodation power of the lens. The lens makes fine adjustments to bring such sharp focus on retina.

Question 62.
Where is pituitary located?
Answer:
Pituitary is located inside a depression called sella turcica which is present in sphenoid bone of the skull.

Question 63.
What is the difference between Lorain dwarf and Frohlic dwarf?
Answer:
Lorain dwarf is mentally normal while Frohlic dwarf is mentally abnormal.

Question 64.
How is Simmond’s disease caused?
Answer:
Simmond’s disease is caused due to hyposecretion of GH during adulthood.

Question 65.
How is Addison’s disease caused?
Answer:
Addison’s disease is caused by the hyposecretion of ACTH that leads to adrenal failure.

Question 66.
How is Cushing’s disease caused?
Answer:
Hypersecretion of corticoids causes Cushing’s disease.

Question 67.
What is the main difference between diabetes mellitus and diabetes insipidus?
Answer:
Diabetes mellitus is caused due to deficiency of insulin while diabetes insipidus is caused due to deficiency of ADH.

Question 68.
Which cells secrete coherin? What is the function of coherin?
Answer:
Coherin is the hormone secreted by hypothalamic neurons that brings about prolonged, rhythmic integrated contractions of the jejunum.

Question 69.
What is the difference between Gull’s disease (Myxoedema) and Grave’s disease?
Answer:
Gull’s disease is caused due to hyposecretion of thyroxine while Grave’s disease is caused due to hypersecretion of thyroxine.

Question 70.
In which part thyroid gland stores its hormones?
Answer:
The lumen of thyroid follicles store the thyroid hormones in the form of thyroglobulins.

Question 71.
What are thymosins?
Answer:
Thymosins are hormones secreted by thymus gland which promote the production of antibodies.

Question 72.
Which hormone is secreted by the heart?
Answer:
Heart secretes ANF or Atrial Natriuretic Factor.

Name the following

Question 1.
Name the region consisting of nerve fibres that connects cerebrum and medulla oblongata.
Answer:
Pons Varolii.

Question 2.
Give the names of cranial nerve number VIth and VIIth.
Answer:
VTth cranial nerve is Abducens and Vllth cranial nerve is Facial.

Question 3.
Name the three sulci present on the cerebral hemispheres.
Answer:
Central sulcus, lateral sulcus and parieto¬occipital sulcus.

Question 4.
Name the band of nerve fibres that connect cerebrum, cerebellum and spinal cord.
Answer:
Crura cerebri.

Question 5.
Name the second largest part of the brain.
Answer:
Cerebellum.

Question 6.
Name the three branches of trigeminal nerve.
Answer:
Ophthalmic, Maxillary and Mandibular.

Question 7.
Name the nerve which arises from ventral side of medulla and supplies the tongue.
Answer:
Hypoglossal.

Question 8.
Name of emergency hormones secreted by sympathetic nervous system.
Answer:
Adrenaline and nor-adrenaline are the emergency hormones secreted by sympathetic nervous system.

Question 9.
Name the disorders caused by hyposecretion of thyroxine in children and adults.
Answer:

1. Hyposecretion of thyroxine in children : Cretinism.
2. Hyposecretion of thyroxine in adults : Myoxedema.

Question 10.
Name the dual exocrine as well as endocrine gland. Name the hormones secreted by it.
Answer:
Pancreas is the dual gland, exocrine as well as endocrine, it secretes hormones like insulin, glucagon and somatostatin.

Question 11.
Name the four peptide hormones secreted by endocrine cells of alimentary canal.
Answer:
Gastrin, secretin, cholecystokinin and GIPT or Gastric inhibitory peptide.

Question 12.
Name the disorder caused by the under secretion of thyroxine in children.
Answer:
Cretinism is the disorder seen in children due to under secretion of thyroxine.

Give functions of the following

Question 1.
Meninges and CSF.
Answer:
Functions of meninges:

1. Meninges give protection to the brain and the spinal cord.
2. They are also nutritive in function.
3. Cerebrospinal fluid acts as a shock absorber. It protects the brain from mechanical injuries and from desiccation.
4. It also maintains constant pressure inside and outside the CNS and regulates the temperature.

Functions of CSF:

1. CSF helps in exchange of nutrients and wastes between the blood and the brain tissue.
2. CSF supplies oxygen to the brain.

Question 2.
The functions of forebrain.
Answer:
(A) Functions of cerebrum:

1. The cerebrum controls the voluntary activities.
2. The cerebrum perceives various sensory stimuli received through vision, taste, smell, sound, touch, speech, etc.
3. It is the centre of memory, will-power, intelligence, reasoning and learning.
4. The cerebrum is the centre for emotions, thoughts and feelings, pain, pleasure, fear, fatigue, pressure, temperature, etc.
5. It is also the centre for micturition, defecation, weeping, laughing, etc

(B) Function of olfactory lobes: Sensation of smell.

(C) Functions of diencephalon:

1. Diencephalon acts as a relay centre for motor and sensory impulses between spinal cord, brainstem and various areas of cerebral cortex.
2. Diencephalon consists of epithalamus, thalami and hypothalamus. Therefore it acts as a centre for homeostasis and higher centre of autonomous nervous system.
3. Hypothalamic nuclei secrete neurohormones which influence the pituitary gland.
4. Diencephalon regulates heartbeats, blood pressure and water balance.
5. Anterior choroid plexus which is located in the diencephalon secretes cerebrospinal fluid.
6. Hypothalmic regions control many involuntary functions such as hunger, thirst, thermo-regulation, fear, anger, sleep, sexual desire, etc.

Question 3.
Write various functions of hindbrain.
Answer:
Functions of hindbrain:
(i) Cerebellum:

1. Cerebellum is a primary centre for the control of equilibrium, posture, balancing and orientation.
2. Neuromuscular activities are regulated by the cerebellum.
3. Coordination of walking, running, speaking, etc. is under the control of hindbrain.

(ii) Pons:

1. Activities of two cerebellar hemispheres are coordinated by pons.
2. Nerve fibres cross over in this area and thus the right side of the brain controls the left part of the body and vice versa.
3. Pons controls the consciousness of the brain.
4. Breathing centre is located in pons along with medulla.

(iii) Medulla oblongata:

1. Medulla oblongata controls all the involuntary activities such as heartbeats, respiration, vasomotor activities.
2. Peristalsis and reflex actions such as coughing, sneezing, swallowing, etc. are also under the control of medulla oblongata.
3. Medulla oblongata is essential for all the vital functions of the body.

Question 4.
Spinal cord.
Answer:
Functions of spinal cord:

1. Spinal cord is the main pathway for conduction of sensory and motor nerve impulses.
2. The sensory impulses travel from the body
   to the brain and the motor impulses travel from the brain to the body.
3. Spinal reflexes are controlled by spinal cord.
4. The spinal cord reduces the load on the brain by taking appropriate actions in a reflex way.

Question 5.
Pituitary gland.
Answer:
Functions of pituitary:

1. Pituitary secretes seven main hormones viz. ACTH, TSH. GH or STH, LTH or Prolactin, MSH and gonadotropins such as FSH, LH or ICSH.
2. These hormones are secreted upon receiving proper releasing factor from hypothalamus.
3. These hormones bring about many coordinating functions in the body. Almost all endocrine glands are under the control of these hormones.
4. The neurohypophysis part stores two hormones, viz. vasopressin and oxytocin which are secreted by hypothalamus.
5. Important functions such as growth and reproductive processes, pregnancy, childbirth, lactation, etc. are under the control of pituitary.
6. Neurohypophysis acts as a neurohaemal organ and stores the hormones for some time.

Question 6.
Functions of hormones secreted by the thyroid gland.
Answer:

1. Thyroxine is the main metabolic hormone in the body.
2. Thyroxine maintains basal metabolic rate (BMR) by increasing glucose oxidation. It brings about calorigenic effect by energy production.
3. It also controls normal protein synthesis.
4. The physical growth, development of gonads and development of mental faculties is under the control of thyroxine.
5. It controls tissue differentiation during metamorphosis particularly in amphibia.
6. Body weight, respiration rate, heart rate, blood pressure, temperature, digestion, etc. is regulated by thyroxine.
7. Another hormone of thyroid, i.e. calcitonin regulates calcium metabolism of the body.

Question 7.
Give significance of relaxin and inhibin.
Answer:
1. Relaxin : Relaxin relaxes the cervix of the pregnant female and ligaments of pelvic girdle during parturition.
2. Inhibin : Inhibin inhibits the FSH and GnRH production.

Question 8.
Enlist hormones secreted by GI tract and state their role.
Answer:
Gastrointestinal tract:
In the gastrointestinal mucosa, certain ells are endocrine in function. These cells produce hormones which play vital role in digestive processes and flow of digestive juices.

1. Gastrin : This hormone stimulates gastric glands to produce gastric juice.
2. Secretin : This hormone is responsible for secretion of pancreatic juice and bile from pancreas and liver.
3. Cholecystokinin CCK/Pancreozymin PZ : This hormone stimulates gall bladder to release bile and stimulates the pancreas to release its enzymes.
4. Entero-gastrone/Gastric inhibitory peptide (GIP) : It slows gastric contractions and inhibits the secretion of gastric juice.

Distinguish between the following

Question 1.
Electrical and chemical synapses:
Answer:

Electrical synapse	Chemical synapse
1. The gap between the successive neurons in electrical synapse is very less [3.8 nm],	1. The gap between two successive neurons in chemical synapse is larger than electrical synapse [10-20 nm].
2. Transmission across the gap is faster in electrical synapse.	2. Transmission across the gap in chemical synapse is relatively slower than electrical synapse.
3. Electrical synapse is less common.	3. Chemical synapse is more common.
4. Electrical synapse is found in those places of the body requiring instant response.	4. Chemical synapse is found almost everywhere and connects neuron to neuron, muscles or glands.

Question 2.
Sympathetic and parasympathetic nervous system.
Answer:

Sympathetic nervous system	Parasympathetic nervous system
1. Sympathetic nervous system is formed by 22 pairs of sympathetic ganglia, 2 sympathetic cords which run parallel to vertebral column.	1. Parasympathetic nervous system has nerve fibres which run along with cranial and spinal nerves.
2. Sympathetic nervous system works through neurotransmitter, adrenaline.	2. Parasympathetic nervous system works through release of acetylcholine.
3. Sympathetic nervous system enhances all the involuntary functions.	3. Parasympathetic nervous system retards all the involuntary functions.
4. It brings about fight, fright and flight responses.	4. It brings about relaxation, comfort, pleasure, etc.
5. The pre-ganglionic nerve fibres are short and the post-ganglionic nerve fibres are long in sympathetic nervous system.	5. The pre-ganglionic nerve fibres are long and the post-ganglionic nerve fibres are short in parasympathetic nervous system.

Question 3.
Unconditional reflexes and Conditional reflexes
Answer:

Unconditional reflexes	Conditional reflexes
1. Unconditional reflexes are inborn.	1. Conditional reflexes are not inborn, they require training.
2. Unconditional reflexes are permanent.	2. Conditional reflexes are temporary.
3. They never disappear and need no previous experience.	3. They may disappear after sometime and need proper training for developing it.
4. Unconditional reflexes are heritable.	4. Conditional reflexes are non-heritable.
5. Sneezing, coughing, blinking of eye, etc. are unconditional reflexes.	5. Cycling, driving, playing games, etc. are due to conditional reflexes.

Question 4.
Epithalamus and hypothalamus.
Answer:

Epithalamus	hypothalamus
1. Epithalamus is the roof of diencephalon.	1. Hypothalamus is the floor of diencephalon.
2. Epithalamus shows pineal stalk to which pineal gland is attached.	2. Hypothalamus shows infundibulum to which pituitary gland is attached.
3. Epithalamus is non-nervous in nature.	3. Hypothalamus is the higher centre of autonomous nervous system.
4. Epithalamus has anterior choroid plexus which secretes cerebrospinal fluid.	4. Hypothalamus has neurons which secrete two endocrine hormones.
5. Epithalamus controls biological rhythm.	5. Hypothalamus controls homeostasis of the body.

Question 5.
Dura mater and pia mater.
Answer:

Dura mater	pia mater
1. Dura mater is the outermost meninx.	1. Pia mater is the innermost meninx.
2. Dura mater lies on the innermost side of skull or cranium.	2. Pia mater lies on outermost side of the brain.
3. Dura mater is tough, thick and fibrous.	3. Pia mater is thin and highly vascular.
4. Dura mater is mainly protective in function.	4. Pia mater is mainly nourishing in nature.
5. Below dura mater is subdural space.	5. Above pia mater is sub-arachnoidal space.

Question 6.
Cerebrum and cerebellum
Answer:

Cerebrum	Cerebellum
1. The cerebrum is the larger part forming 85% of the brain. It has four lobes.	1. The cerebellum is the smaller part forming 11%. of the brain. It has three lobes.
2. The cerebrum coordinates the functions of the sensory and motor areas.	2. The cerebellum coordinates the equilibrium of muscular movements during walking and running.
3. The cerebrum plays an important role in receiving the sensory impulses such as touch, pain, heat, cold, etc.	3. The cerebellum plays an important role in maintaining the posture and balance of the body.
4. The cerebrum is concerned with higher mental faculties such as memory, will and intelligence.	4. The cerebellum is concerned with muscular mechanism.

Question 7.
Cranial nerves and Spinal nerves
Answer:

Cranial nerves	Spinal nerves
1. Nerves arising from the brain are cranial nerves.	1. Nerves arising from the spinal cord are spinal nerves.
2. There are 12 pairs of cranial nerves.	2. There are 31 pairs of spinal nerves.
3. Cranial nerves are of three types, viz sensory, mixed and motor.	3. All spinal nerves are of mixed type.
4. Cranial nerves are responsible for cerebral reflexes.	4. Spinal nerves are responsible for spinal reflexes.

Question 8.
Extero and entero receptors.
Answer:

Exteroceptors	Interoceptors
1. Receptors receiving stimuli from outer environment of the body are called exteroceptors.	1. Receptors receiving stimuli from inside the body are called interoceptors.
2. These are somatic receptors.	2. These are visceral receptors.
3. Exteroceptors keep the body informed about . changes in the environment like temperature, pressure, touch, etc.	3. Interoceptors keep the homeostasis in the body by receiving stimuli from inside the body.
4. Mechanoreceptors, thermoreceptors, chemical receptors, photoreceptors and statoacoustic receptors are the different types of exteroceptors.	4. Propioceptors, enteroceptors, baroceptors are the different types of interoceptors.

Question 9.
Adenohypophysis and neurohypophysis.
Answer:

Adenohypophysis	Neurohypophysis
1. Adenohypophysis is the anterior lobe of pituitary.	1. Neurohypophysis is the posterior lobe of pituitary.
2. There is portal system between adenohypo-physis and hypothalamus which has blood sinusoids.	2. There is axonal knobs and blood vessels that connect neurohypophysis and hypothalamus.
3. Adenohypophysis forms 75% of pituitary.	3. Neurohypophysis forms 25% of pituitary.
4. Adenohypophysis has three parts, pars tuberalis, pars distalis and pars intermedia.	4. Neurohypophysis has three parts, median eminence, infundibulum and pars nervosa
5. Adenohypophysis has chromophil (acidophil and basophil) and chromophobe cells.	5. Neurohypophysis has axonic fibres and pituicytes.
6. Adenohypophysis secretes seven different hormones after receiving an appropriate message from hypothalamus through releasing factors.	6. Neurohypophysis does not produce hormones on its own. It is a neurohaemal organ as it receives and stores two hormones from hypothalamus.

Question 10.
FSH and LH
Answer:

FSH	LH
1. FSH is follicle stimulating hormone essential for the development of ovary.	1. LH is luteinizing hormone responsible for ovulation in females.
2. FSH stimulates ovary (follicular cells) to produce estrogen.	2. LH stimulates ovary (corpus luteum) to produce progesterone.
3. FSH in males is responsible for the spermatogenesis.	3. LH in females is responsible for the development of corpus luteum.
4. Negative feedback mechanism exists between amounts of FSH and estrogen in females.	4. Negative feedback mechanism exists between amounts of LH and progesterone in females.
5. FSH is indirectly responsible for the development of secondary sexual characters in females.	5. LH is indirectly responsible for maintenance of pregnancy in females.

Question 11.
Glucocorticoids and mineralcorticoids
Answer:

Glucocorticoids	Mineralocorticoids
1. Glucocorticoids control carbohydrate metabolism.	1. Mineralocorticoids regulate mineral concentration.
2. These are secreted by the cells of zona fasciculata.	2. These are secreted by the cells of zona glomerulosa.
3. These also regulate protein and fat metabolism.	3. These regulate salt-water balance.
4. Cortisol is the main glucocorticoid.	4. Aldosterone is the main mineralocorticoid.

Give scientific reasons

Question 1.
Number of gyri is related to the degree of intelligence.
Answer:

1. Gyri are the ridges in the folds present on the cerebral cortex.
2. The number of folds increase the surface area of the cerebral cortex.
3. Cerebral cortex has sensory and motor areas such as Wernicke’s area, Broca’s area, etc. If the surface area is greater, the neurons in these areas are also more in number.
4. Greater the number of neurons greater is the intelligence.
5. The number of gyri therefore is said to be related to the degree of intelligence.

Question 2.
A drunken person cannot maintain balance of the body.
Answer:

1. Cerebellum is the primary centre for controlling equilibrium and balance of the body.
2. Alcohol has an adverse effect on the neurons of cerebellum.
3. Consciousness of brain is also controlled by cerebellum.
4. When a person is drunk, the alcohol in his or her blood affects the activities of cerebellum and hence the person cannot maintain the balance of the body.

Question 3.
We are able to understand the smell of the first showers of rain or the sudden changes in the climate.
Answer:

1. We are able to understand any smell because of our olfactory mucosa and olfactory lobes of the brain.
2. Volatile substances are received by the olfacto-receptors in the nose.
3. Nerve impulse generated are carried by olfactory nerve and transmitted to brain where the impulse is interpreted.
4. The characteristic earthy smell is due to a compound ‘geosmin’.
5. Geosmin is produced by some species of Streptomyces [ gram positive soil bacterium],
6. Similarly, sudden change in the climate is easily noticed in the form of temperature change in the surrounding.
7. This change is detected by caloreceptors of the skin. From these receptors the signal is transmitted to CNS where the change is perceived.

Question 4.
We are able to hear the chirping of the birds and recognize the sound of the bird.
Answer:

1. The phonoreceptors of the body receive the sound waves and transfer the nerve impulses to the auditory areas of the brain.
2. The interpretation, of the sound is a combined effort of sensory and association areas [auditory areas] of temporal lobes of the brain.
3. This is how we are able to hear the chirping of the birds and recognize the sound of the bird.

Question 5.
We can see and enjoy the beautiful colours of the nature after the sunrise.
Answer:

1. The light receptor are present in the retina of eyes.
2. These are photosensitive rod and cone cells. Cone cells are sensitive to bright light-and colours.
3. On receiving the light rays, these cells convert them into nerve impulses.
4. The cones are of three types, which contain their own characteristic photo-pigments that respond to red, green and blue lights.
5. Various combinations of these cones and their photo pigments produce sensation of different colours.
6. These impulses are carried to the visual cortex of the occipital lobe where the image is interpreted.
7. In this manner, we can see and enjoy the beautiful colours of the nature after the sunrise.

Question 6.
Cerebellum is well developed in humans.
Answer:

1. Our posture is upright and mode of locomotion is bipedal.
2. While standing, walking and running, our body has to be in a state of balance.
3. Cerebellum controls balancing, posture, body equilibrium and orientation.
4. Thus to control static as well as dynamic equilibrium of the body, cerebellum is well developed.

Question 7.
Mr. Sharma suffered from a stroke and the right side of his body was paralysed. However his response was normal for knee jerk reflex with either leg. Explain how and why?
Answer:

1. Stroke is damage to the brain due to interruption of its blood supply.
2. Due to this brain functions and cerebral reflexes are severely affected.
3. However, spinal reflexes remain largely unaffected.
4. In the above case. Mr. Sharma suffered from a stroke and developed paralysis. But his response to the knee jerk was normal as it is controlled by spinal cord.

Question 8.
The word hormone is a misnomer.
Answer:

1. Hormone is adapted from the Greek word ‘hormein’ means to excite.
2. But the hormones secreted by the endocrine glands and cells carry out two fold functions, viz. excitatory and inhibitory.
3. The word hormone is therefore a misnomer.

Question 9.
Pituitary may be considered as the coordinator of endocrine orchestra but not master endocrine gland.
Answer:

1. The pituitary gland was formerly considered as a master endocrine gland, because all other endocrine glands are under the control of pituitary hormones.
2. But now it is known that the pituitary gland itself is under the control of hypothalamus through hypo thalamo – hypophysial axis.
3. Through various releasing factors and release inhibiting factors, the secretions of pituitary are regulated by hypothalamus.
4. Pituitary in turn controls growth, secretion and maintenance of glands such as adrenal cortex, thyroid and gonads.

Question 10.
Pituitary gland plays an important role in pregnancy and childbirth.
Answer:
Pituitary gland plays an important role in pregnancy and childbirth for the following reasons:

1. Luteinizing hormone secreted by the pituitary gland stimulates ovulation, formation of corpus luteum and the synthesis of progesterone which are necessary for pregnancy.
2. Lactogenic hormone secreted by the pituitary gland promotes breast development during pregnancy and stimulates the secretion of milk after childbirth.
3. Oxytocin secreted by hypothalamus and sent by pituitary gland stimulates the contraction of the uterus during childbirth.

Question 11.
Fall and rise in blood calcium stimulates secretion of parathyroid.
Answer:

1. Normal calcium level in the blood is called calcemia.
2. Increase in the blood calcium is called hypercalcemia while decrease in the level is called hypocalcemia.
3. Decrease in the blood calcium, stimulates the parathyroid glands to secrete PTH.
4. PTH stimulates osteoclasts of bone to start bone resorption. This will help to cause increase in the blood calcium level.
5. Increase in the blood calcium above normal will retard the secretion of PTH. Hence further increase in the blood calcium is stopped.

Question 12.
Pancreas is both exocrine as well as endocrine gland.
Answer:

1. Pancreas is heterocrine gland i.e. both exocrine and endocrine gland.
2. The exocrine part is pancreatic acini. The cells of acini secrete pancreatic juice containing digestive enzymes like trypsinogen, chymotrypsinogen, etc.
3. Endocrine part of pancreas is made up of groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

Alpha (α) cells (20%) secrete glucagon.
Beta (β)) cells (70%) secrete insulin.
Delta (δ) cell (5%) secrete somatostatin. PP cells or F cells (5%) secrete pancreatic polypeptide (PP).
Hence, pancreas is said to be exocrine as well as endocrine gland.

Question 13.
Patient suffering from hypothyroidism shows increased level of TSH. Why?
Answer:

1. Hypothyroidism means decrease in the secretion of T₃ and T₄
2. Decrease in T₄, i.e. thyroxine in the blood triggers negative feedback mechanism.
3. The hormone receptors in hypothalamus detect this change and secrete TRE
4. TRF stimulates the pituitary to secrete TSH
5. Thus, Patient suffering from hypothyroidism shows increased level of TSH due to negative feedback control.

Question 14.
We are advised to use iodized salt.
OR
Why do we use iodized salt?
Answer:

1. Iodine is needed for synthesis of thyroid hormone.
2. If there is deficiency of iodine in the diet, it causes enlargement of thyroid gland leading to simple goitre.
3. This disease is common in hilly areas hence it is also called endemic goitre.
4. Addition of iodine to table salt prevents this disease.
5. Therefore we must use iodised salt.

Write short notes on the following

Question 1.
Nervous system of Hydra.
Answer:

1. Hydra, a cnidarian shows the diffused nervous system in the form of nerve net.
2. It is the most primitive nervous system.
3. There are two nerve nets in the mesoglea-one connected towards the epidermis and second towards the gastro-dermis.
4. Hydra lacks sensory organs, but the sensory cells scattered in the body wall.
5. The nerve impulse shows no polarity or direction. As all neurons are interconnected the response is seen throughout the body.

Question 2.
Nervous system of Planaria.
Answer:

1. Planaria is a flatworm and belongs to the phylum Platyhelminthes.
2. It is the most primitive animal with a Central Nervous System (CNS) located on the ventral side of body.
3. Nervous system consists of a mass of cerebral or cephalic ganglion appearing like an inverted U-shaped brain.
4. Ventrally from below the ganglia arise a pair of Ventral Nerve Cords (VNC) or long nerve cords. These are interconnected to each other by transfer nerve or commissure in a ladder like manner.
5. The PNS include sensory cells arranged in lateral cords in the body.

Question 3.
Resting potential of a nerve fibre.
Answer:

1. The neurons have a property of excitability.
2. The transmission of the nerve impulse along the long nerve fibre is a result of electrical changes across the neuronal membrane during conduction of an excitation.
3. The plasma membrane separates the outer [extra cellular fluid] and inner solutions [intra cellular fluid/cytoplasm] of different chemical compounds.
4. The external tissue fluid has both Na⁺ and K⁺
5. On the inside there is an excess of K⁺ along with large amount of negatively charged protein molecules and nucleic acid.
6. This condition of a resting nerve is also called a polarised state and it is established by maintaining an excess of Na+ on the outer side.

Question 4.
Electrical synapse.
Answer:

1. In this type of synapse gap between the neighbouring neurons is very narrow.
2. This electrical link is formed between the pre-and post-synaptic neurons.
3. At the gap junction, the two cells are within almost 3.8 nm distance of each other.
4. Transmission across the gap is faster.
5. Electrical synapses are found in those places of the body requiring fastest response as in the defence reflexes.

Question 5.
Velocity of nerve impulse.
Answer:

1. The rate of transmission of impulse is higher in long and thick nerves.
2. It is higher in warm blooded animals [homeotherms] than in cold blooded animal [poikilotherms].
3. The velocity of transmission is higher in voluntary fibres (100- 120 m/second in man) as opposed to autonomic or involuntary nerves (10-20 m/second).
4. It is faster in medullated nerve fibre [up to 150 m/s], as the impulse has to jump from one node of Ranvier to the next as compared with non-medullated nerve fibre [10-25 m/s]

Question 6.
Meninges of CNS.
Answer:

1. Meninges are the connective tissue membranes that cover the brain and the spinal cord. There are three meninges, viz. dura mater, arachnoid mater and pia mater that cover the Central nervous system.
2. The outermost tough, thick and fibrous meninx is dura mater. It is protective in function as it is attached to the inner side of the cranium.
3. The middle, thin and vascular membrane formed of reticular connective tissue is called arachnoid mater. It carries out nutritive function and also gives protection to the brain.
4. The innermost highly vascular and thin membrane is pia mater. It lies in contact with CNS. It is nutritive in function.
5. There is subdural space between the dura mater and arachnoid mater. It is filled with serous fluid.
6. Besides, there is a sub-arachnoidal space lying between arachnoid mater and pia mater. It is filled with cerebrospinal fluid.

Question 7.
Ventricles in human brain.
Answer:

1. Ventricles are the cavities present in different parts of the brain.
2. There are four ventricles in the human brain. All the ventricles are connected with each other.
3. They are filled with cerebrospinal fluid.
4. Paracoel or lateral ventricles-I and-II are present inside the cerebral hemispheres.
5. The diencephalon has ventricle-III.
6. Ventricle-III is in connection with lateral ventricles by foramen of Monro.

Question 8.
Sympathetic nervous system.
Answer:

1. Sympathetic Nervous System is formed by 22 pairs of sympathetic ganglia. These ganglia are linearly arranged on two sympathetic cords. Sympathetic nerve cords run on either side of the vertebral column.
2. Sympathetic nerve cords are connected to CNS by rami communicans of spinal nerve fibres.
3. This system works during stress, pain, anger, fear or emergency. It is supposed to bring about fight, flight or fright reactions.
4. Action of sympathetic nervous system is dependent on adrenaline or noradrenaline. This neurotransmitter is secreted by sympathetic nervous system as an emergency hormone.

Question 9.
Parasympathetic nervous system.
Answer:

1. Parasympathetic nervous system consists of nerve fibres of some cranial nerves, sacral nerves and parasympathetic ganglia.
2. These parasympathetic ganglia are present on the sides of visceral organs like heart, lungs, stomach, kidney, etc.
3. Parasympathetic ganglia gives out parasympathetic fibres which innervates these involuntary organs.
4. Parasympathetic nervous system works through release of acetylcholine which acts as neurotransmitter. It is an inhibiting neurotransmitter which affects visceral organs.
5. This system works during rest and brings about relaxation, comfort, pleasure, etc.

Question 10.
Parkinson’s disease.
Answer:

1. Degeneration of dopamine-producing neurons in the CNS causes Parkinson’s disease.
2. Symptoms develop gradually over the years.
3. Symptoms are tremors, stiffness, difficulty in walking, balance and coordination.
4. Seen in old age and is incurable.

Question 11.
Alzheimer’s disease.
Answer:

1. Alzheimer’s disease is the most common form of dementia.
2. Its incidence increases with the age.
3. Symptoms include the loss of cognitive functioning, thinking, remembering, reasoning and behavioural abilities. It interferes with the person’s daily life and activities.
4. It occurs due to loss of cholinergic and other neurons in the CNS and accumulation of amyloid proteins.
5. There is no cure for Alzheimer’s, but treatment slows down the progression of the disease and may improve the quality of life.

Question 12.
Types of reflex actions.
Answer:

1. The reflex actions are of two types, viz. cerebral and spinal.
2. Cerebral reflex actions are controlled by the brain.
3. Spinal reflex actions are controlled by the spinal cord.
4. In man, most of the reflex actions are controlled by the spinal cord.

Question 13.
Pavlov’s experiment about conditional reflex.
Answer:

1. Conditional reflex was demonstrated by Pavlov while performing experiments with dogs.
2. Pavlov offered some food to dog and noticed that the dog starts salivating after smelling and seeing the food.
3. Simultaneously he rang the bell so that the dog associated the food with the sound of bell. This experiment was repeated many a times by him.
4. Later he only rang the bell and did not give any food to the dog. But still the dog salivated.
5. This shows that the dog was conditioned to the sound of bell. The dog learnt that there is a relation between food and sound of bell. This is called conditional response.

Question 14.
Mechanism of vision.
Answer:

1. The light rays of visible wavelength pass through the cornea and the lens and are focused on the retina of the eye.
2. The sight is possible due to conjugated proteins present in the rods and the cones.
3. These are photo pigments which are composed of opsin (a protein) and retinal (Vitamin A derivative).
4. The light induces dissociation of retinal from the opsin, which causes a change in the structure of the opsin.
5. This causes the change in the permeability of the retinal cells.
6. It generates action potential which is carried via bipolar cells and ganglion cells and further conducted by the optic nerves to the visual cortex (vision centre) of the brain.
7. The neural impulses are analyzed and the image formed on the retina is thus recognized.

Question 15.
Mechanism of hearing.
Answer:

1. The external ear receives the sound waves and sends to the tympanic membrane. The tympanum vibrates transmitting the vibrations to the chain of three ossicles and then to the oval window.
2. The vibrations are further passed on to the fluid of cochlea.
3. The waves in the perilymph and endolymph induces movements in the basilar membrane.
4. The hair cells of organ of Corti bend and are pressed against the tectorial membrane.
5. Due to this pressure, the nerve impulses are generated and are sent to the afferent nerve fibres.
6. The impulses are carried by the auditory nerves to the auditory centre of the brain, where the impulses are analyzed and the sound is perceived.

Question 16.
Types of endocrine systems.
Answer:
There are 3 types of endocrine systems.

1. Discrete endocrine system : The glands exclusively endocrine in function are called discrete endocrine glands.
2. Mixed endocrine system : The glands endocrine as well as exocrine in function are called mixed endocrine glands.
3. Diffused endocrine system : Some endocrine cells scattered in a particular region/gland form diffused endocrine system.

Question 17.
Thymus.
OR
Functions of Thymosin.
Answer:

1. Thymus is located on the dorsal side of the heart and the aorta.
2. It consists of many lobules.
3. The thymus plays major role in the development of the immune system.
4. Thymus secretes thymosin which plays an important role in the differentiation of T-lymphocytes. These cells built cell mediated immunity.
5. The thymosin also promotes the production of antibodies providing humoral immunity.
6. The degeneration of thymus gland occur in old individuals leading to decreased production of thymosin thereby weakening of immune response.

Question 18.
The role of heart and kidney in hormone secretion.
Answer:
(I) Kidney:

1. Kidney produces renin, erythropoietin and calcitriol (calcitriol is the active form of vitamin cholecalciferol (D3).
2. Renin along with angiotensin helps in maintaining the blood pressure in the renal artery by vasoconstriction.
3. Erythropoietin stimulates erythropoiesis.
4. Calcitriol helps in absorbing calcium from the stomach.

(II) Heart:

1. Heart walls secrete Atrial natriuretic hormone /ANE
2. ANF increases sodium excretion [natriuresis] along with water.
3. It acts along with kidneys and reduces blood pressure by lowering blood volume.

Question 19.
Hormones secreted by adrenal gland.
Answer:

1. The adrenal cortex secretes corticoids. Corticoids is a group of several hormones that control several vital body functions.
2. Corticoids are of two types, viz. mineralocorticoids and glucocorticoids.
3. Small amounts of androgenic steroids are also secreted by the adrenal cortex which have the role in the growth of axial and pubic hairs and facial hairs during puberty.
4. The mineralocorticoids regulate the electrolyte balance while the glucocorticoids are involved in carbohydrate metabolism.
5. Adrenal medulla secretes adrenaline or epinephrine and noradrenaline or norepinephrine.

Question 20.
Role of mineralocorticoids.
Answer:

1. The mineralocorticoids regulate ionic and osmotic balance, by regulating the amounts of electrolyte and water.
2. Aldosterone is the main mineralocorticoid that acts on the renal tubules.
3. Aldosterone stimulates the re-absorption of Na⁺ and water and excretion of K⁺ and phosphate ions.
4. The aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

Question 21.
Secretions of adrenal medulla and their role.
Answer:

1. The adrenal medulla secretes two catecholamine hormones, viz. adrenaline (epinephrine) and noradrenaline (nor-epinephrine).
2. Adrenaline and noradrenaline increase alertness, dilation of pupils, piloerection, sweating, etc.
3. Both the hormones increase the rate of heartbeat, strength of heart contraction and rate of respiration.
4. These hormones also stimulate the breakdown of glycogen, lipids and proteins thereby increasing blood glucose level.
5. All the above reactions are useful for survival during emergency situations and in stress condition. Therefore, they are called emergency hormones or 3 F hormones of fright, fight or flight.

Question 22.
Cortisols and their role.
Answer:
I. Cortisol : Cortisol is the main glucocorticoid hormone.

II. Role of cortisol:

1. Cortisol stimulates many metabolic reactions such as gluconeogenesis, lipolysis and proteinolysis.
2. It inhibits cellular uptake and utilization of amino acids.
3. Cortisol also plays an important role in maintaining the cardiovascular system and kidney functions.
4. It is also involved in anti-inflammatory reactions and suppresses the immune response.
5. Cortisol stimulates the RBC production.

Question 23.
Disorders of adrenal cortical hormones.
Answer:

1. Disorders of adrenal cortical secretions are of two types, viz. hyposecretion and hypersecretion.
2. Hyposecretion of corticosteroids causes Addison’s disease.
3. The symptoms of Addison’s disease are general weakness, weight loss, low body temperature, feeble heart action, low BR acidosis, excessive loss of Na+ and Cl in urine, impaired kidney functioning and kidney failure, etc.
4. Hypersecretion of corticoids causes Cushing’s disease.
5. The symptoms of Cushing’s disease are alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, etc.

Question 24.
Hormones of adenohypophysis.
Answer:
I. Somatotropic Hormone (STH) or Growth Hormone (GH):

1. The secretion of GH is under dual control of hypothalamus through GHRF (Growth hormone releasing factor) and GHIF (Growth hormone inhibiting factor).
2. The GH brings about general growth of the body.
3. The principal actions of GH Eire promotion of linear growth in the skeleton, increase in the size of the muscles and connective tissue.
4. GH enhances the protein synthesis. The lipolysis in adipose tissue to release more fatty acids is also stimulated by GH.
5. The growth of bones by absorption of calcium takes place due to GH.

II. Thyroid Stimulating Hormone (TSH) or Thyrotropin:

1. TSH is regulated by TRF (Thyrotropin releasing factor) from hypothalamus’
2. For inhibition there is a negative feedback between thyroxine level in the blood and secretion of TSH.
3. TSH stimulates thyroid glands to increase uptake of iodine for synthesis of thyroxine. It brings breakdown of colloid to release thyroxine.

III. Adrenocorticotropic Hormone (ACTH) Corticotropin :

1. ACTH stimulates growth of adrenal cortex Eind stimulates it to secrete glucocorticoids Eind mineralocorticoids.
2. The regulation of ACTH secretion is under the control of hypothalamic CRF (Corticotropin releasing factor) and the negative feedback mechanism between plasma level of cortisol and ACTH.

IV. Prolactin (PRL) or Luteotropic hormone (LTH):
1. Secretion of prolactin is under duad control by hypothalamus by two factors such as PRF (Prolactin releasing factor) and PIF (Prolactin inhibiting factor) of hypothalamus.

2. Prolactin performs many functions therefore it has many terms as follows :

• Development of mammary glands (Mammo tropin).
• Milk secretion by mammary glands (Lactogenic hormone).
• Maintenance of corpus luteum so that- it keeps on secreting progesterone during pregnancy (Luteotropin).

3. It may be inhibiting the chances of pregnancy during lactation period.

V. Gonadotropic Hormones (GTH) or Gonadotropins:
1. There are two types of gonadotropins, viz. FSH and LH or ICSH.

2. The secretions of gonadotropin are regulated by gonadotropin releasing factor (GHRF) of hypothalamus.

(3) They are regulated by negative feedback by sex hormones such as testosterone and estrogen.
(a) Follicle Stimulating Hormone (FSH):

• FSH in female stimulates development of Graafian follicles. It helps in the formation of ovum by stimulating oogenesis.
• It also stimulates ovarian follicular cells for secretion of female sex hormones, estrogen.
• Under influence of estrogen, development of secondary sexual characters occurs in female.
• In males, FSH stimulates germinal epithelium of seminiferous tubules for spermatogenesis and helps in the production and maturation of sperms.
• Deficiency of FSH leads to infertility in both the sexes.

(b) Luteinizing hormone (LH) in females and Interstitial cell stimulating hormone (ICSH) in males:

• LH brings about ovulation, i.e. rupture and release of ovum from the mature Graafian follicle. The empty Graafian follicle is transformed into corpus luteum.
• Corpus luteum is a secondary endocrine source which secretes gestational hormone progesterone. Progesterone is a pregnancy stabilizing hormone.
• In males, ICSH stimulates interstitial cells of Leydig which in turn secretes male sex hormone, the testosterone.
• Testosterone develops secondary sexual characters in males.
• High level of progesterone in female signals negative feedback, to pituitary to stop secretion of LH.
• In males, high level of testosterone in blood gives negative feedback signal to Inhibit the secretion of FSH.

Short answers questions

Question 1.
Give one point of distinction between nervous coordination and hormonal coordination.
Answer:

1. The activity of nervous coordination is quick, immediate and fast as it sends the electrical signals.
2. Hormonal coordination is slow and long lasting as it takes place through the action of hormones.

Question 2.
Write about types of nerves.
Answer:
Nerve is a group of neurons enclosed in a connective tissue sheath epineurium. It is classified as:

1. Sensory nerve : A nerve having all sensory neurons is called sensory nerve. It carries information from sense organs to CNS. It is also called afferent nerve.
2. Motor nerve : A nerve having all motor neurons is called motor nerve. It carries information from CNS to effector organs. It is also called efferent nerve.
3. Mixed nerve : A nerve having with both sensory and motor neurons is called a mixed nerve. Sensory neurons in it carry nerve impulses from sense organs to CNS while motor neurons carry nerve impulses from CNS to effector organs.

Question 3.
What is the composition of neural tissue?
Answer:
Neural tissue is derived from embryonic ectoderm. It consists of two types of cells.

1. Neurons or nerve fibres : These are structural and functional units of nervous tissue. They conduct nerve impulses and coordinate various body activities.
2. Neuroglial cells : These are supportive cells which protect neurons throughout CNS and PNS. They perform other functions like secretion of myelin sheath, phagocytosis, production of CSF, etc. e.g Schwann cells, astrocytes, satellite cells, etc.

Question 4.
Explain the structure of cyton.
Answer:

1. Cyton is the main body of a neuron or nerve fibre.
2. The cyton has a distinct central nucleus with a nucleolus and neuroplasm.
3. Cytoplasm surrounds the nucleus around which there are neurofibrils, Nissl’s granules and other cell organelles.
4. Nissl’s granules are rich in ribosomes and proteins.
5. Neurofibrils play an important role in transmission of nerve impulse.
6. The cytons are generally found inside the brain, spinal cord (CNS) and in the ganglia.
7. Cytons within CNS form ‘nuclei’ while those present outside CNS in nerves form ‘ganglia’.

Question 5.
Enlist the various connective tissue layers in a nerve along with their location.
Answer:
Connective tissue layers in a nerve are:

1. Endoneurium : Covers each nerve fibre.
2. Perineurium : Covers each nerve bundle having a number of neurons.
3. Epineurium : Covers many nerve bundles to form a peripheral nerve.

Question 6.
What is a synapse?
Answer:

1. Synapse is a microscopic functional gap between two successive neurons.
2. In this telodendrites of pre-synaptic neuron are in close proximity with dendrites of post-synaptic neuron.
3. This gap is also called synaptic cleft.
4. During transmission of nerve impulse, the synapse get filled with neurotransmitters like acetyl choline.

Question 7.
What are the three divisions of nervous system? What are their chief functions?
Answer:

1. The three divisions of nervous system are central nervous system, peripheral nervous system and autonomous nervous system.
2. The central nervous system (CNS) consists of brain and spinal cord. The brain and spinal cord are the coordinators for all nervous functions.
3. The peripheral nervous system (PNS) is constituted by several nerves given by the central nervous system to all the body parts. All these nerves carry impulses to the CNS and bring back the responses from them. They are divided into cranial nerves and spinal nerves.
4. The autonomous nervous system controls all the internal organs and is not under voluntary control.

Question 8.
Does this CSF remain enclosed inside the ventricles? What can be the outcome of such a situation?
Answer:

1. CSF is present within the CNS as well as around it.
2. This fluid communicates with each other on the roof of medulla oblongata through 3 apertures, viz. Foramina of Luschka and foramen of Magendie.
3. This communication ensures constant pressure of CSF within as well as outside the CNS.
4. In the absence of this communication, there would be a pressure difference within as well as outside the CNS which will result in disturbances in the activities of CNS. Moreover, intercranial pressure would rise.

Question 9.
Enlist the different parts of the brain.
Answer:

1. There are three divisions of the brain, viz. forebrain (prosencephalon), midbrain mesencephalon) and hindbrain (rhombencephalon).
2. Forebrain is divided into cerebrum (telencephalon) and diencephalon (thalamencephalon). Underdeveloped of factory lobes (rhinencephalon) can also be seen in the anterior region.
3. Midbrain consists of corpora quadrigemina and crura cerebri.
4. Hindbrain has cerebellum (metencephalon) and brain stem. It is divided into pons varolii and medulla oblongata (myelencephalon).

Question 10.
Describe functional areas of cerebral cortex.
Answer:
Functional areas of cerebrum:

1. There are three functional areas in cerebrum viz., sensory, association and motor area.
2. In sensory area, sensory receptors bring the sensory inputs. These inputs are analysed in sensory area.
3. The sensory speech area is located in parietal lobe. It is called Wernicke’s area.
4. Association area forms the major portion of the cerebrum. It processes, analyses and stores the information given by the inputs. Power of reasoning, will, understanding, memory, etc. are the faculties present in the cerebral cortex.
5. Motor area is present in the frontal lobe lying anterior to the premotor area. In the lower part of the motor area just above the lateral sulcus lies the Broca’s area or motor speech area. The Broca’s area controls the movements necessary for speech.

Question 11.
Explain in detail the regions associated with the diencephalon.
Answer:
Diencephalon is a part present between forebrain and midbrain. It has three regions epithalamus, thalami and hypothalamus.
(1) Epithalamus : Epithalamus is the roof of diencephalon. It is highly vascular and non- nervous. It forms anterior choroid plexus that secretes cerebrospinal fluid. Pineal body is attached to epithalamus with the help of pineal stalk. Pineal body secretes serotonin and melatonin.

(2) Thalami : The lateral parts of diencephalon which are interconnected with the habencular commissure are called thalami. From thalami all sensory impulses (except olfactory impulses) pass on to the cerebrum.

3. Hypothalamus : Hypothalamus is the floor of the brain. Pituitary gland is attached to this floor by an infundibular stalk. Hypothalamus has many hypothalamic nuclei which are scattered in the white matter.

Question 12.
What is EBG? What information can be obtained from the EEG?
Answer:

1. EEG stands for electroencephalography.
2. It refers to the recordings of the brain’s spontaneous electrical activities in certain period of time.
3. These are recorded using multiple electrodes.
4. EEG is non-invasive method and measures voltage fluctuations resulting from ionic current within the neurons.
5. The basic concepts involved in this are similar to ECG.
6. It is used to diagnose conditions like epilepsy, sleep disorders, encephalopathies, coma, etc.

Question 13.
Find out how different functional areas of the brain can be mapped?
Answer:
Functional areas and status of the brain can be mapped by several imaging techniques available such as-

1. MRI : Magnetic Resonance Imaging
2. CT : Computed Tomography
3. PET : Positron Emission Tomography

Question 14.
Which are silent areas of the brain?
Answer:

1. Silent areas of the brain refer to association areas of the brain.
2. One such area is in the prefrontal cortex of brain.
3. These are the areas of the brain in which pathogenic conditions may occur without producing symptoms.
4. Injury to these areas is not accompanied by symptoms related to sensory and motor functions.

Question 15.
Is nervous tissue without lymphatic vessels?
Answer:

1. CSF is the lymph of CNS.
2. CSF is continuously generated by the ependymal cells lining the ventricles and central canal and simultaneously drained out of the brain into the blood stream.
3. There are no lymphatic vessels in the nervous system.
4. But the CSF is drained into peripheral blood circulation with the help of lymph vessels associated with meninges mainly the dura mater.

Question 16.
Explain the structure of spinal cord.
Answer:
Structure of spinal cord:

1. Spinal cord is a 42 to 45 cm long, 2 cm thick and hollow tube, extending from medulla oblongata to lumbar region.
2. It lies in the neural canal of vertebral column.
3. At the other end, it tapers down and is called conus medullaris. The posterior most end is called filum terminale which appears as a thread-like structure.
4. Beyond the second lumbar vertebra, it forms a horse tail-like structure called cauda equina. Cauda equina is a bunch of dorsal and ventral roots of last pair of spinal nerves.
5. There are two swellings on the spinal cord. The upper is cervical swelling and lower is lumbar swelling. Accordingly there are two plexuses, the cervical plexus supplying nerves to hands and the lumbar plexus supplying nerves to legs.
6. 31 pairs of spinal nerves arise from spinal cord.

Question 17.
What is the significance of reflex action?
Answer:
Significance of reflex action :

1. Reflex action helps the animals to adjust quickly to the changing environment.
2. Reflex action is for quick actions necessary for survival. The life may have been in danger in the absence of reflex action.
3. Most of the reflexes are spinal reflexes, i.e. reflexes controlled by spinal cord. Thus the brain is not involved in these actions. This prevents overloading of the brain and brain fatigue.
4. Some reflexes are inborn and hence training or learning is not required for these.

Question 18.
During extraction of a tooth, the dentist gives an injection of Anaesthesia to the patient before extraction. Is the action potential generated? How does the local anaesthesia work? What is the effect of pain killer on the nervous system?
Answer:

1. Teeth are innervated by branches of trigeminal nerve [Vth cranial nerve]
2. Extraction of tooth stimulates this nerve which then carries the impulse [action potential) to the pain centre of the brain where the pain is perceived.
3. To avoid this, dentists give anaesthesia, to numb the nerve.
4. Action potential is not generated due to anaesthesia.
5. Hence the pain is not perceived.
6. Similarly, some common pain killers act on the nerve endings and pain centres of the brain, preventing generation of action potential.

Question 19.
Give a list of psychological disorders.
Answer:

1. Autism spectrum disorder.
2. Bipolar disorder.
3. Depression.
4. Anxiety disorder.
5. ADHD (Attention Deficit Hyperactivity Disorder).
6. Stress related disorders.

Question 20.
What are endocrine glands?
Answer:

1. Endocrine glands are ductless glands which are capable of secreting hormones.
2. The hormones are poured directly into the bloodstream as the endocrine glands do not have duct.
3. Hormones regulate the function of target tissue or organ.
4. They either have excitatory effect or have an inhibitory effect.

Question 21.
What are the main endocrine glands in human body?
Answer:
The main endocrine glands in human body are as follows:

1. Pituitary or hypophysis
2. Hypothalamus
3. Thyroid
4. Parathyroid
5. Adrenal or suprarenal
6. Islets of Langerhans in pancreas
7. Endocrine parts of gonads, i.e. testis and ovary.
8. Pineal gland and thymus are also endocrine glands of less importance.

Question 22.
What are the common properties of hormones?
OR
State properties of hormones.
Answer:

1. Hormones are specifically produced in response to a certain stimulus.
2. Depending on nature and intensity of the stimulus, the rate of secretion of a hormone varies from low to very high.
3. Hormones are produced in one organ and show their effect on distant ‘target’ organ. The source and the target region may be distantly located.
4. Hormones are directly poured in blood circulation and always carried through blood.
5. Hormones are always bound to specific carrier proteins while being transported through the blood.
6. Hormones have a high degree of target specificity.
7. Every hormone acts basically by modifying some aspect of cellular metabolism.
8. The excessive secretions or deficiencies- of hormones may lead to serious disorders. Such disorders are called hyper – and hypo- disorders, respectively.

Question 23.
What are the disorders caused due to hyposecretion and hypersecretion of GH or STH?
Answer:
(1) Hypersecretion is excessive secretion. In children, the hypersecretion of GH causes gigantism. In adults, it causes Acromegaly.

(2) Hyposecretion, i.e. lesser secretion of GH in children cause dwarfism. The person is also referred to as midget. There are two types of dwarfs, viz. Frohlic dwarf who are mentally abnormal and Lorain dwarf who are mentally normal.

(3) Hyposecretion of GH in adults cause Simmond’s disease.

Question 24.
What are the disorders caused due to hyposecretion and hypersecretion of ACTH?
Answer:

1. Hyposecretion of ACTH leads to Addison’s disease, i.e. adrenal failure. This results in affected carbohydrate metabolism leading to weakness and fatigue.
2. The hypersecretion leads to excessive growth of adrenal cortex. This causes Cushing’s disease.

Question 25.
Write an account of hormones secreted by the thyroid gland.
Answer:

1. Thyroid secretes triiodothyronine or T₃, tetraiodothyronine or thyroxine or T₄ and thyr ocalcitonin.
2. The thyroid gland synthesize, store and discharge these hormones.
3. T₃ and T₄ are iodinated derivatives of amino acid tyrosine which are secreted by thyroid follicular cells and stored in follicles. They have similar function. The secretion of T₃ and T₄ are regulated by Thyroid stimulating hormone (TSH) or thyrotropin of pituitary gland in negative feedback manner. T₃ is more active and T₄ is more potent hormone.
4. Thyrocalcitonin is secreted by the parafollicular cells.
5. Thyrocalcitonin regulates blood calcium level. It stimulates bones to take up Ca⁺⁺ from the blood and deposit it in the form of calcium phosphates in the bones, thereby decreasing blood Ca⁺⁺ level. Increased calcium level of blood stimulates ‘C’ cells to secrete thyrocalcitonin and vice versa.

Question 26.
Describe adrenal glands with respect to morphology, histology and secretions.
Answer:

1. A pair of adrenal or suprarenal glands are located just on the upper border of kidneys.
2. The adrenal gland shows two distinct regions, viz. thicker outer cortex and thinner inner medulla.
3. The adrenal cortex consists of three distinct regions. The outer zona glomerulosa, the middle zona fasciculata and inner zona reticularis.
4. Adrenal cortex produces corticoids. Corticoids is a collective term for many hormones, such as glucocorticoids, mineralocorticoids and steroid sex hormones.
5. Adrenal medulla secretes adrenaline or epinephrine and noradrenaline or norepinephrine.

Question 27.
Why are reproductive organs called dual in function?
Answer:

1. A pair of testes in males and a pair of ovaries in female both secrete hormones which are essential for sexual characters and function.
2. Besides this, they also produce male and female gametes respectively. Therefore they are said to be dual in function.

Question 28.
What are male hormones? What is their source and functions?
OR
Write a short note on the functions of androgens.
Answer:

1. Androgens are male hormones. The most significant androgen is the testosterone.
2. Interstitial cells of Leydig present in the testis of mature man produce androgen. Androgens are steroid in chemical nature.
3. Androgens regulate and stimulate the development, maturation and functions of the male reproductive organs like seminiferous tubules, epididymis, vas deferens, seminal vesicles, prostate glands and urethra.
4. Androgens are made sex hormones. They produce secondary sexual characteristics. Low pitch of voice is produced due to changes in the vocal cords which take place due to testosterone, etc. They stimulate muscular growth and growth of facial and axillary hairs.
5. The mental make up of a man like aggressiveness is due to testosterone.
6. They stimulate seminiferous tubules for the process of spermatogenesis.

Question 29.
What are female sex hormones? What role do they play?
Answer:
(1) The ovaries secrete two steroid hormones viz., estrogen and progesterone.

(2) The estrogen is secreted by the developing ovarian follicles. It has many roles in stimulation of female reproductive functions and growth of ovaries, fallopian tubes, uterus and vagina.

(3) It also controls female secondary sexual characteristics like high pitch of voice, development of mammary glands, broadening of pelvis, growth of pubic hairs and deposition of subcutaneous fats to produce feminine stature.

(4) The estrogen also regulates female sexual behaviour.

(5) The empty Graafian follicle after ovulation is converted into a structure called corpus luteum which secretes a hormone known as progesterone. Progesterone is a gestational hormone which is essential for maintaining the pregnancy. It also acts on the mammary glands and stimulates them for lactation, milk synthesis and ejection.

Question 30.
Why is pancreas called a dual gland?
Answer:

1. Pancreas is called a dual gland because it is exocrine as well as endocrine in nature.
2. The exocrine pancreas secretes digestive enzymes through acini.
3. The endocrine pancreas secretes hormones through its endocrine cells called Islets of Langerhans.

Question 31.
What are the hormones of pancreas? Describe the functions of pancreatic hormones.
OR
Pancreas plays an important role in controlling diabetes mellitus. Explain.
Answer:

1. Islets of Langerhans consists of three types of cells known as a-cells, β-cells and δ-cells.
2. α-cells secrete glucagon while β-cells secrete insulin. δ-cells secrete somatostatin.
3. The glucagon is a hyperglycemic hormone. It is a peptide hormone which acts mainly on the liver cells. Here it stimulates hepatocytes for glycogenolysis (i.e. breakdown of glycogen) leading to increased level of blood glucose (i.e. hyperglycemia).
4. It also stimulates gluconeogenesis (i.e. formation of glucose from non-carbohydrate sources). This in turn brings rise in blood glucose level or hyperglycemia.
5. Glucagon reduces the cellular glucose uptake and utilisation.
6. Insulin is also a peptide hormone, which plays a major role in maintenance of blood glucose level.
7. Insulin stimulates hepatocytes and adipocytes for cellular glucose uptake and utilization.

Therefore glucose from the blood decreases causing hypoglycemia. This hormone helps in the conversion of glucose to glycogen (i.e. glycogenesis) that occurs in target cells.

Question 32.
How is blood glucose level maintained?
Answer:
The blood glucose level is maintained by the joint but antagonistic action of insulin and glucagon.
Insulin is hypoglycemic hormone while glucagon is hyper/glycemic hormone.

When there is excess sugar in the blood, more insulin is secreted by the pancreatic islets. This causes the conversion of blood glucose into glycogen. This process is known as glycogenesis. This causes decline in the level of glucose in the blood.

When there is less blood glucose level then the glucagon is secreted. It causes stored glycogen to be converted into glucose. This process is called glycogenolysis.

Question 33.
What happens when there is insufficiency or deficiency of insulin in the body?
Answer:

1. Due to insufficiency, of insulin level there is prolonged hyperglycemia. This leads to diabetes mellitus.
2. In this diabetic condition cells are unable to utilize glucose. Therefore, in a diabetic person blood glucose levels are high. The glucose is excreted in urine.
3. The harmful compounds like ketone bodies are formed leading to ketosis.
4. Diabetes can be treated by taking insulin injections or tablets (insulin therapy) or with hypoglycemic drugs.

Question 34.
Where are parathyroid glands located? What are their functions?
OR
Write a short note on the functions of Parathyroid hormone (PTH).
Answer:

1. Parathyroid glands are located on the back side of the thyroid gland.
2. There are two pairs of parathyroid glands. One pair of parathyroid is in each lobe of thyroid.
3. Parathyroid glands secrete a peptide hormone known as parathromone or parathyroid hormone (PTH).
4. The level of Ca⁺⁺ in the blood regulates the secretion of PTH.
5. PTH is hypercalcemic hormone, it increases blood calcium level. Thus the calcium balance is maintained by TCT and PTH.

Question 35.
What are the gastrointestinal hormones? Explain the function of each.
Answer:

1. There are scattered endocrine cells in different parts of alimentary canal.
2. These cells secrete four peptide hormones which are gastrin, secretin, cholecystokinin (CCK) and gastric inhibitory peptide (GIP).
3. Gastrin stimulates gastric glands for the secretion of hydrochloric acid and pepsinogen.
4. The secretin acts on exocrine pancreas and stimulates secretion of water and bicarbonate ions to form pancreatic juice.
5. CCK acts on pancreas and gall bladder and stimulates the secretion of pancreatic enzymes and bile juice respectively.
6. GIP inhibit gastric secretion and motility.

Question 36.
Name the hormone secreted by the heart. What is its function?
Answer:

1. The atrial wall of the heart secrete a peptide hormone known as atrial natriuretic factor (ANF).
2. When the blood pressure increases, ANF hormone is secreted.
3. It causes dilation of the blood vessels.
4. Blood then can easily flow with lesser resistance and hence BP decreases.

Question 37.
What are the hormones of kidney? What function do they carry out ?
Answer:

1. The juxtaglomerular cells of the kidney produce a peptide hormone known as erythropoietin.
2. Erythropoiesis stimulates bone marrow for the production of RBCs. It thus stimulates the process of erythropoiesis.
3. Hormone calcitriol from kidney helps in the formation of bones.

Question 38.
Give importance of hypothalamus.
Answer:

1. Hypothalamus is the controlling centre for hypophysis.
2. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of hypophysis. E.g. Adrenocorticotropin hormone releasing factor or CRF; Thyrotropin releasing factor or TSHRF; GHRF and GHRIE i.e. Growth hormone releasing and release inhibiting factor, etc.
3. Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.
4. Hormones like vasopressin and oxytocin are secreted by neurosecretory cells of hypothalamus.
5. Hypothalamus can register the internal changes in the body as it is a part of diencephalon and thus it accordingly brings about coordination in the body through endocrine system.

Question 39.
Write a brief account of releasing factors secreted by hypothalamus.
Answer:
(1) In the hypothalamus, there are several groups of neurosecretory cells which form different nuclei.

(2) These hypothalamic nuclei are supraoptic, paraventricular, dorso-median and ventromedian nuclei. These neurosecretory cells produce releasing and inhibiting factors.

(3) The hypothalamic neurohormones regulating the release of pituitary hormones are called releasing factors. The following are some of J the important releasing factors:

• CRF or corticotropin releasing factor or ACTH releasing factor releases secretion of Adrenocorticotropin hormone (ACTH).
• TRF or TSHRF (Thyroid stimulating hormone releasing factor) stimulates release of TSH.
• FSH RF (Follicle stimulating hormone ; releasing factor) stimulates release of FSH.
• GHRF (Growth hormone releasing factor) GHRIF (Growth hormone release inhibiting factor) act on release and regulation of growth hormone.
• PRF (Prolactin releasing factor) and PRIF ; (Prolactin release inhibiting factor) act on release and regulation of prolactin.
• MSHRF (Melanocyte stimulating hormone releasing factor) and MSH RIF (Melanocyte stimulating hormone release inhibiting factor) act on release and regulation of MSH.

Question 40.
Hormones are called chemical messengers and regulators. Explain.
Answer:

1. Hormones bring about coordination in the body with the help of nervous system.
2. Endocrine system and nervous system together form neuro-endocrine system.
3. This system works in tune with the external and internal environmental changes.
4. The hormones are either excitatory or inhibitory. They bring about the actions accordingly to keep the body in homeostasis or equilibrium.
5. Almost all endocrine glands are controlled by negative feedback inhibition. Some glands are auto-regulatory. Therefore, the concentration of hormones cannot be in excess or in deficiency.
6. Almost all the functions such as metabolism, growth, reproduction, etc. are under the control of hormones. Therefore hormones are called regulators and messengers.

Chart based/Table based questions

Question 1.
Draw a flow chart of – steps in generation and conduction of a nerve impulse.
Answer:

Question 2.
Enlist the names of following cranial nerves: I, II, VII, XII
Answer:

Number	NAME	NATURE
I	Olfactory	Sensory
II	Optic	Sensory
VII	Facial	Mixed
XII	hypoglossal	Motor

Question 3.
Enlist the names of following cranial nerves : III, W VI, XI
Answer:

Number	NAME	NATURE
III	Occulomotor	Motor
IV	Trochlear/Pathetic	Motor
VI	Abducens	Motor
XI	Spinal accessory	Motor

Question 4.
Complete the table.

Number	Type	No. of Pairs	Region
————	Cervical	————-	—————
T1 – T12	————-	12 pairs	————-
L1 – L5	————-	5 pairs	Lower back
————-	Sacral	————-	Pelvic
————–	Coccygeal	—————	Tall region

Answer:

Number	Type	No. of Pairs	Region
C1 – C8	Cervical	8 pairs	Neck
T1-T12	Thoracic	12 pairs	Thorax / Upper back
L1 -L5	Lumber	5 pairs	Lower back
S – S5	Sacral	5 pairs	Pelvic
Co1	Coccygeal	1 pair	Tall region

Question 5.
Write types of neuroglial cells of CNS and PNS in tabular form.
Answer:

CNS – glial cells	PNS – glial cells	Functions
Oligodendrocytes [cells with few branches]	Schwann cells	Secrete myelin sheath
Astrocytes [star-shaped and most abundant glial cells in CNS]	Satellite cells	Protect, cushion and supply nutrients to nearby neurons. Help in maintaining blood-brain barrier.
Microglia [small cells with few branches]	Macrophages	Phagocytosis
Ependymal cells lining the ventricles of brain [mostly columnar]	Ependymal cells lining central canal of spinal cord	Secrete cerebrospinal fluid

Question 6.
Enlist the various receptors found at various location in the body.
Answer:

Receptors	Types	locations
Mechanoreceptors	Thermoreceptors	Skin
	Tango [touch and pressure] receptors	Skin
	Tactile [light touch] receptors	Skin
Chemoreceptors	Gustato receptors	tongue
	Olfacto receptors	Olfactory mucosa
Photoreceptors	Rod and cone cells	Retina of eye
Phonoreceptors	Organ of Corti	Cochlea of internal ear
Stato receptors	Cristae and maculae	Semicircular canals, utricle, saccule of internal ear

Question 7.
Sketch the concept maps for mechanism of vision and mechanism of hearing.
Answer:

Question 8.
Complete the table given below by putting a tick (✓) in the boxes where applicable.

Action	Reflex	Voluntary
1. Touching a hot object	———-	———–
2. Releasing saliva on smelling food	———–	————
3. Applying a car’s brakes in an emergency	————	————–
4. Blinking of eyes when a small insect touches the eye	———–	————

Answer:

Action	Reflex	Voluntary
1. Touching a hot object		✓
2. Releasing saliva on smelling food	✓
3. Applying a car’s brakes in an emergency	can be a conditioned feclex too.	✓
4. Blinking of eyes when a small insect touches the eye	✓	————

Question 9.
Complete the following table

Action	Reflex	Voluntary
1. Optic nerve	———–	———–
2. Facial	———–	————
3. Hypoglossal	————	————–
4. Trigeminal	———–	————
5. Auditory
6. Glosso-pharyngeal

Answer:

Action	Reflex	Voluntary
1. Optic nerve	Sensory	Sense of vision and light
2. Facial	Mixed	Facial expression, movement of neck, tongue, etc. and saliva secretion
3. Hypoglossal	Motor	Movement of tongue
4. Trigeminal	Mixed	Sensation of touch, taste and jaw movements
5. Auditory	Sensory	Hearing and equilibrium
6. Glosso-pharyngeal	Mixed	Taste, pharyngeal contractions ‘and saliva secretion

Diagram based questions

Question 1.
Sketch and label – nerve net of Hydra.
Answer:

Question 2.
Sketch and label – nervous system of Planaria
Answer:

Question 3.
Sketch and label – depolarization and repolarization along nerve.
Answer:

Question 4.
Sketch and label ultrastructure of synapse.
Answer:

Question 5.
Sketch and label – lateral view of brain.
Answer:

Question 6.
Sketch and label – functional areas of Brain?
Answer:

Question 7.
Sketch and label – ventral view of human brain.
Answer:

Question 8.
Sketch and label – ventricles of brain.
Answer:

Question 9.
Sketch and label T.S. of spinal cord.
Answer:

Question 10.
Sketch and label – formation of spinal nerve.
Answer:

Question 11.
Sketch and label – mechanism of hormone action.
Answer:

Question 12.
Sketch and label – V.S. of pituitary.
Answer:

Question 13.
Sketch and label – morphology of thyroid
Answer:

Question 14.
Sketch and label – histology of thyroid
Answer:

Question 15.
Sketch and label – parathyroid glands
Answer:

Question 16.
Sketch and label thymus.
Answer:

Question 17.
Sketch and label – adrenal gland.
Answer:

Long answer questions

Question 1.
Explain transmission of nerve impulse across a synapse.
OR
Explain how is impulse transmitted through a synapse?
Answer:

1. The nerve impulse travels along the axon of the pre-synaptic neuron to the axon terminal.
2. Pre-synaptic neurons or axons have several synaptic knobs at their ends or terminals.
3. These knobs have membranous sacs, called synaptic vesicles having neurotransmitter molecules.
4. When an impulse reaches a synaptic knob, voltage sensitive Ca⁺⁺ channels open and calcium ions (Ca⁺⁺) diffuse inward from the extracellular fluid.
5. The increased calcium concentration inside the cells, initiates a series of events that help to fuse the synaptic vesicles with the cell membrane of pre-synaptic neuron, where they release their neurotransmitters by exocytosis.
6. The neurotransmitters bind to the receptors of the post-synaptic cell,
7. This action is either excitatory (stimulating) or inhibitory (slowing down/stopping).
8. Once the impulse has been transferred across the synapse, the enzyme like acetyl cholinesterase destroys the
9. neurotransmitter and the synapse is ready to receive a new impulse.

Question 2.
Explain transmission of nerve impulse along the axon.
OR
Describe the conduction of a nerve impulse in the neuron.
Answer:
1. Before conduction of nerve impulse, the cell membrane is in the polarized state.

2. When a stimulus is applied at a site on the polarised membrane, the membrane at that site becomes freely permeable to Na⁺.

3. This leads to a rapid influx of Na⁺ followed by the reversal of the polarity at that site, i.e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged.

4. The polarity of the membrane at that site [site A] is thus reversed and hence depolarised. The electrical potential difference across the plasma membrane at the site of stimulation is called the action potential, which is in fact termed as a nerve impulse.

5. At sites immediately ahead [site B], the axon membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site A to site B.

6. On the outer surface current flows from site B to site A to complete the circuit of current flow. Hence, the polarity at the site is reversed, and an action potential is generated at site B. Thus, the impulse (action potential) generated at site A arrives at site B.

7. The sequence is repeated along the length of the axon and consequently the impulse is conducted.

8. The rise in the stimulus-induced permeability to Na⁺ is extremely short-lived. It is quickly followed by a rise in permeability to K⁺.

9. Within a fraction of a second, K⁺ diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation and the fibre becomes once more responsive to further stimulation.

Question 3.
Explain the structure of cerebrum. Structure of cerebrum
Answer:
Answer: Structure of cerebrum:

1. Cerebrum is the largest part of the brain. It forms 80-85% volume of the brain.
2. A median longitudinal fissure divides the cerebrum into two cerebral hemispheres. These hemispheres are interconnected by a thick band of transverse nerve fibres called corpus callosum.
3. The outer part of cerebrum is called cerebral cortex while the inner part is called cerebral medulla.
4. The roof of cerebrum is called pallium. Pallium is highly folded forming sulci and gyri. Sulci are depressions while gyrl are ridges. The gyri increase the surface area of cerebral cortex.
5. The ventro-lateral walls of cerebrum are thickened and are called corpora striata.
6. The cerebral cortex has three deep sulci, the central, lateral and parieto-occipital.
7. These sulci divide the cerebral hemisphere into four lobes. These are frontal, parietal, occipital and temporal lobes. A fifth median lobe called insula or insular cortex is folded deep within the lateral sulcus.
8. The central sulcus separates frontal and parietal lobes, the lateral sulcus separates parietal and temporal lobes and the parieto¬occipital sulcus separates parietal and occipital lobe.

Question 4.
Describe the structure and function of midbrain.
Answer:
1. Structure of midbrain:

• Midbrain is the middle part of the brain situated between the forebrain and the hindbrain. It is present between the pons varolii and diencephalon.
• It has two distinct regions : Corpora quadrigemina and crura cerebri.
• Corpora quadrigemina consists of two pairs of lobes, viz., superior colliculi and inferior colliculi. These are located in the thick wall of midbrain.
• Crura cerebri are thick bands of longitudinal nerve fibres, present on the floor of midbrain.

2. Functions of midbrain:

• Inferior colliculi control and coordinate auditory reflexes.
• Superior colliculi control head and eye movements.
• Crura cerebri connect the cerebrum to cerebellum and spinal cord.

Question 5.
Give an account of structure of hindbrain.
Answer:
Structure of hindbrain:

1. Hindbrain includes cerebellum, pons varolii and medulla oblongata.
2. Cerebellum is 11% of the total brain and is the second largest part of the brain.
3. It has three lobes, median vermis and lateral two cerebral hemispheres. It has outer grey and inner white matter.
4. Cerebral cortex shows sulci and gyri. The inner white matter of cerebellar medulla shows arbor vitae or branching tree-like processes.
5. Pons is the part that connects the two cerebral hemispheres. It has outer white and inner grey matter. Pons is made up of nerve fibres which form bridges between cerebrum and medulla oblongata.
6. Medulla oblongata is the last part of the hindbrain which continues further as a spinal cord. It has outer white and inner grey matter.
7. Its roof shows has posterior choroid plexus.
8. Eight pairs of cranial nerves arise from medulla oblongata.

Question 6.
Describe T.S. of spinal cord.
Answer:

1. Externally there are three meninges covering spinal cord Duramater, arachnoid membrane and pia mater.
2. Dorsoventrally there are two fissures, the shallow dorsal or posterior fissure and the deeper ventral or anterior fissure.
3. From dorsal fissure a dorsal septum extends inside.
4. Neurocoel or central canal is situated in the centre of spinal cord.
5. The central canal is filled with cerebro¬spinal fluid and is lined by cuboidal epithelial cells called layer of ependyma.
6. There is inner grey and outer white matter in the spinal cord. This grey matter is in the shape of ‘H’ with two dorsolateral horns and two ventro-lateral horns.
7. Dorsal horns form dorsal roots and ventral horns form ventral roots.
8. White matter is divided into three columns, viz., the dorsal funiculi, ventral funiculi and lateral funiculi on either side.
9. Ascending and descending tracts of nerve fibres arise from dorsal and ventral roots of the spinal cord. Ascending tracts are sensory while descending tracts are motor in nature.

Question 7.
What are the different types of reflexes?
Answer:
1. Based on the location of their action : Thereflexes are divided into somatic reflexes and visceral reflexes.

• When effector is located in body structures such as skeletal muscles, it is called a somatic reflex.
• When the effector is located in the visceral organs such as glands or smooth muscles then it is called a visceral reflex.

2. Based on the basis of number of neurons : Reflexes are of two types, viz. monosynaptic reflexes and polysynaptic reflexes.

• Simple or monosynaptic reflexes are those in which one sensory and one motor neuron are involved in the reflex action.
• Polysynaptic or complex reflexes are those when more than two neurons are involved in the reflex action.

3. Based on inheritance and experience of learning : The reflexes are subdivided into unconditional or inborn and conditional or acquired.

• Unconditional or inborn reflexes are inborn or hereditary. They are permanent, never disappear and need no previous experience, e.g. blinking of eyes, suckling, swallowing, knee jerk, sneezing, coughing, etc.
• Conditional or acquired reflexes are acquired during life by experience or learning. They are based on individual learning or experience. These are not heritable, temporary and may disappear or reappear, e.g. driving, cycling, etc.

Question 8.
Explain the mechanism of reflex action.
Answer:

1. Mechanism of reflex action: There are series of sequential events in which reflex action is completed:
2. Stimulus is picked up by any receptors, e.g. pricking of a needle in the hand, causes skin to be a receptor.
3. Sensory impulse is formed in grey matter of spinal cord. It receives sensory impulse, interprets it and generates motor impulse.
4. The cyton of motor neuron present in the ventral horn of grey matter and axon conducts motor impulse from spinal cord to effector organ. This is further carried by dendrites innervating the skin.
5. Impulse is carried to the association neuron by axon of sensory neuron, when impulse reaches the end of the axon there is a synapse.
6. Transmission takes place by releasing acetylcholine from the synaptic buttons at the end of the axon.
7. It fills the synaptic gap and helps in chemical transmission of the impulse from axon of one neuron to dendron of the other neuron. Once the impulse reaches the dendrites of association neuron; axonic button releases an enzyme, acetylcholine esterase which neutralizes the acetylcholine and again a synaptic gap is formed. This mechanism helps to receive new impulse or avoid the mixing of different impulses.
8. The association neuron receives sensory impulse, interprets it, analyses it and generates motor impulse. Motor impulse again travels through synapse between association neuron and motor neuron.
9. Impulse travels through motor neuron and reaches the effector organ like skeletal muscles or the gland. The effector organ gives a proper response like contraction of the muscles or secretion by the gland.

Question 9.
Define receptors. Enlist different types of receptors.
Answer:
1. Receptors : Receptors are specialized cells, tissues or organs present in the body which receive different stimuli.

2. Types of receptors:

• Receptors are of two types, viz. exteroceptors and interoceptors.-
• Exteroceptors receive stimuli directly from the external environment. They are somatic in nature.
• Interoceptors are located inside the body and are visceral in nature. They respond to internal changes in the body.
• The various types of exteroceptors and interoceptors, their location and functions have been summarized in the table given below:
  

  Types of Exteroceptors	Location	Function
  1. Mechanoreceptors	Touch corpuscles in skin	Tangoreceptors Pressure Tactile receptors- Touch
  2. Thermoreceptors	Skin	Frigido receptors (cold) Heat receptors (warmth)
  3. Chemoreceptors	Tongue, nasal mucosa	Gustatoreceptors – Taste Olfactoreceptors- Smell
  4. Statoacoustic receptors	Internal ear	Cochlea – Hearing Semicircular canals-Balance and equilibrium
  5. Photoreceptors	Retina of the eye	Rods and cones interpret images Rods-black and white image. Sensitive to dim light. Cones – Coloured image. Sensitive to bright light.

Question 10.
Describe the different parts of human eye.
OR
Describe briefly the structure of eye.
Answer:

1. Human eyes are a pair of organs located in sockets of the skull called orbits.
2. Eyeball is spherical and has three layers.
3. Sclera is the outer layer of dense connective tissue with anterior transparent cornea.
4. Choroid is the middle layer. It is bluish in colour containing many blood vessels. The anterior region is thick and forms the ciliary body. Posterior 2/3rd region is thinner.
5. Iris is the forward segment of the ciliary body which is pigmented and opaque. This part is the visible coloured portion of the eye.
6. Lens is present anteriorly inside the iris and is held in position by the ligaments of ciliary body.
7. The aperture surrounded by the iris in front of the lens is known as pupil. The movement of the pupil is regulated by the muscle fibres of iris.
8. The innermost layer of the eye is the retina having three sub-layers formed by ganglion cells, bipolar cells and photoreceptor cells, which are sensitive to light.
9. There are two types of photoreceptor cells, viz. rods and cones containing light sensitive proteins. They are termed as photo pigments, rhodopsin which is a derivative of vitamin A (in rods) and iodopsin (in cones).
10. The cones are responsible for daylight or photopic vision and colour vision. The rods function in dim light giving scotopic vision.
11. The cones are of three types, each containing its own characteristic photopigments that respond to red, green and blue lights.
12. The optic nerve leaves the eye at a point slightly away from the median posterior pole of the eyeball. In this region, the rods and cones are absent therefore this region is known as blind spot. Macula lutea, a yellowish pigmented spot is present lateral to the blind spot.
13. Fovea is a central pit present beside it. Fovea is a thinned out portion of the retina where only the cones are densely packed and therefore have greatest visual acuity (resolution).
14. A space between the cornea and the lens is called aqueous chamber. It contains a thin watery fluid known as aqueous humor.
    

Question 11.
Describe the internal structure of human ear.
OR
Ear is one of the important sense organs known for its role in hearing and balancing. Describe those structures present in the internal ear which helps in these functions.
Answer:

1. The ears are the auditory sensory organs, also involved in maintaining equilibrium of the body.
2. The ear is composed of three divisions namely the outer ear, middle ear and internal ear.
3. The external ear consists of the pinna and external auditory meatus (canal). The pinna is for the collection of sound waves coming from the environment. The external auditory canal is the circular tube leading inside up to the eardrum or tympanic membrane.
4. The tympanic membrane or eardrum is formed of connective tissues having outer skin cover and inner mucus membrane.
5. The middle ear consists of chain of three ossicles called malleus, incus and stapes, The malleus is attached to the tympanic membrane and the stapes is connected to the oval window of the internal ear. They help in the transmission of sound waves from external auditory canal to internal ear.
6. Connecting middle ear with the pharynx is eustachian tube which helps in equalizing the air pressure on either side of the tympanic membrane.
7. The internal ear is fluid filled structure called labyrinth. It has two parts, bony and the membranous labyrinth.
8. The outer bony labyrinth is formed by the series of channels in which the membranous abyrinth containing endolymph fluid is present.
9. The membranes consist of coiled cochlea, the reissner’s membrane and basilar membranes. These membranes divide the surrounding perilymph filled bony labyrinth into an upper scala vestibule and a lower scala tympani.
10. The space within cochlea which is known as scala media is filled with endolymph. The scala vestibule ends at the oval window at the base of cochlea.
11. The scala tympani terminates at the round window which opens to the middle ear. The organ of corti is located on the basilar membrane. It contains the hair cells which act as auditory receptors.
12. The hair cells are columnar cells present in rows. The basal ends of the hair cells are in close contact with the afferent nerve fibres while their apical end contains numerous cilia. A thin elastic membrane projects above the rows of the hair cells called tectorial membrane.
13. Above the cochlea, the internal ear also contains vestibular apparatus. It consists of three semicircular canals and the otolith organ formed of the sacculus and utriculus. The semicircular canals lie in different plane at right angles to each other and are suspended in the perilymph.
14. The bases of canals are swollen and are called ampullae, which contain a projecting ridge known as crista ampullaris which contain hair cells.
15. The sacculus and utriculus also have projecting ridge called macula. The crista and macula are the specific receptors of vestibular apparatus. They are responsible for maintenance of body posture and the balance.
    

Question 12.
Write an account of position and structure of pituitary gland.
Answer:
Pituitary gland (Hypophysis):
I. Position : Pituitary or Hypophysis is located on the ventral side of brain below the hypothalamus. Infundibulum or hypophyseal stalk attaches pituitary to hypothalamus just behind the optic chiasma. It is well protected in sella turcica which is a depression of the sphenoid bone of the skull.

II. Morphological structure of pituitary- gland: The pituitary gland shows two distinct regions : Anterior lobe or adenohypophysis and posteriorlobe or neuro-hypophysis.
(1) Adenohypophysis or Anterior lobe : It is the largest lobe of the gland and forms about 75% of pituitary gland. It develops as an outgrowth called Rathke’s pouch from the roof of embryonic buccal cavity. It has three divisions, viz. pars tuberalis, pars distalis and pars intermedia.

(i) Pars tuberalis : Tubular region present below the hypothalamus is known as pars tuberalis. It is like a collar around the infundibulum. It is non-secretory in nature.
(ii) Pars distalis : The largest anterior region which is secretory in nature is called pars distalis. It is made up of loose cords of epitheloid secretory cells which are separated by reticular connective tissue containing blood sinusoids. It is connected to the hypothalamus by portal system formed by blood sinusoids.
(iii) Pars intermedia : The narrow cleft between the pars distalis and pars nervosa of neuro – hypophysis is called the intermediate part or pars intermedia. It is reduced, less developed and non-functional in human being.

(2) Neuro-hypophysis or Posterior lobe : The posterior lobe of the pituitary which is attached to hypothalamus by infundibular stalk is called neuro-hypophysis. It is smaller and constitute 25% of pituitary. It has the following three parts:

1. Median eminence : The swollen median part of the hypothalamus where infundibulum gets attached is called median eminence.
2. Infundibulum : Infundibulum is the hypophyseal stalk that helps in attachment of pituitary gland to the hypothalamus. It contains mainly the axonic fibres of neurosecretory cells present in hypothalamus. It forms the major connection of hypothalamo-hypophysis axis.
3. Pars nervosa : The lowermost, larger region of neuro-hypophysis that contains axons is called pars nervosa. It acts as a neurohaerhal organ and contains specialized cells called pituicytes.

Question 13.
In a person, Pars distalis part of the Pituitary gland is not producing hormones in sufficient quantity. Explain the effects it will produce with respect to the different hormones.
OR
Give names and functions of hormones secreted by adenohypophysis.
Answer:
Pars distalis of the pituitary gland produces following hormones:
GH, ACTH, TSH, FSH, LH/ICSH, LTH and MSH. If these hormones are produced in deficient quantities, following disorders are seen.
1. GH:
(a) Hyposecretion of GH in childhood leads to dwarfism. Frohlich dwarf or Lorain dwarf may be produced based on mental capacity. Hyposecretion in adulthood causes Simmonds’s disease.
(b) Hypersecretion of GH in childhood causes gigantism and in adulthood it causes acromegaly.

2. TSH :
(a) Hyposecretion of TSH leads to thyroid atrophy.
(b) Hypersecretion of TSH causes excessive functioning of thyroid gland.

3. ACTH :
(a) Hyposecretion of ACTH causes Addisons’ disease, in which adrenal gland shows failure of functions.
(b) Hypersecretion of ACTH causes Cushing’s disease, in which the adrenal cortex undergoes excessive growth.

4. FSH : Hyposecretion of FSH leads to infertility in both the sexes. Hypersecretion of FSH in females cause disturbances in menstrual cycle.

5. LH/ICSH : Hyposecretion of LH in females will cause lack of ovulation. Hyposecretion of ICSH in males cause reduction in masculinity. Sperm production may be affected. Hypersecretion of LH/ICSH can cause disturbances in reproductive cycles.

6. LTH : Corpus luteum is not maintained due to lesser amount of LTH. Lactogenesis will also hamper if there is hyposecretion of LTH.

Question 14.
Describe the hormones of neuro¬hypophysis.
Answer:
Hormones of neuro-hypophysis : Neuro-hypophysis does not secrete any hormone itself but stores the hormones which are secreted by hypothalamic neurons. It stores and releases the following hormones, viz. ADH, Oxytocin and coherin.
1. Anti Diuretic Hormone (ADH) or Vasopressin:

• ADH brings about anti-diuretic action and also increases blood pressure.
• It is a regulatory hormone which plays a major role in osmoregulation.
• It increases the permeability of distal convoluted tubule or collecting tubules of uriniferous tubules of kidney.
• Higher ADH levels decrease the urine output and helps for water conservation. It helps in the absorption of water from the ultrafiltrate thus regulates the water balance of body fluids.
• ADH also controls constriction of arterioles and increases blood pressure in kidney which facilitates ultra filtration. Therefore it is also called vasopressin.
• ADH is regulated by increase or decrease of osmotic pressure of blood in a feedback manner.
• The osmotic pressure is detected by osmoreceptors in the hypothalamus.

2. Oxytocin (Birth hormone):

• Oxytocin helps in parturition.
• It is a powerful stimulant of contraction of uterine myometrium at the end of gestation due to which the labour is initiated.
• It also stimulates myoepithelial cells of mammary glands for milk ejection during lactation.
• It also helps in fertilization by powerful contractions of the uterine musculature to drive the sperms upward towards fallopian tubes.
• Oxytocin also excites musculature of gallbladder, ureters, urinary bladder, intestine, etc. for proper functioning of these organs.

3. Coherin : Coherin induces prolonged, rhythmic integrated contractions of the jejunum.

Question 15.
Describe the morphology of thyroid gland.
Or
With the help of a suitable diagram describe the structure of thyroid gland.
Answer:
Morphology of thyroid gland:

1. Thyroid is the largest endocrine gland in the body.
2. It weighs about 25 to 30 g and measures about 5 cm in length and 3 cm in width.
3. It is located in the neck region anteriorly just below the larynx and situated ventrolaterally to the trachea.
4. The thyroid is derived from the endoderm of the embryo.
5. The thyroid can vary in size as per age, sex and diet.
6. It is reddish brown, bilobed and highly vascular
7. The two lobes are joined by connective tissue called isthmus which is located at 2nd to 4th tracheal cartilage.
8. Therefore, the right and left lobe of thyroid are seen on both sides of the trachea.
9. The gland is H-shaped having butterfly-like appearance.
10. The structural and functional units of thyroid gland are the thyroid follicles.
11. From the outer surface there lies a connective tissue capsule. A number of septa arise from the connective tissue which are called trabeculae. They divide the gland into lobules. Each lobule has follicles which store hormone. The number of follicles are about three million.

Question 16.
Describe neurohormonal regulation of pituitary and thyroid gland.
Answer:
Pituitary gland is directly under the influence of neurohormones of hypothalamus while thyroid is indirectly influenced.
I. Neurohormonal regulation of pituitary:

1. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
2. Hypothalamus forms the hypothalamohypophysial axis through which transportation of neuro-hormones take place.
3. Pituitary secretes a variety of hormones which influence other endocrine glands of the body. E.g. GH, PRL, TSH, ACTH, Gonadotropins

II. Neurohormonal regulation of thyroid :

1. Hypothalamus secretes TRF [Thyrotropin releasing factor] which influences the anterior pituitary to release TSH.
2. TSH in turn stimulates thyroid follicles to produce and release two thyroid hormones – T₃ and T₄. (Thyroxin)
3. Increase in T₃ and T₄ triggers negative feedback mechanism that stops the secretion of TRF.

(4) As the pituitary does not get the signal in the form of TRF TSH secretion is stopped.

Question 17.
Name the hormones secreted by the adrenal cortex and state their role.
Answer:
Adrenal cortex secretes 3 types of corticoids – mineralocorticoids, glucocorticoids and sex corticoids.
I. Mineralocorticoids:

1. The mineralocorticoids regulate ionic and osmotic balance, by regulating the amounts of electrolyte and water.
2. Aldosterone is the main mineralocorticoid that acts on the renal tubules.
3. Aldosterone stimulates the re-absorption of Na+ and water and excretion of K⁺ and phosphate ions.
4. The aldosterone helps in the maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure.

II. Glucocorticoids:

1. Cortisol is the main glucocorticoid. Cortisol stimulates many metabolic reactions such as gluconeogenesis, lipolysis and proteinolysis.
2. It inhibits cellular uptake and utilization of amino acids.
3. Cortisol also plays an important role in maintaining the cardiovascular system and kidney functions.
4. It is also involved in anti-inflammatory reactions and suppresses the immune response.
5. Cortisol stimulates the RBC production.

III. Sex corticoids (Gonadocorticoids).

1. Sex corticoids, Androgens and estradiols are produced by the adrenal cortex.
2. In males, they have a role in development and maintenance of external sex characters.
3. Excess sex corticoids in female causes adrenal virilism and hirsutism (excess hair on face)
4. Excess sex corticoids in males causes gynaecomastia i.e. enlarged breast.

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 5 Gravitation Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 5 Gravitation

Question 1.
Mention some main characteristics of gravitational force.
Answer:
Characteristics of gravitational force:

1. Every massive object in the universe experiences gravitational force.
2. It is the force of mutual attraction between any two objects by virtue of their masses.
3. It is always an attractive force with infinite range.
4. It does not depend upon the intervening medium.
5. It is much weaker than other fundamental forces. Gravitational force is 10 times weaker than strong nuclear force.

Question 2.
State and explain Kepler’s law of orbits.
Answer:
Statement:
All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.
Explanation:

1. The figure M shows the orbit of a around the planet P Sun.
   
2. Here, S and S’ are the foci of the ellipse and the Sun is situated at S.
3. P is the closest point along the orbit from S and is called perihelion.
4. A is the farthest point from S and is called aphelion.
5. PA is the major axis whose length is 2a. PO and AO are the semimajor axes with lengths ‘a’ each.
   MN is the minor axis whose length is 2b. MO and ON are the semiminor axes with lengths ‘b’ each.

Question 3.
State and prove Kepler’s law of equal areas.
Answer:
Statement:

The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
Explanation:
i)Kepler observed that planets move faster when they are nearer to the Sun while they move slower when they are farther from the Sun.

ii) Suppose the Sun is at the origin. The position of planet is denoted by \(\vec{r}\) and its momentum is denoted by \(\vec{p}\) (component ⊥ \(\vec{r}\)).

vi) For central force the angular momentum is conserved. Hence, from equations (4) and (5),
\(\frac{\overrightarrow{\Delta \mathrm{A}}}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{L}}}{2 \mathrm{~m}}\) = constant
This proves the law of areas.

Question 4.
What is a central force?
Answer:
A central force on an object is a force which is always directed along the line joining the position of object and a fired point usually taken to the origin of the coordinate system.

Question 5.
State and explain Kepler’s law of periods.
Answer:
Statement:
The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.
Explanation:
If r is length of semi major axis then, this law states that.
T² × r³ or \(\frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}\) = constant

Solved Examples

Question 6.
What would be the average duration of year if the distance between the Sun and the Earth becomes
i) thrice the present distance.
ii) twice the present distance.
Solution:
i) Consider r₁ be the present distance between the Earth and Sun
We know, T₁ = 365 days.
When the distance is made thrice, r₂ = 3r₁
According to Kepler’s law of period,

i) The duration of the year would be 1897 days when distance is made thrice.
ii) The duration of the year would be 1032 days when distance is made twice.

Question 7.
What would have been the duration of the year if the distance between the Earth and the Sun were half the present distance?
Solution:
Given: r₂ = \(\frac{\mathrm{r}_{1}}{2}\)
∴ \(\frac{r_{2}}{r_{1}}=\frac{1}{2}\)
To find. Time period (T₂)
Formula: \(\left(\frac{T_{2}}{T_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{3}\)
Calculation: We know. T₁ = 365 days
From formula,

= 365 × 0.3536
= 129.1
T₂ = 129.1 days
The duration of the year would be 129.1 days.

Question 8.
Calculate the period of revolution of Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of the Earth’s orbit is 5.
(Period of revolution of the Earth is 1 year.)
Solution:
Given: \(\frac{\mathrm{r}_{\mathrm{J}}}{\mathrm{r}_{\mathrm{E}}}=\frac{5}{\mathrm{l}}\), T_E = 1 year
To find: Period of revolution of Jupiter (T_J)
Formula: T² ∝ r³
Calculation: From formula,
\(\left(\frac{T_{J}}{T_{E}}\right)^{2}=\left(\frac{r_{J}}{r_{E}}\right)^{3}\)
∴ T_J = 5^3/2
= 5 x \(\sqrt {5}\)
= 11.18 years.
Period of revolution of Jupiter around the Sun is 11.18 years.

Question 9.
A Saturn sear is 29.5 times the Earth’s year. How far is the Saturn from the Sun if the Earth is 1.50 × 10⁸ km away from the Sun?
Solution:
Given: T_S = 29.5 T_E,
r_E = 1.50 × 10⁸ km
To find: Distance of Saturn form the Sun (r_S)
Formula: T² ∝ r³
Calculation: From formula,

∴ r_S = 14.32 × 10⁸ km
Saturn is 14.32 × 10⁸ km away from the Sun.

Question 10.
The distances of two planets from the Sun are 10¹³ m and 10¹² m respectively. Find the ratio of time periods of the two planets.
Solution:
Given: r₁ = 10¹³ m, r₂ = 10¹² m

Question 11.
Heavy and light objects are released from same height near the Earth’s surface. What can we conclude about their acceleration?
Answer:
Heavy and light objects, when released from the same height, fall towards the Earth at the same speed., i.e., they have the same acceleration.

Question 12.
Explain how Newton concluded that gravitational force F ∝ = \(\frac{\mathrm{Mm}}{\mathrm{r}^{2}}\)
Answer:
Before generalising and stating universal law of gravitation, Newton first studied the motion of moon around the Earth.

1. The known facts about the moon were,
   • the time period of revolution of moon around the Earth (T) = 27.3 days.
   • distance between the Earth and the moon (r) = 3.85 × 10⁵ km.
   • the moon revolves around the Earth in almost circular orbit with constant angular velocity ω.
2. Thus, the centripetal force experienced by moon (directed towards the centre of the Earth) is given by,
   F = mrω² …………… (1)
   Where, m = mass of the moon
3. From Newton’s laws of motion,
   F = ma
   ∴ a = rω² ……………… (2)
4. 
5. This is the acceleration of the moon directed towards the centre of the Earth.
6. This acceleration is much smaller than the acceleration felt by bodies near the surface of the Earth (while falling on Earth).
7. The value of acceleration due to Earth’s gravity at the surface is 9.8 m/s².
   
8. Newton therefore concluded that the acceleration of an object towards the Earth is inversely proportional to the square of distance of object from the centre of the Earth.
   ∴ a ∝ \(\frac{1}{r^{2}}\)
   x. As, F = ma
   Therefore, the force exerted by the Earth on an object of mass m at a distance r from it is
   F ∝ \(\frac{\mathrm{m}}{\mathrm{r}^{2}}\)
   Similarly, an object also exerts a force on the Earth which is
   F_E ∝ \(\frac{\mathrm{M}}{\mathrm{r}^{2}}\)
   Where M is the mass of the Earth. .
9. According to Newton’s third law of motion, F = F_E. Thus, F is also proportional to the mass of the Earth. From these observations, Newton concluded that the gravitational force between the Earth and an object of mass m is F ∝ \(\frac{\mathrm{Mm}}{\mathrm{r}^{2}}\)

Question 13.
Discuss the vector form of gravitational force between two masses with the help of diagram.
Answer:

1. Consider two point masses m₁ and m₂ having position vectors \(\overrightarrow{\mathrm{r}_{1}}\) and \(\overrightarrow{\mathrm{r}_{2}}\) from origin O respectively as shown in figure.
   
2. The position vector of mass m₂ with respect to m₁ is given by, \(\vec{r}_{2}-\vec{r}_{1}=\vec{r}_{21}\)
3. Similarly, position vector of mass m₁ with respect to m₂ is, \(\overrightarrow{\mathrm{r}}_{1}-\overrightarrow{\mathrm{r}_{2}}=\overrightarrow{\mathrm{r}_{12}}\)
4. Let \(\left|\overrightarrow{\mathrm{r}_{12}}\right|=\left|\overrightarrow{\mathrm{r}_{21}}\right|\) Then, the force acting on mass m₂ due to mass m₁ will be given as,
   \(\overrightarrow{\mathrm{F}_{21}}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\left(-\hat{\mathrm{r}}_{21}\right)\)
   where, \(\hat{\mathbf{r}}_{21}\) is the unit vector from m₁ to m₂.
   The force \(\overrightarrow{\mathrm{F}_{21}}\) is directed from m₂ to m₁.
5. Similarly, force experienced by m₁ due to m₂ is given as, \(\overrightarrow{\mathrm{F}}_{12}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\left(-\hat{\mathrm{r}}_{12}\right)\)
   ∴ \(\overrightarrow{\mathrm{F}}_{12}=-\overrightarrow{\mathrm{F}}_{21}\)
   [Note: As \(\hat{\mathbf{r}}_{21}\) is defined as unit vector from m₁ to m₂, conceptually force \(\overrightarrow{\mathrm{F}}_{21}\) is directed from m₂ lo m₁.]

Question 14.
Why Is the law of gravitation known as universal law of gravitation?
Answer:
The law of gravitation is applicable to all material objects in the universe. Hence it is known as the universal law of gravitation.

Question 15.
Give formula for the gravitational force due to a collection of masses and represent it diagrammatically.
Answer:

For a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by all the other point masses.
For n particles, force on i^th mass \(\overrightarrow{\mathrm{F}}_{\mathrm{i}}=\sum_{\mathrm{j}=1 \atop \mathrm{j} \neq \mathrm{i}}^{n} \overrightarrow{\mathrm{F}}_{\mathrm{ij}}\)
where, \(\vec{F}_{\mathrm{ij}}\) is the force on i^th particle due to j^th particle.

Question 16.
Discuss qualitative idea for the gravitational force of attraction due to a hollow, thin spherical shell of uniform density on a point mass situated inside it.
Answer:

1. Let us consider the case when the point mass A, is at the centre of the hollow thin shell.
2. In this case as every point on the shell is equidistant from A, all points exert force of equal magnitude on A but the directions of these forces are different.
3. Consider the forces on A due to two diametrically opposite points on the shell.
4. The forces on A due to them will be of equal magnitude but will be in opposite directions and will cancel each other.
5. Thus, forces due to all pairs of points diametrically opposite to each other will cancel and there will be no net force on A due to the shell.
6. When the point object is situated elsewhere inside the shell, the situation is not symmetric. However, gravitational force varies directly with mass and inversely with square of the distance.
7. When the point object is situated elsewhere inside the shell, the situation is not symmetric. However, gravitational force varies directly with mass and inversely with square of the distance.
8. Thus, some part of the shell may be closer to point A, but its mass is less. Remaining part will then have larger mass but its centre of mass is away from A.
9. In this way, mathematically it can be shown that the net gravitational force on A is still zero, so long as it is inside the shell.
10. Hence, the gravitational force at any point inside any hollow closed object of any shape is zero.

Question 17.
Discuss qualitative idea for the gravitational force of attraction between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside the shell / sphere.
Answer:

• Gravitational force caused by different regions of shell can be resolved into components along the line joining the point mass to the centre and along a direction perpendicular to this line.
• The components perpendicular to this line cancel each other and the resultant force remains along the line joining the point to the centre.
• Mathematical calculations show that, this resultant force on this point equals the force that would get exerted by the shell whose entire mass is situated at its centre.

Solved Problems

Question 18.
The gravitational force between two bodies is 1 N. If distance between them is doubled, what will be the gravitational force between them?
Solution:
Let m₁ and m₂ be masses of the given two bodies. If they are r distance apart initially, then the force between them will be,

The force between two bodies reduces to 0.25 N.

Question 19.
Calculate the force of attraction between two metal spheres each of mass 90 kg, if the distance between their centres is 40 cm. (G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: m₁ = 90 kg, m₂ = 90 kg,
r = 40 cm = 40 × 10^-2 m.
G = 6.67 × 10^-11 N m²/kg²
To find: Force of attraction (F)
Formula: F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
Calculation.
From formula,
F = \(\frac{6.67 \times 10^{-11} \times 90 \times 90}{\left(40 \times 10^{-2}\right)^{2}}\)
= \(\frac{6.67 \times 8100}{1600} \times 10^{-7}\)
∴ F = 3.377 × 10^-6 N
The force of attraction between the two metal spheres is 3.377 × 10^-6 N.

Question 20.
Three particles A, B, and C each having mass m are kept along a straight line with AB = BC = 1. A fourth particle D is kept on the perpendicular bisector of AC at a distance ¡ from B. Determine the gravitational force on D.
Solution:

From figure,
distance of particle D, from particles A and C is.
ADCD = \(\sqrt{\mathrm{AB}^{2}+\mathrm{BD}^{2}}\)
= \(\sqrt{(l)^{2}+(l)^{2}}=l \sqrt{2}\)
Gravitational force on particle D is the vector sum of forces due to particles A, B, and C. Gravitational force due to A,
F_A = \(\frac{\mathrm{Gmm}}{(l \sqrt{2})^{2}}=\frac{\mathrm{Gm}^{2}}{2 l^{2}}\) = along \(\overrightarrow{\mathrm{DA}}\)
This force can he resolved into horizontal and vertical components using rectangular unit vectors as shown in figure.
From figure,


Negative sign indicates net force is acting along DB.
The net force acting on particle D is \(\frac{\mathbf{G m}^{2}}{l^{2}}\left(\frac{1}{\sqrt{2}}+1\right)\) directed along \(\overrightarrow{\mathbf{D B}}\).

[Note: When force, in given case is resolved into its components, its horizontal component contains cos argument while vertical component contains sine argument.]

Question 21.
Three balls of masses 5 kg each are kept at points P(1, 2, 0) Q(2, 3, 0) and R(2, 2, 0). Find the gravitational force exerted on the ball at point R.
Solution:
The net force acting on ball placed at R will be vector sum of forces acting due to balls at P and Q.

= 2.358 × 10^-9 N
The net force acting on the ball at point R will be 2.358 × 10^-9 N.

Question 22.
For what purpose Cavendish balance is used?
Answer:
Cavendish balance is used to find the magnitude of the gravitational constant G.

Question 23.
Describe the construction of Cavendish balance with the help of neat labelled diagram.
Answer:

1. The Cavendish balance consists of a light rigid rod. It is supported at the centre by a fine vertical metallic fibre about 100 cm long.
2. Two small spheres, s₁ and s₂ of lead having equal mass m and diameter about 5 cm are mounted at the ends of the rod and a small mirror M is fastened to the metallic fibre as shown in figure.
   
3. The mirror can be used to reflect a beam of light onto a scale and thereby measure the angel through which the wire will be twisted.
4. Two large lead spheres L₁ and L₂ of equal mass M and diameter of about 20 cm are kept close to the small spheres on opposite side as shown in figure.

Question 24.
Describe the working of the experiment performed to measure the value of gravitational constant.
Answer:

1. In the experiment performed to find the magnitude of gravitational constant (G), Cavendish balance is used.
2. The large spheres in the balance attract the nearby smaller spheres by equal and opposite force \(\overrightarrow{\mathrm{F}}\). Hence, a torque is generated without exerting any net force on the bar.
3. Due to the torque the bar turns and the suspension wire gets twisted till the restoring torque due to the elastic property of the wire becomes equal to the gravitational torque.
4. If r is the initial distance of separation between the centres of the large and the neighbouring small sphere, then the magnitude of the force between them is, F = G\(\frac{\mathrm{mM}}{\mathrm{r}^{2}}\)
5. If length of the rod is L, then the magnitude of the torque arising out of these forces is
   τ = FL = G\(\frac{\mathrm{mM}}{\mathrm{r}^{2}}\)L
6. At equilibrium, it is equal and opposite to the restoring torque.
   ∴ G\(\frac{\mathrm{mM}}{\mathrm{r}^{2}}\)L = Kθ
   Where, K is the restoring torque per unit angle and θ is the angle of twist.
7. By knowing the values of torque τ₁ it and corresponding angle of twist a, the restoring torque per unit twist can be determined as K = τ₁/α.
8. Thus, in actual experiment measuring θ and knowing values of τ, m, M and r, the value of G can be calculated from equation (2).

Question 25.
Derive the expression for the acceleration due to gravity on the surface of the Earth.
Answer:

1. The Earth is an extended object and can be assumed to be a uniform sphere.
2. If the mass of the Earth is M and that of any point object is m, the distance of the point object from the centre of the Earth is r then the force of attraction between them is given by,
   F = \(\text { G } \frac{M m}{r^{2}}\) …. (1)
3. If the point object is not acted upon by any other force, it will be accelerated towards the centre of the Earth under the action of this force. Its acceleration can be calculated by using Newton’s second law,
   F = ma … (2)
4. From equations (1) and (2),
   ma = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
   ∴ Acceleration due to the gravity of the Earth
   \(=\frac{\mathrm{GMm}}{\mathrm{r}^{2}} \times \frac{1}{\mathrm{~m}}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}\)
   This is denoted by g.
5. If the object is close to the surface of the Earth, r ≈ R, then,
   g_{Earth’s surface} = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 26.
Explain why the Earth doesn’t appear to move even though the object of mass m (m << M) kept on the Earth exerts equal and opposite gravitational force on it.
Answer:

1. An object of mass m (much smaller than the mass of the Earth) is attracted towards the Earth and falls on it.
2. At the same time, the Earth is also attracted by the equal and opposite force towards the mass m.
3. However, its acceleration towards m will be,
   
4. As m << M, a_Earth << g and is nearly zero. As, a result, practically only the mass m moves towards the Earth and the Earth doesn’t appear to move.

Solved Examples

Question 27.
Calculate mass of the Earth from given data, Acceleration due to gravity g = 9.81 m/s², Radius of the Earth R_E = 6.37 × 10⁶ m, G = 6.67 × 10^-11 N m²/kg²
Solution:
Given: g = 9.81 m/s², R_E = 6.37 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²
To find: Mass of the Earth (M_E)
Formula: g = \(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^{2}}\)
Calculation: From formula,

= antilog {0.9912 + 0.8041 + 0.8041 – 0.8241) × 10²³
= antilog {1.7753} × 10²³
= 59.61 × 10²³
= 5.961 × 10²⁴ kg
Mass of the Earth is 5.961 × 10²⁴ kg.

Question 28.
Calculate the acceleration due to gravity at the surface of the Earth from the given data. (Mass of the Earth = 6 × 10²⁴ kg, Radius of the Earth = 6.4 × 10⁶ m, G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given. M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10^-11 N m²/kg²
Tofind. Acceleration due to gravity (g)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,

The acceleration due to gravity at the surface of the Earth is 9.77 m/s².

Question 29.
Calculate the acceleration due to gravity on the surface of moon if mass of the moon is 1/80 times that of the Earth and diameter of the moon is 1/4 times that of the Earth (g = 9.8 m/s²)
Solution:
Given: M_m = \(\frac{\mathrm{M}_{\mathrm{E}}}{80}\), R_m = \(\frac{\mathrm{R}_{\mathrm{E}}}{4}\), g = 9.8 m/s²
To find: Acceleration due to gravity on the surface of moon (g_m)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: For moon, from formula,

Acceleration due to gravity on surface of the planet is 0.245 m/s².

Question 30.
Find the acceleration due to gravity on a planet that is 10 times as massive as the Earth and with radius 20 times of the radius of the Earth (g = 9.8 m/s²).
Solution:
Given: M_P = 10 × Mass of the Earth = 10 M_E,
R_P = 20 × radius of the Earth = 20 R_E, g = 9.8 m/s²
To find: Acceleration due to gravity on surface of the planet (g_P)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,

Acceleration due to gravity on surface of the planet is 0.245 m/s².

Question 31.
Acceleration due to gravity on the Earth is g. A planet has mass and radius half that of the Earth. How much will be percentage change in the acceleration due to gravity on the planet?
Solution:

The percentage change in acceleration due to gravity between the planet and the Earth will be 100%.

Question 32.
Explain the graph showing variation of acceleration due to gravity with altitude and depth.
Answer:
The value of acceleration due to gravity is calculated to be maximum at the surface of the Earth. The value goes on decreasing with
i) increase in depth below the Earth’s surface. [varies linearly with (R – d) = r]

ii) increase in height above the Earth’s surface. [varies inversely with (R + h)² = r²].

Graph of g, as a function of r, the distance from the centre of the Earth, is plotted as shown in figure.
For r< R,
g_d = g\(\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
if r = R – d,
g(r) = g\(\left(\frac{r}{R}\right) \Rightarrow g(r) \propto r\)
Hence, the graph shows a straight line passing through origin and having slope \(\frac{\mathrm{g}}{\mathrm{R}}\).

which is represented in the graph.

Question 33.
Why does the weight of a body of a finite mass m is zero at the centre of the Earth?
Answer:
Acceleration at depth d due to gravity is given
by,
g_d = g\(\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)\)
As, at centre of the Earth d = R ⇒ g_d = 0.
Hence, the weight of a body of a finite mass m is zero at the centre of the Earth.

Question 34.
Discuss the variation of acceleration due to gravity at poles and equator due to latitude of the Earth.
Answer:

1. Effective acceleration due to gravity at P is given as,
   g’ = g – Rω²cos²θ.
2. As the value of θ increases, cos θ decreases. Therefore g’ will increase as we move away from equator towards any pole due to the rotation of the Earth.
3. At equator θ = 0°
   ∴ cos θ = 1
   ∴ g’_e = g – Rω²
   The effective acceleration due to gravity (g’_e) is minimum at equator, as here it is reduced by Rω²
4. At poles θ = 90° cos θ = 0
   ∴ g’_p = g – Rω² cos θ
   = g – 0
   = g
   There is no reduction in acceleration due to gravity at poles, due to the rotation of the Earth as the poles are lying on the axis of rotation and do not revolve.

Question 35.
If g = 9.8 m/s² on the surface of the Earth, find its value at h = \(\frac{\mathbf{R}}{\mathbf{2}}\) from the surface of the Earth.
Solution:
Given: g = 9.8 m/s², h = \(\frac{\mathrm{R}}{2}\)
To find: Acceleration due to gravity (g_h)
Formula: \(\frac{\mathrm{g}_{\mathrm{h}}}{\mathrm{g}}\) = \(\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}\)
Calculation:
From formula,

At h = \(\frac{\mathrm{R}}{2}\) from the surface of the Earth, the value of g is 4.35 m/s².

Question 36.
At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface? (Radius of Earth = 6400 km)
Solution:
Given: g_h = 90% of g i.e., \(\frac{g_{h}}{g}\) = 0.9,
R = 6400 km = 6.4 × 10⁶ m
To find: Distance above the surface of the Earth (h)

Question 37.
Find the altitude at which the acceleration due to gravity is 25% of that at the surface of the Earth.
(Radius of the Earth = 6400 km)
Solution:
Given: g_h = 25% of g = \(\frac{25}{100} \times \mathrm{g}=\frac{\mathrm{g}}{4}\), R = 6400 km = 6.4 × 10⁶ m
To find: Height (h)
Formula: g_h = g\(\left(\frac{R}{R+h}\right)^{2}\)
Calculation: From formula,
\(\frac{\mathrm{g}}{4}\) = g\(\left(\frac{R}{R+h}\right)^{2}\)
(R + h)² = 4R²
R + h = 2R
∴ h = 2R – R
∴ h = R
∴ h = 6400 km

Question 38.
A hole is drilled half way to the centre of the Earth. A body is dropped into the hole. How much will it weigh at the bottom of the hole if the weight of the body on the Earth’s surface is 350 N?
Solution:
Given: W = mg = 350 N, d = \(\frac{\mathrm{R}}{2}\)
To find: Weight at certain depth (W_d)
Formula: g_d = \(\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]\)
Calculation: Since W_d = mg_d,
from formula,

Question 39.
Assuming the Earth to be a homogeneous sphere, determine the density of the Earth from following data. (g = 9.8 m/s², G = 6.673 × 10^-11 N m²/kg², R = 6400 km)
Solution:
Given g = 9.8 m/s²,
G = 6.673 × 10^-11 N m²/kg², R = 6400 km = 6.4 × 10⁶ m
To find: Density (ρ)
Formula: g = \(\frac{4}{3} \pi \mathrm{R} \rho \mathrm{G}\)
Calculation: From formula,

Question 40.
If the Earth were a perfect sphere of radius 6.4 × 10⁶ m rotating about its axis with the period of one day (8.64 × 10⁴ s), what is the difference in acceleration due to gravity from poles to equator?
Solution:
Given: R = 6.4 × 10⁶ m, T = 8.64 × 10⁴ s
To find: Difference in acceleration due to gravity (g_P – g_E)
Formula: g’ = g – Rω² cos²θ
Calculation: Since ω = \(\frac{2 \pi}{\mathrm{T}}\)
∴ ω = \(\frac{2 \times 3.14}{8.64 \times 10^{4}}\) = \(\frac{6.28}{8.64 \times 10^{4}}\)
= 0.7268 × 10^-4
ω = 7.268 × 10^-5 rad/s
At poles, θ = 90°
From formula,
g_P = g – Rω²cos² (90°)
= g – 0 ….(∵ cos 90° = 0)
∴ g_P = g …. (i)
At equator, θ = 0°,
∴ g_E = g – Rω²cos² 0°
g_E = g – Rω² …. (ii)
Subtracting equation (ii) from equation (i), we have,
g_P – g_E = g – (g – Rω²)
∴ g_P – g_E = Rω²
= 6.4 × 10⁶ × (7.268 × 10^-5)²
= 6.4 × 10⁶ × 52.82 × 10^-10
= 338 × 10^-4
∴ g_P – g_E = 3.38 × 10^-2 m/s²

Question 41.
The Earth is rotating with angular velocity ω about its own axis. R is the radius of the Earth. If Rω² = 0.03386 m/s², calculate the weight of a body of mass 100 gram at latitude 25°. (g = 9.8 m/s²)
Solution:
Given: Rω² = 0.03386 m/s², θ = 25°,
m = 0.1 kg, g = 9.8 m/s²
To find: Weight (W)

Formulae:
i) g’ = g – Rω² cos² θ
ii) W = mg

Calculation:
From formula (i),
g’ = 9.8 – [0.03386 – cos² (25°)]
∴ g’ = 9.8 – [0.03386 × (0.9063)²]
∴ g’ = 9.8 – 0.02781
∴ g’ = 9.772 m/s²
From formula (ii),
W = 0.1 × 9.772
∴ W = 0.9772 N

Question 42.
If the angular speed of the Earth is 7.26 × 10^-5 rad/s and radius of the Earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.
Solution:
Given: R = 6.4 × 10⁶ m, ω = 7.26 × 10^-5 rad/s
To find: Change in weight (∆W)
Formulae:
i) ∆g = g_p – g_eq = Rω²
ii) ∆W = m∆g

Calculation: From formula (i) and (ii),
∆W = m(Rω²)
= 1 × 6.4 × 10⁶ × (7.26 × 10^-5)²
= 3373 × 10^-5 N

Question 43.
Define potential energy.
Answer:
Potential energy is the work done against conservative force (or forces) in achieving a certain position or configuration of a given system.

Question 44.
Explain with examples the universal law which states that “Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy”.
Answer:
Example 1:

1. A spring in its natural state, possesses minimum potential energy. Whenever we stretch it or compress it, we perform work against the conservative force.
2. Due to this work, the relative distances between the particles of the system change i.e., configuration changes and potential energy of the spring increases.
3. The spring finally regains its original configuration of minimum potential energy on removal of the applied force.
   This explains that the spring always try to rearrange itself in order to attain minimum potential energy.

Example 2:

1. When an object is lying on the surface of the Earth, the system of that object and the Earth has minimum potential energy.
2. This is the gravitational potential energy of the system as these two are bound by the gravitational force. While lifting the object to some height, we do work against the conservative gravitational force.
3. In its new position, the object is at rest due to balanced forces. However, now, the object has a capacity to acquire kinetic energy, when allowed to fall.
4. This increase in the capacity is the potential energy gained by the system. The object falls on the Earth to achieve the configuration of minimum potential energy on dropping it from the new position.

Question 45.
Obtain an expression for change in gravitational potential energy of any object displaced from one point to another.
Answer:
i) Work done against gravitational force in displacing an object through a small displacement, stored in the system in the form of increased potential energy of the system.
∴ dU = –\(\overrightarrow{\mathrm{F}}_{\mathrm{g}} \cdot \overrightarrow{\mathrm{dr}}\)
Negative sign appears because dU is the work done against the gravitational force \(\overrightarrow{\mathrm{F}_{\mathrm{g}}}\).

ii) For displacement of the object from an initial position \(\overrightarrow{\mathrm{r}_{\mathrm{i}}}\) to the final position \(\overrightarrow{\mathrm{r}_{\mathrm{f}}}\), the change in potential energy ∆U, can be obtained by integrating dU.

 

iii) Gravitational force of the Earth, \(\overrightarrow{\mathrm{F}}_{\mathrm{g}}\) = –\(\frac{\mathrm{GMm}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\) where \(\hat{r}\) is the unit vector in the direction of \(\overrightarrow{\mathrm{r}}\).
Negative sign appears here because \(\overrightarrow{\mathrm{F}_{\mathrm{g}}}\) is directed towards centre of the Earth and opposite to \(\overrightarrow{\mathrm{r}}\).

iv) For Earth and mass system,

Hence, change in potential energy corresponds to the work done against conservative forces.

Question 46.
Using expression for change in potential energy, show that gravitational potential energy of the system of object of mass m and the Earth with separation of r is, –\(\frac{\text { GMm }}{\text { r }}\)
Answer:

1. Change in P.E. for a system of Earth and mass is given by,
   ∆U = GMm\(\left(\frac{1}{r_{i}}-\frac{1}{r_{f}}\right)\)
2. For gravitational force, point of zero potential energy is taken to be at r = ∞.
3. Hence, U(r_i) = 0 at r_i = ∞
   Final point r_f is the point where the potential energy of the system is to be determined.
4. At r_f = r
   
   This is gravitational potential energy of the system of object of mass m and Earth of mass M having separation r (between their centres of mass).

Question 47.
Derive the formula for increase in gravitational potential energy of a Earth – mass system when the mass is lifted to a height h provided h << R.
Answer:

1. If the object is on the surface of Earth, r = R
   U₁ = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
   If the object is lifted to height h above the surface of Earth, the potential energy becomes _ GMm 12 ~~ R+h
   U₂ = –\(\frac{G M m}{R+h}\)
2. Increase in the potential energy is given by
   
3. If g is acceleration due to gravity on the surface of Earth. GM = gR²
   ∴ ∆U = mgh\(\left(\frac{R}{R+h}\right)\) … (1)
4. Equation (1) gives the work to be done to raise an object of mass rn to a height h, above the surface of the Earth.
5. If h << R, we can use R + h ≈ R.
   ∴ ∆U = mgh
   Thus, mgh is increase in the gravitational potential energy of the Earth – mass system if an object of mass m is lifted to a height h, provided h << R.

Question 48.
Write a short note on gravitational potential.
Answer:
The gravitational potential energy of the system of Earth and any mass m at a distance r from the centre of the Earth is given by,

The factor –\(\frac{\mathrm{GM}}{\mathrm{r}}\) = (V_E)_r is defined as the
gravitational potential of Earth at distance r from its centre.

As the potential depends only upon mass of the Earth and location of the object, it is same for any mass m bound to the Earth.

Question 49.
Explain the relation between the gravitational potential energy and the gravitational potential.
Answer:

1. In terms of potential, we can write the potential energy of the Earth-mass system as, Gravitational potential energy (U) = Gravitational potential (V_r) × mass (m)
2. Thus, gravitational potential is gravitational potential energy per unit mass.
   ∴ V_r = \(\frac{\mathrm{U}}{\mathrm{m}}\)
3. For any conservative force field, the concept of potential can be defined on similar lines.
4. Gravitational potential difference between any two points in gravitational field can be written as,
   V₂ – V₁ = \(\frac{U_{2}-U_{1}}{m}\)
   = \(\frac{\mathrm{dW}}{\mathrm{m}}\)
   This is the work done (or change in potential energy) per unit mass.
5. Therefore, in general, for a system of any two masses m₁ and m₂, separated by distance r, we can write,
   U = –\(\frac{\mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}}\)
   = (V₁)m₂
   = (V₂)m₁
   Here V₁ and V₂ are gravitational potentials at r due to m₁ and m₂ respectively.

Solved Exmaples

Question 50.
What will be the change in potential energy of a body of mass m when it is raised from height R_E above the Earth’s surface to 5/2 R_E above the Earth’s surface? R_E and M_E are the radius and mass of the Earth respectively.
Solution:
Change in potential energy (P.E.) of a body of mass m is given by,
∆U = GM_Em\(\left(\frac{1}{r_{i}}-\frac{1}{r_{f}}\right)\)
Here, r_i = R_E + R_E = 2 R_E

[Note: Answer calculated above is in accordance with textual methods of calculation.]

Question 51.
What will be the change in potential energy of a body of mass m when it is placed on the surface of the Earth from height R above the Earth’s surface?
Solution:
Change in potential energy (P.E.) of a body of mass m is given by,
∆U = GM_Em\(\frac{1}{r_{i}}-\frac{1}{r_{r}}\)
Here, r_i = R + R = 2R
Similarly, r_f = R
∴ ∆U = GM_Em \(\left[\frac{1}{2 R}-\frac{1}{R}\right]\) = GM_Em\(\left(-\frac{1}{2 R}\right)\)
∴ ∆U = –\(\frac{\mathbf{G} \mathbf{M} \mathbf{m}}{\mathbf{2 R}}\)
Negative sign indicates that potential energy is decreasing.

Question 52.
Determine the gravitational potential of a body of mass 80 kg whose gravitational potential energy is 5 × 10⁹ J on the surface of the Earth.
Solution:
Given: m = 80 kg, U = 5 × 10⁹ J
To find: Gravitational potential (V)
Formula: V = \(\frac{\mathrm{U}}{\mathrm{m}}\)
Calculation: From formula,
V = \(\frac{5 \times 10^{9}}{80}=\frac{25}{4}\) × 10⁷
= 6.25 × 10⁷ J kg^-1
Potential of the body at the surface of the Earth is 6.25 × 10⁷ J kg^-1.

Question 53.
Calculate the escape velocity of a body from the surface of the Earth.
(Average density of Earth = 5.5 × 10³ kg/m³, G = 6.67 × 10^-11 N m²/kg², radius of Earth R = 6.4 × 10⁶ m)
Solution:
Given: ρ = 5.5 × 10³ kg/m³, R = 6.4 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²
To find: Escape velocity (v_e)

= 2 × 6.4 × 10⁶ × 8.759 × 10^-4
∴ v_e = 11.21 × 10³ m/s = 11.21 km/s
The escape velocity of a body is 11.21 km/s.

Question 54.
What is a satellite?
Answer:
An object which revolves in an orbit around a planet is called as satellite.
Example:

• Moon is a natural satellite of the Earth.
• INSAT is an artificial satellite of the Earth.

Question 55.
Write a short note on Polar satellites.
Answer:

1. Polar Satellites are placed in lower polar orbits.
2. They are at low altitude 500 km to 800 km.
3. Period of revolution of polar satellite is nearly 85 minutes, so it can orbit the Earth 16 time per day.
4. They go around the poles of the Earth in a north-south direction while the Earth rotates in an east-west direction about its own axis.
5. The polar satellites have cameras fixed on them. The camera can view small stipes of the Earth in one orbit. In entire day the whole Earth can be viewed strip by strip.
6. Polar region and equatorial regions close to it can be viewed by these satellites.
7. Polar satellites are used for weather forecasting and meteorological purpose. They are also used for astronomical observations and study of Solar radiations.

Question 56.
Derive the expression for the critical velocity of a satellite revolving close to the surface of the Earth in terms of acceleration due to gravity.
Answer:

1. When the satellite is revolving close to the surface of the Earth, the height is very small as compared to the radius of the Earth.
2. Hence the height can be neglected and radius of the orbit is nearly equal to R, i.e., R + h ≈ R
3. The critical speed of the satellite then becomes,
   v_c = \(\sqrt{\frac{G M}{R}}\)
4. G is related to acceleration due to gravity by the relation,
   g = \(\frac{G M}{R^{2}}\)
   ∴ GM = gR²
5. Thus, critical speed in terms of acceleration due to gravity, neglecting the air resistance, can be obtained as,
   v_c = \(\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}\)

Question 57.
From an inertial frame of reference, explain the apparent weight for a person standing in a lift having zero acceleration.
Answer:

1. A passenger inside a lift experiences only two forces:
   • Gravitational force mg directed vertically downwards and
   • normal reaction force N directed vertically upwards, exerted by the floor of the lift.
2. As these forces are oppositely directed, the net force in the downward direction will be F = ma – N.
3. Though the weight of a passenger is the gravitational force acting upon it, the person experiences his weight only due to the normal reaction force N exerted by the floor.
4. A lift has zero acceleration when the lift is at rest or is moving upwards or downwards with constant velocity.
5. Thus, a net force acting on the passenger inside the lift will be,
   F = 0 = mg – N
   ∴ mg = N
   Hence, in this case the passenger feels his normal weight mg.

Question 58.
What happens to the apparent weight of the person inside the lift moving with net upward acceleration?
Answer:

1. The lift is said to be moving with net upward acceleration in two possible conditions:
   • when the lift just starts moving upwards or
   • is about to stop at a lower floor during its downward motion.
2. As the net acceleration is upwards, the upward force must be greater.
   ∴ F = ma = N – mg
   ∴ N = mg + ma
   ∴ N > mg
3. Thus, for a passenger inside this lift, his apparent weight is more than his actual weight when the lift was not accelerated.

Question 59.
Why does a passenger feel lighter when the lift is about to stop at a higher floor during its upward motion?
Answer:

1. When the lift is about to stop at a higher floor during its upward motion it has a net downward acceleration.
2. As the net acceleration is downwards, the downward force must be greater.
   ∴ F = ma = mg – N
   ∴ N = mg – ma
   i.e., N < mg
   Hence, a passenger feels lighter when the lift is about to stop at a higher floor during its upward motion.

Question 60.
When does a weighing machine will record zero for a passenger in a lift?
Answer:
If the cables of the lift are cut, the downward acceleration of the lift, a_d = g. In this case, we get,
N = mg – ma_d = 0
Thus, there will not be any feeling of weight and the weighing machine will record zero.

Question 61.
Define time period of a satellite.
Obtain an expression for the period of a satellite in a circular orbit round the Earth. Show that the square of the period of revolution of a satellite is directly proportional to the cube of the orbital radius.
Answer:
Definition:
The time taken by a satellite to complete one revolution around the Earth is called its time period.
Expression for time period:
i) Consider, m = mass of satellite, h = altitude of satellite. Thus, r = R + h = radius of orbit of the satellite.
ii) In one revolution, distance traced by satellite is equal to circumference of its circular orbit.
iii) If T is the time period of satellite, then

Since π², G and M are constant,
∴ T² ∝ r³
Hence, square of the period of revolution of a satellite is directly proportional to the cube of the radius of its orbit.

vi) Taking square roots on both the sides of equation (4), we get,
T = 2π\(\sqrt{\frac{\mathrm{r}^{3}}{\mathrm{GM}}}\)
T = 2π\(\sqrt{\frac{(R+h)^{3}}{G M}}\)
This is the required expression for period of satellite orbiting around the Earth in circular path.

Question 62.
For an orbiting satellite very close to surface of the Earth, show that T = 2π \(\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\).
Answer:

1. Time period of an orbiting satellite at certain height is given by, T = 2π \(\sqrt{\frac{(R+h)^{3}}{G M}}\)
2. If satellite is orbiting very close to the Earth’s surface, then h ≈ 0
   

Solved Examples

Question 63.
Show that the critical velocity of a body revolving in a circular orbit very close to the surface of a planet of radius R and
mean density ρ is 2R\(\sqrt{\frac{G \pi \rho}{3}}\).
Solution:
Since the body is revolving very close to the surface of a planet,
∴ h << R
R = radius of planet
ρ = mean density of planet
Critical velocity of a body very close to Earth is given by,

Question 64.
Find the orbital speed of the satellite w hen satellite is revolving round the Earth in circular orbit at a distance 9 × 10⁶ m from its centre. (Given: Mass of Earth = 6 × 10²⁴ kg, G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: r = 9 × 10⁶ m, M = 6 × 10²⁴kg,
G = 6.67 × 10^-11 N m²/kg²
To find: Orbital speed (v_c)
Formula: v_c = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\)
Calculation: From formula,

∴ v_c = 6.668 × 10³ m/s
The orbital speed of the satellite is 6.668 × 103 m/s.

Question 65.
Taking radius of the Earth as 6400 km and g at the Earth’s surface as 9.8 m/s², calculate the speed of revolution of a satellite orbiting close to the Earth’s surface.
Solution:
Given: R = 6400 km = 6.4 × 10⁶ m, g = 9.8 m/s²
To find: Critical velocity (v_c)
Formula: v_c = \(\sqrt{\mathrm{gR}}\)
Calculation: From formula,
v_c = \(\sqrt{9.8 \times 6.4 \times 10^{6}}\)
= \(\sqrt{98 \times 64 \times 10^{4}}\)
= 7\(\sqrt{2}\) × 8 ×10²
= 7.92 × 10³ m/s
The speed of revolution of the satellite orbiting close to the Earth’s surface is 7.92 × 10³ m/s.

Question 66.
The critical velocity of a satellite revolving around the Earth is 10 km/s at a height where g_h = 8 m/s ². Calculate the height of the satellite from the surface of the Earth. (R = 6.4 × 10⁶ m)
Solution:
Given: v_c = 10 km/s = 10 × 10³ m/s,
g_h = 8 m/s³, R = 6.4 × 10⁶ m
To find: Height of the satellite (h)
Formula: v_c = \(\sqrt{g_{\mathrm{h}}(R+h)}\)
Calculation: From formula,
10 × 10³ = \(\sqrt{8 \times(\mathrm{R}+\mathrm{h})}\)
Squaring both the sides, we get,
100 × 10⁶ = 8(R + h)
∴ 8(R + h) = 100 × 10⁶
∴ R + h = \(\frac{100}{8}\) × 10⁶
∴ h = 12.5 × 10⁶ – R
= 12.5 × 10⁶ – 6.4 × 10⁶
= 6.1 × 10⁶m
∴ h = 6100 km
The height of the satellite from the surface of the Earth is 6100 km.

Question 67.
An artificial satellite revolves around a planet in circular orbit close to its surface. Obtain the formula for period of the satellite in terms of density p and radius R of planet.
Solution:
Time period of a satellite revolving around the planet at certain height is given by,

Question 68.
Calculate the period of revolution of a polar satellite orbiting close to the surface of the Earth. Given R = 6400 km, g = 9.8 m/s².
Solution:
Given: For satellite close to Earth surface,
R + h ≈ R
R = 6400 km = 6.4 × 10⁶ m, g = 9.8 m/s²
To find: Time period of satellite (T)
Formula: T = 2π \(\sqrt{\frac{R}{g}}\)
Calculation: From Formula,
T = 2 × 3.14 × \(\sqrt{\frac{6.4 \times 10^{6}}{9.8}}\)
= 6.28 × 8.081 × 10²
= 5.075 × 10³ sec
≈ 85 min
The time period of satellite very close to the Earth’s surface is nearly 85 minute.

Question 69.
A satellite orbits around the Earth at a height equal to R of the Earth. Find its period. (R = 6.4 × 10⁶ m, g = 9.8 m/s²)
Solution:
Given: h = R = 6.4 × 10⁶m, g = 9.8 m/s²
To find: Time period (T)

The time period of the satellite is 1.435 × 10⁴ s.

Question 70.
Calculate the height of the communication satellite. (Given: G = 6.67 × 10^-11 N m²/kg², M = 6 × 10²⁴ kg, R = 6400 km)
Solution:
For communication satellite, T = 24 × 60 × 60 s,
Given: M = 6 × 10²⁴ kg,
G = 6.67 × 10^-11 N m²/kg²,
R = 6400 km = 6.4 × 10⁶m
To find: Height (h)

The height of the communication satellite is 35910 km.

Question 71.
How will you ‘weigh the Sun’, that is estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 × 10⁸ km.
Solution:
Given: r = 1.5 × 10⁸ × 10³m,
T = 365 days = 365 × 24 × 60 × 60 s
To find: Mass (M)
Formula: T = 2π \(\sqrt{\frac{r^{3}}{G M}}\)
Calculation:
From formula,
M = \(\frac{4 \pi^{2} r^{3}}{G T^{2}}\)
= \(\frac{4 \times(3.14)^{2}\left(1.5 \times 10^{11}\right)^{3}}{\left(6.67 \times 10^{-11}\right)(365 \times 24 \times 60 \times 60)^{2}}\)
∴ M = 2.01 × 10³⁰kg
The mass of the Sun is 2.01 × 10³⁰ kg.
[Trick: To ‘weigh the Sun’, i.e., estimate its mass, one needs to know the period of one of its planets and the radius of the planetary orbit.]

Question 72.
Calculate the B.E. of a satellite of mass 2000 kg moving in an orbit very close to the surface of the Earth. (G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: m = 2 × 10³ kg, R = 6.4 × 10⁶ m,
R = 6.4 × 10⁶ m, M = 6 × 10²⁴kg,
G = 6.67 × 10^-11 N m²/kg²,
M = 6 × 10²⁴ kg
To find: Binding Energy (B.E.)
Formula: For satellite very close to Earth,
B.E. = \(\frac{1}{2} \times \frac{\mathrm{GMm}}{\mathrm{R}}\)
Calculation: From formula,
B.E. = \(\frac{1}{2} \times \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{3}}{6.4 \times 10^{6}}\)
∴ B.E. = 6.25 × 10¹⁰ joule
The binding energy of the satellite is 6.25 × 10¹⁰ joule.

Question 73.
Find the binding energy of a body of mass 50 kg at rest on the surface of the Earth. (Given: G = 6.67 × 10^-11 N m²/kg², R = 6400 km, M = 6 × 10²⁴ kg)
Solution:
Given: G = 6.67 × 10^-11 N m²/kg²,
R = 6400 km = 6.4 × 10⁶m,
M = 6 × 10²⁴ kg, m = 50 kg
To find: Binding energy (B.E.)
Formula: B.E. = \(\frac{\text { GMm }}{\mathrm{R}}\)
Calculation: From formula,
B.E. = \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 50}{6.4 \times 10^{6}}\)
= \(\frac{2001}{6.4}\) × 10⁷
∴ B.E. = 3.127 × 10⁹ J
The binding energy of the body 3.127 × 10⁹ J.

Question 74.
Find the total energy and binding energy of an artificial satellite of mass 1000 kg orbiting at height of 1600 km above the Earth’s surface.
(Given: G = 6.67 × 10^-11 N m²/kg², R = 6400 km, M = 6 × 10²⁴ kg)
Solution:
Given: h = 1600 km = 1.6 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²,
R = 6400 km = 6.4 × 10⁶m,
m = 1000 kg, M = 6 × 10²⁴kg
To find:
i) Total Energy (T.E.)
ii) Binding Energy (B.E.)
Formulae: i. T.E. = –\(\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
ii. B.E. = -T.E.

Calculation: From formula (i),
T.E = – \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2(6.4+1.6) \times 10^{6}}\)
= –\(\frac{40020 \times 10^{7}}{2 \times 8}\)
∴ T.E. = -2.501 × 10¹⁰J
From formula (ii),
B.E. = 2.501 × 10¹⁰ J
i) The total energy of the artificial satellite is -2.501 × 10¹⁰ J.
ii) The binding energy of the artificial satellite is 2.501 × 10¹⁰ J.

Question 75.
Determine the binding energy of satellite of mass 1000 kg revolving in a circular orbit around the Earth when it is close to the surface of Earth. Hence find kinetic energy and potential energy of the satellite. (Mass of Earth = 6 × 10²⁴ kg, radius of Earth = 6400 km; gravitational constant G = 6.67 × 10^-11 N m²/kg²)
Solution:
Given: m = 1000 kg, M = 6 × 10²⁴ kg,
R = 6400 km, G = 6.67 × 10^-11 N m²/kg²
To find:
i) Binding Energy (B.E.)
ii)Kinetic Energy (K.E.)
iii) Potential Energy (P.E.)

Formulae: For satellite very close to Earth,
i) B.E. = \(\frac{1}{2} \times \frac{\mathrm{GMm}}{\mathrm{R}}\)
ii) K.E.= B.E.
iii) P.E. = -2 K.E.

Calculation: From formula (i),
B.E. = \(\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000}{2 \times 6.4 \times 10^{6}}\)
∴ B.E. = 3.1265 × 10¹⁰ J
From formula (ii),
K.E. = 3.1265 × 10¹⁰
∴ K.E. = 3.1265 × 10¹⁰ J
From formula (iii),
P.E. = -2(3.1265 × 10¹⁰)
∴ P.E. = -6.2530 × l0¹⁰J
i) The binding energy of the satellite is 3.1265 × 10¹⁰ J.
ii) The kinetic energy of the satellite is 3.1265 × 10¹⁰ J.
iii) The potential energy of the satellite is -6.2530 × 10¹⁰J.

Apply Your Knowledge

Question 76.
How are Kepler’s law of periods and Newton’s law of gravitation related?
Answer:
Consider a planet of mass m revolving around the Sun of mass M in a circular orbit.
Let,
r = radius of the circular orbit of the planet.
T = Time period of revolution of planet around the Sun.
ω = angular velocity of planet.
F = Centripetal force exerted by the Sun on the planet.
Centripetal force is given by,
F = mrω²;
But ω = \(\frac{2 \pi}{T}\)
∴ F = mr (\(\frac{2 \pi}{\mathrm{T}}\))²
∴ F = \(\frac{4 \pi^{2} \mathrm{mr}}{\mathrm{T}^{2}}\) …(i)
According to Kepler’s third law,
T² ∝ r³
T² = Kr³ ……….. (where, K = constant) (ii)
Substituting equation (ii) in equation (i),
F = \(\frac{4 \pi^{2} \mathrm{mr}}{\mathrm{Kr}^{3}}\)
∴ F = \(\frac{4 \pi^{2}}{\mathrm{~K}} \frac{\mathrm{m}}{\mathrm{r}^{2}}\)
∴ F ∝ \(\frac{\mathrm{m}}{\mathrm{r}^{2}}\) ….(∵ \(\frac{4 \pi^{2}}{\mathrm{~K}}\) is a constant quantity)
Since, the gravitational attraction between the Sun and the planet is mutual, force exerted by the planet on the Sun will be proportional to the mass M of the Sun.
∴ F ∝ \(\frac{\mathrm{Mm}}{\mathrm{r}^{2}}\)
∴ F = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
The above equation represents Newton’s law of gravitation. In this manner, Newton’s law of gravitation is derived from Kepler’s law of periods.

Question 77.
Represent graphically the variation of total energy, kinetic energy and potential energy of a satellite with its distance from the centre of the Earth.
Answer:
For a satellite,
Potential energy (U) = \(\)
Kinetic energy (K) = \(\) and
Total energy (E) = \(\), where, r = R + h
Also, U and E remain negative whereas K remains positive.
Hence, the graph will be:

Multiple Choice Questions

Question 1.
Kepler’s law of equal areas is an outcome of
(A) conservation of energy
(B) conservation of linear momentum
(C) conservation of angular momentum
(D) conservation of mass
Answer:
(C) conservation of angular momentum

Question 2.
Amongst given statements, choose the correct statement.
(I) Kepler derived the laws of planetary motion.
(II) Newton provided the reason behind the laws of planetary motion.
(A) (I) is correct.
(B) (II) is correct.
(C) Both (I) and (II) are correct.
(D) Neither (I) nor (II) is correct.
Answer:
(B) (II) is correct.

Question 3.
The figure shows the motion of a planet satellite in terms of mean density of Earth. around the Sun in an elliptical orbit with Sun at the focus. The shaded areas A and B are also shown in the figure which can be assumed to be equal. If t₁ and t₂ represent the time for the planet to move from a to b and d to c respectively, then

(A) t₁ < t₂
(B) t₁ >t₂
(C) t₁ = t₂
(D) t₁ ≤ t₂
Answer:
(C) t₁ = t₂

Question 4.
A planet is revolving around a star in a circular orbit of radius R with a period T. If the gravitational force between the planet and the star is proportional to R^-3/2, then
(A) T^{2 ∝} R^5/2
(B) T² ∝ R^-7/2
(C) T² ∝ R^3/2
(D) T² ∝ R⁴
Answer:
(A) T² ∝ R^5/2

Question 5.
Time period of revolution of a satellite around a planet of radius R is T. Period of revolution around another planet whose radius is 3R is
(A) T
(B) 9T
(C) 3T
(D) 3\(\sqrt{3}\) T
Answer:
(D) 3\(\sqrt{3}\) T

Question 6.
Newton’s law of gravitation is called universal law because
(A) force is always attractive.
(B) it is applicable to lighter and heavier bodies.
(C) it is applicable at all times,
(D) it is applicable at all places of universe for all distances between all particles.
Answer:
(D) it is applicable at all places of universe for all distances between all particles.

Question 7.
If the mass of a body is M on the surface of the Earth, the mass of the same body on the surface of the moon is
M
(A) 6M
(B) \(\frac{M}{6}\)
(C) M
(D) Zero
Answer:
(C) M

Question 8.
Which of the following statements about the gravitational constant is true?
(A) It has no units.
(B) It has same value in all systems of units.
(C) It is a force.
(D) It does not depend upon the nature of medium in which the bodies lie.
Answer:
(D) It does not depend upon the nature of medium in which the bodies lie.

Question 9.
The gravitational force between two bodies is ______
(A) attractive at large distance only
(B) attractive at small distance only
(C) repulsive at small distance only
(D) attractive at all distances large or small
Answer:
(D) attractive at all distances large or small

Question 10.
Mass of a particle at the centre of the Earth is
(A) infinite.
(B) zero.
(C) same as at other places.
(D) greater than at the poles.
Answer:
(C) same as at other places.

Question 11.
Which of the following is not a property of gravitational force?
(A) It is an attractive force.
(B) It acts along the line joining the two bodies.
(C) The forces exerted by two bodies on each other form an action-reaction pair.
(D) It has a very finite range of action.
Answer:
(D) It has a very finite range of action.

Question 12.
If the distance between Sun and Earth is made two third times of the present value, then gravitational force between them will become
(A) \(\frac{4}{9}\)times
(B) \(\frac{2}{3}\)times
(C) \(\frac{1}{3}\)times
(D) \(\frac{9}{4}\) times
Answer:
(D) \(\frac{9}{4}\) times

Question 13.
The gravitational constant G is equal to 6.67 × 10^-11 N m²/kg² in vacuum. Its value in a dense matter of density 10¹⁰ g/cm³ will be
(A) 6.67 × 10^-1 N m²/kg²
(B) 6.67 × 10^-11 N m²/kg²
(C) 6.67 × 10^-10 N m²/kg²
(D) 6.67 × 10^-21 N m²/kg²
Answer:
(B) 6.67 × 10^-11 N m²/kg²

Question 14.
Acceleration due to gravity above the Earth’s surface at a height equal to the radius of the Earth is ______
(A) 2.5 m/s²
(B) 5 m/s²
(C) 9.8 m/s²
(D) 10 m/s²
Answer:
(A) 2.5 m/s²

Question 15.
If R is the radius of the Earth and g is the acceleration due to gravity on the Earth’s surface, the mean density of the Earth is

Answer:
(D) \(\frac{3 \mathrm{~g}}{4 \pi \mathrm{RG}}\)

Question 16.
Variation of acceleration due to gravity (g) with distance x from the centre of the Earth is best represented by (R → Radius of the Earth)

Answer:
(D)

Question 17.
Which of the following statements is not correct for the decrease in the value of acceleration due to gravity?
(A) As we go down from the surtce of the Earth towards its centre.
(B) As we go up from the surface of the Earth.
(C) As we go from equator to the poles on the surface on the Earth.
(D) As the rotational velocity of the Earth is increased.
Answer:
(C) As we go from equator to the poles on the surface on the Earth.

Question 18.
Calculate angular velocity of Earth so that acceleration due to gravity at 60° latitude becomes zero. (Radius of Earth = 6400 km, gravitational acceleration at poles = 10 m/s², cos60° = 0.5)
(A) 7.8 × 10^-2 rad/s
(B) 0.5 × 10^-3 radis
(C) 1 × 10^-3 radis
(D) 2.5 × 10^-3 rad/s
Answer:
(D) 2.5 × 10^-3 rad/s

Question 19.
The gravitational potential energy per unit mass at a point gives ________ at that point.
(A) gravitational field
(B) gravitational potential
(C) gravitational potential energy
(D) gravitational force
Answer:
(B) gravitational potential

Question 20.
A satellite is orbiting around a planet at a constant height in a circular orbit. If the mass of the planet is reduced to half, the satellite would
(A) fall on the planet.
(B) go to an orbit of smaller radius.
(C) go to an orbit of higher radius,
(D) escape from the planet.
Answer:
(D) escape from the planet.

Question 21.
How does the escape velocity of a particle depend on its mass?
(A) m²
(B) m
(C) m⁰
(D) m^-1
Answer:
(C) m⁰

Question 22.
Escape velocity on a planet is ve. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes
(A) 4v_e
(B) 2v_e
(C) v_e
(D) 0.5 v_e
Answer:
(B) 2v_e

Question 23.
If the escape velocity of a body on Earth is 11.2 km/s, the escape velocity of the body thrown at an angle 45° with the horizontal will be
(A) 11.2 km/s
(B) 22.4 km/s
(C) \(\frac{11.2}{\sqrt{2}}\)km/s
(D) 11.2 \(\sqrt{2}\) km/s
Answer:
(A) 11.2 km/s

Question 24.
Potential energy of a body in the gravitational field of planet is zero. The body must be
(A) at centre of planet.
(B) on the surface of planet.
(C) at infinity.
(D) at distance equal to radius of Earth.
Answer:
(C) at infinity.

Question 25.
If gravitational force of Earth disappears, what will happen to the satellite revolving round the Earth?
(A) Satellite will come back to Earth.
(B) Satellite will continue to revolve.
(C) Satellite will escape in tangential path.
(D) Satellite will start falling towards centre.
Answer:
(C) Satellite will escape in tangential path.

Question 26.
If v_e and v_o represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R respectively, then
(A) v_e = v_o
(B) \(\sqrt{2}\)v_o = v_e
(C) v_e = \(\frac{1}{\sqrt{2}}\)v_o
(D) v_e and v_o are not related
Answer:
(B) \(\sqrt{2}\)v_o = v_e

Question 27.
If the kinetic energy of a satellite is 2 × 10⁴ J, then its potential energy will be
(A) – 2 × 10⁴ J
(B) 4 × 10⁴ J
(C) -4 × 10⁴ J
(D) -10⁴J
Answer:
(C) -4 × 10⁴ J

Competitive Corner

Question 1.
A body weighs 200 N on the surface of the Earth. How much will it weigh half way down to the centre of the Earth?
(A) 250 N
(B) 100 N
(C) 150 N
(D) 200 N
Answer:
(B) 100 N
Hint:
Acceleration due to gravity at depth d,
g_d = g (1 – \(\frac{\mathrm{d}}{\mathrm{R}}\))
= g(1 – \(\frac{1}{2}\)) …(∵ d = \(\frac{1}{2}\))
∴ g_d = \(\frac{\mathrm{g}}{2}\)
Weight of the body at depth d = R/2,
W_d = mg_d = m × g/2 = \(\frac{1}{2}\) × 200
∴ W_{d = 100 N}

Question 2.
The work done to raise a mass m from the surface of the Earth to a height h, which is equal to the radius of the Earth, is:
(A) \(\frac{1}{2}\) mgR
(B) \(\frac{3}{2}\) mgR
(C) mgR
(D) 2mgR
Answer:
(A) \(\frac{1}{2}\) mgR
Hint:
Initial potential energy on Earth’s surface,
U_i = \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
Final potential energy at height h = R
U_f = \(\frac{-\mathrm{GMm}}{2 \mathrm{R}}\)
Work done, W = U_f – U_i

∴ W = \(\frac{1}{2}\) mgR

Question 3.
The time period of a geostationary satellite is 24 h, at a height 6R_E (R_E is radius of Earth) from surface of Earth. The time period of another satellite whose height is 2.5 R_E from surface will be,
(A) \(\frac{12}{2.5}\)h
(B) 6\(\sqrt{2}\) h
(C) 12\(\sqrt{2}\) h
(D) \(\frac{24}{2.5}\)h
Answer:
(B) 6\(\sqrt{2}\) h
Hint:
By Kepler’s third law,

Question 4.
Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final – initial) of an object of mass m, when taken to a height h from the surface of Earth (of radius R), is given by,
(A) \(\frac{\text { GMm }}{R+h}\)
(B) – \(\frac{\text { GMm }}{R+h}\)
(C) \(\frac{\text { GMmh }}{R(R+h)}\)
(D) mgh
Answer:
(C) \(\frac{\text { GMmh }}{R(R+h)}\)
Hint:
Potential energy of object of mass m on the surface of Earth,
P.E = \(\frac{-\mathrm{GMm}}{\mathrm{R}}\)
Potential energy of object of mass m at a height h from the surface of the Earth,
P.E.’ = \(\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\)
∴ Change in potential energy
= P.E.’ – P.E.

Question 5.
A body mass ‘m’ is dropped from height \(\), from Earth’s surface, where ‘R’ is the radius of Earth. Its speed when it will hit the Earth’s surface is (v_e = escape velocity from Earth’s surface)

Answer:
(B) \(\frac{\mathbf{v}_{\mathrm{e}}}{\sqrt{3}}\)
Hint:

Question 6.
According to Kepler’s Law, the areal velocity of the radius vector drawn from the Sun to any planet always
(A) decreases.
(B) first increases and then decreases.
(C) remains constant.
(D) increases.
Answer:
(C) remains constant.

Question 7.
A body is thrown from the surface of the Earth with velocity ‘u’ m/s. The maximum height in m above the surface of the Earth upto which it will reach is (R = radius of Earth, g = acceleration due to gravity)

Answer:
(A) \(\frac{u^{2} R}{2 g R-u^{2}}\)
Hint:
(T.E.) on surface = (T.E.) at height ‘h’
∴ (K.E.)₁ + (P.E.)₁ = (K.E.)₂ + (P.E.)₂

Question 8.
A satellite is revolving in a circular orbit at a height ‘h’ above the surface of the Earth of radius ‘R’. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the Earth. The relation between ‘h’ and ‘R’ is
(A) h = 2R
(B) h = 3R
(C) h = 5R
(D) h = 7R
Answer:
(D) h = 7R
Hint:

Question 9.
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
(A) keep floating at the same distance between them.
(B) move towards each other.
(C) move away from each other.
(D) will become stationary.
Answer:
(B) move towards each other.