Class 11 – Page 7 – Maharashtra Board Solutions

11th Physics Chapter 1 Exercise Units and Measurements Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 1 Units and Measurements Textbook Exercise Questions and Answers.

Units and Measurements Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 1 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 1 Exercise Solutions

1. Choose the correct option.

Question 1.
[L¹M¹T^-2] is the dimensional formula for
(A) Velocity
(B) Acceleration
(C) Force
(D) Work
Answer:
(C) Force

Question 2.
The error in the measurement of the sides of a rectangle is 1%. The error in the measurement of its area is
(A) 1%
(B) \(\frac{1}{2}\)%
(C) 2%
(D) None of the above.
Answer:
(C) 2%

Question 3.
Light year is a unit of
(A) Time
(B) Mass
(C) Distance
(D) Luminosity
Answer:
(C) Distance

Question 4.
Dimensions of kinetic energy are the same as that of
(A) Force
(B) Acceleration
(C) Work
(D) Pressure
Answer:
(C) Work

Question 5.
Which of the following is not a fundamental unit?
(A) cm
(B) kg
(C) centigrade
(D) volt
Answer:
(D) volt

2. Answer the following questions.

Question 1.
Star A is farther than star B. Which star will have a large parallax angle?
Answer:

i). ‘b’ is constant for the two stars
∴ θ ∝ \(\frac{1}{D}\)

ii) As star A is farther i.e., D_A > D_B
⇒ θ_A < θ_B
Hence, star B will have larger parallax angle than star A.

Question 2.
What are the dimensions of the quantity l \(\sqrt{l / g}\), l being the length and g the acceleration due to gravity?
Answer:

[Note: When power of symbol expressing fundamental quantity appearing in the dimensional formula is not given, ills taken as 1.]

Question 3.
Define absolute error, mean absolute error, relative error and percentage error.
Answer:
Absolute error:
a. For a given set of measurements of a quantity, the magnitude of the difference between mean value (Most probable value) and each individual value is called absolute error (∆a) in the measurement of that quantity.
b. absolute error = |mean value – measured value|
∆a₁ = |a_mean – a₁|
Similarly, ∆a₂ = |a_mean – a₂|
. . . ..
. . . .
. . . .
∆a_n = |a_mean – a_n|

Mean absolute error:
For a given set of measurements of a same quantity the arithmetic mean of all the absolute errors is called mean absolute error in the measurement of that physical quantity.
∆a_mean = \(\frac{\Delta \mathrm{a}_{1}+\Delta \mathrm{a}_{2}+\ldots \ldots .+\Delta \mathrm{a}_{n}}{\mathrm{n}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{n} \Delta \mathrm{a}_{\mathrm{i}}\)

Relative error:
The ratio of the mean absolute error in the measurement of a physical quantity to its arithmetic mean value is called relative error.
Relative error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\mathrm{mean}}}\)

Percentage error:
The relative error represented by percentage (i.e., multiplied by 100) is called the percentage error.
Percentage error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\mathrm{mean}}}\) × 100%
[Note: Considering conceptual conventions question is modified to define percentage error and not mean percentage error.]

Question 4.
Describe what is meant by significant figures and order of magnitude.
Answer:
Significant figures:

Significant figures in the measured value of a physical quantity is the sum of reliable digits and the first uncertain digit.
OR
The number of digits in a measurement about which we are certain, plus one additional digit, the first one about which we are not certain is known as significant figures or significant digits.
Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true.
If one uses the instrument of smaller least count, the number of significant digits increases.

Rules for determining significant figures:

All the non-zero digits are significant, for example if the volume of an object is 178.43 cm³, there are five significant digits which are 1,7,8,4 and 3.
All the zeros between two nonzero digits are significant, eg., m = 165.02 g has 5 significant digits.
If the number is less than 1, the zero/zeroes on the right of the decimal point and to the left of the first nonzero digit are not significant e.g. in 0.001405, significant. Thus the above number has four significant digits.
The zeroes on the right hand side of the last nonzero number are significant (but for this, the number must be written with a decimal point), e.g. 1.500 or 0.01500 both have 4 significant figures each.
On the contrary, if a measurement yields length L given as L = 125 m = 12500 cm = 125000 mm, it has only three significant digits.

Order of magnitude:
The magnitude of any physical quantity can be expressed as A × 10ⁿ where ‘A’ is a number such that 0.5 ≤ A < 5 then, ‘n’ is an integer called the order of magnitude.
Examples:

Speed of light in air = 3 × 10⁸ m/s
∴ order of magnitude = 8
Mass of an electron = 9.1 × 10^-31 kg
= 0.91 × 10³⁰ kg
∴ order of magnitude = -30

Question 5.
If the measured values of two quantities are A ± ∆A and B ± ∆B, ∆A and ∆B being the mean absolute errors. What is the maximum possible error in A ± B? Show that if Z = \(\frac{A}{B}\)
\(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)
Answer:
Maximum possible error in (A ± B) is (∆A + ∆B).
Errors in divisions:
i) A
Suppose, Z = \(\frac{A}{B}\) and measured values of A and B are (A ± ∆A) and (B ± ∆B) then,

∴ Maximum relative error of \(\frac{\Delta Z}{Z}=\pm\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}\right)\)

ii) Thus, when two quantities are divided, the maximum relative error in the result is the sum of relative errors in each quantity.

Question 6.
Derive the formula for kinetic energy of a particle having mass m and velocity v using dimensional analysis
Answer:
Kinetic energy of a body depends upon mass (m) and velocity (v) of the body.
Let K.E. ∝ m^x v^y
∴ K.E. = km^x v^y …….. (1)
where,
k = dimensionless constant of proportionality. Taking dimensions on both sides of equation (1),
[L²M¹T^-2] – [L⁰M¹T⁰]^x [L¹M⁰T^-1]^y
= [L⁰M^xT⁰] [L^yM⁰T^-y]
= [L^0+yM^x+0T^0-y]
[L²M¹T²] = [L^yM^xT^-y] …………. (2)
Equating dimensions of L, M, T on both sides of equation (2),
x = 1 and y = 2 ,
Substituting x, y in equation (1), we have
K.E. = kmv²

3. Solve numerical examples.

Question 1.
The masses of two bodies are measured to be 15.7 ± 0.2 kg and 27.3 ± 0.3 kg. What is the total mass of the two and the error in it?
Answer:
Given: A ± ∆A = 15.7 ± 0.2kg and
B ± ∆B = 27.3 ± 0.3 kg.
To find: Total mass (Z), and total error (∆Z)
Formulae: i. Z = A + B

ii) ±∆Z = ±∆A ± ∆B
Calculation: From formula (i),
Z = 15.7 + 27.3 = 43 kg
From formula (ii),
± ∆Z (± 0.2) + (± 0.3)
=±(0.2 + 0.3)
= ± 0.5 kg
Total mass is 43 kg and total error is ± 0.5 kg.

Question 2.
The distance travelled by an object in time (100 ± 1) s is (5.2 ± 0.1) m. What is the speed and it’s relative error?
Answer:
Given: Distance (D ± ∆D) = (5.2 ± 0.1) m,
time(t ± ∆t) = (100 ± 1)s.
To find: Speed (v), maximum relative error \(\left(\frac{\Delta \mathrm{v}}{\mathrm{v}}\right)\)

Formulae: i. v = \(\frac{\mathrm{D}}{\mathrm{t}}\)
ii. \(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\pm\left(\frac{\Delta \mathrm{D}}{\mathrm{D}}+\frac{\Delta \mathrm{t}}{\mathrm{t}}\right)\)

Calculation: From formula (i),
v = \(\frac{5.2}{100}\) = 0.052 m/s
From formula (ii),
\(\frac{\Delta \mathrm{v}}{\mathrm{v}}=\pm\left(\frac{0.1}{5.2}+\frac{1}{100}\right)\)
= \(\pm\left(\frac{1}{52}+\frac{1}{100}\right)=\pm \frac{19}{650}\)
= ± 0.029 rn/s
The speed is 0.052 m/s and its maximum relative error is ± 0.029 m/s.
[Note: Framing of numerical is modified to make it specific and meaningful.]

Question 3.
An electron with charge e enters a uniform. magnetic field \(\vec{B}\) with a velocity \(\vec{v}\). The velocity is perpendicular to the magnetic field. The force on the charge e is given by
|\(\vec{F}\)| = Bev Obtain the dimensions of \(\vec{B}\).
Answer:
Given: |\(\vec{F}\)| = B e v
Considering only magnitude, given equation is simplified to,
F = B e v

∴ B = [L⁰M¹T^-2I^-1]
[Note: The answer given above is calculated in accordance with textual method considering the given data.]

Question 4.
A large ball 2 m in radius is made up of a rope of square cross section with edge length 4 mm. Neglecting the air gaps in the ball, what is the total length of the rope to the nearest order of magnitude?
Answer:
Volume of ball = Volume enclosed by rope.
\(\frac{4}{3}\) π (radius)³ = Area of cross-section of rope × length of rope.
∴ length of rope l = \(\frac{\frac{4}{3} \pi r^{3}}{A}\)
Given:
r = 2 m and
Area = A = 4 × 4 = 16 mm²
= 16 × 10^-6 m²
∴ l = \(\frac{4 \times 3.142 \times 2^{3}}{3 \times 16 \times 10^{-6}}\)
= \(\frac{3.142 \times 2}{3}\) × 10^-6 m
≈ 2 × 10⁶ m.
Total length of rope to the nearest order of magnitude = 10⁶ m = 10³ km

Question 5.
Nuclear radius R has a dependence on the mass number (A) as R = 1.3 × 10^-16 A^{\(\frac{1}{3}\)} m. For a nucleus of mass number A=125, obtain the order of magnitude of R expressed in metre.
Answer:
R= 1.3 × 10^-16 × A^{\(\frac{1}{3}\)} m
For A = 125
R= 1.3 × 10^-16 × (125)^{\(\frac{1}{3}\)}
= 1.3 × 10^-16 × 5
= 6.5 × 10^-16
= 0.65 × 10^-15 m
∴ Order of magnitude = -15
[Note: Taking the standard value of nuclear radius R = 1.3 × 10^-155 m, the order of magnitude comes to be 10^-14 m.]

Question 6.
In a workshop a worker measures the length of a steel plate with a Vernier callipers having a least count 0.01 cm. Four such measurements of the length yielded the following values: 3.11 cm, 3.13 cm, 3.14 cm, 3.14 cm. Find the mean length, the mean absolute error and the percentage error in the measured value of the length.
Answer:
Given: a₁ = 3.11 cm, a₂ = 3.13 cm,
a₃ = 3.14 cm. a₄ = 3.14cm
Least count L.C. = 0.01 cm.
To find.
i. Mean length (a_mean)
ii. Mean absolute error (∆a_mean)
iii. Percentage error.

Formulae: i. a_mean = \(\frac{a_{1}+a_{2}+a_{3}+a_{4}}{4}\)
ii. ∆a_n = |a_mean – a_n|
iii. ∆a_mean = \(\frac{\Delta \mathrm{a}_{1}+\Delta \mathrm{a}_{2}+\Delta \mathrm{a}_{3}+\Delta \mathrm{a}_{4}}{4}\)
iv. Percentage error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\text {mean }}}\) × 100

Calculation: From formula (i),
a_mean = \(\frac{3.11+3.13+3.14+3.14}{4}\)
= 3.13 cm
From formula (ii),
∆a₁ = |3.13 – 3.11| = 0.02 cm
∆a₂ = |3.13 – 3.13| = 0
∆a₃ = |3.13 – 3.14| = 0.01 cm
∆a₄ = |3.13 – 3.14| = 0.01 cm
From formula (iii),
∆a_mean = \(\frac{0.02+0+0.01+0.01}{4}\) = 0.01 cm
From formula (iii).
% error = \(\frac{0.01}{3.13}\) × 100
= \(\frac{1}{3.13}\)
= 0.3196
……..(using reciprocal table)
= 0.32%

i. Mean length is 3.13 cm.
ii. Mean absolute error is 0.01 cm.
iii. Percentage error is 0.32 %.
[Note: As per given data of numerical, percentage error calculation upon rounding off yields percentage error as 0.32%]

Question 7.
Find the percentage error in kinetic energy of a body having mass 60.0 ± 0.3 g moving with a velocity 25.0 ± 0.1 cm/s.
Answer:
Given: m = 60.0 g, v = 25.0 cm/s.
∆m = 0.3 g, ∆v = 0.1 cm/s
To find: Percentage error in E
Formula: Percentage error in E
\(\left(\frac{\Delta \mathrm{m}}{\mathrm{m}}+2 \frac{\Delta \mathrm{v}}{\mathrm{v}}\right)\) × 100%

Calculation: From formula,
Percentage error in E
= \(\left(\frac{0.3}{60.0}+2 \times \frac{0.1}{25.0}\right)\) × 100%
= 1.3%
The percentage error in energy is 1.3%.

Question 8.
In Ohm’s experiments, the values of the unknown resistances were found to be 6.12 Ω , 6.09 Ω, 6.22 Ω, 6.15 Ω. Calculate the mean absolute error, relative error and percentage error in these measurements.
Answer:
Given: a₁ = 6.12 Ω, a₂ = 6.09 Ω, a₃ = 6.22 Ω, a₄ = 6.15 Ω,

To find:

i) Absolute error (∆a_mean)
ii) Relative error
iii) Percentage error

Formulae:

i) a_mean = \(\frac{a_{1}+a_{2}+a_{3}+a_{4}}{4}\)
ii) ∆a_n = |a_mean – a_n|
iii) ∆a_mean = \(\frac{\Delta \mathrm{a}_{1}+\Delta \mathrm{a}_{2}+\Delta \mathrm{a}_{3}+\Delta \mathrm{a}_{4}}{4}\)
iv) Percentage error = \(\frac{\Delta \mathrm{a}_{\mathrm{mean}}}{\mathrm{a}_{\text {mean }}}\) × 100

From formula (ii),
∆a₁ = |6.145 – 6.12|= 0.025
∆a₂ = |6.145 – 6.09| = 0.055
∆a₃ = |6.145 – 6.22| = 0.075
∆a₄ = |6.l45 – 6.15| = 0.005
From formula (iii),
∆a_mean = \(\frac{0.025+0.055+0.075+0.005}{4}=\frac{0.160}{4}\)
= 0.04 Ω
From formula (iv),
Relative error = \(\frac{0.04}{6.145}\) = 0.0065 Ω
From formula (v).
Percentage error = 0.0065 \frac{0.04}{6.145} 100 = 0.65%

i. The mean absolute error is 0.04 Ω.
ii. The relative error is 0.0065 Ω.
iii. The percentage error is 0.65%.
[Note: Framing of numerical is modified to reach the answer given to the numerical.]

Question 9.
An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. If v depends upon gravitational acceleration g and distance, prove with dimensional analysis that v = k\(\sqrt{g h}\) where k is a constant.
Answer:
Given = v = k\(\sqrt{g h}\)

k being constant is assumed to be dimensionless.
Dimensions of L.H.S. = [v] = [L¹T^-1]
Dimension of R.H.S. = [latex]\sqrt{g h}[/latex]
= [L¹T^-2]^{\(\frac{1}{2}\)} × [L¹]^{\(\frac{1}{2}\)}
= [L²T^-2]^{\(\frac{1}{2}\)}
= [L¹T^-1]
As, [L.H.S.] = [R.H.S.],
=> v = k\(\sqrt{g h}\)is dimensionally correct equation.

Question 10.
v = at + \(\frac{b}{t+c}\) + v₀ is a dimensionally valid equation. Obtain the dimensional formula for a, b and c where v is velocity, t is time and v₀ is initial velocity.
Answer:
Solution: Given: y = at + \(\frac{b}{t+c}\) + + v₀

As only dimensionally identical quantities can be added together or subtracted from each other, each term on R.H.S. has dimensions of L.H.S. i.e., dimensions of velocity.

∴ [LH.S.] = [v] = [L¹T^-1]
This means, [at] = [v] = [L¹T^-1]
Given, t = time has dimension [T^-1]
∴ [a] = \(\frac{\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right]}{[\mathrm{t}]}=\frac{\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right]}{\left[\mathrm{T}^{1}\right]}\) = [L¹T^-2] = L¹M⁰T^-2]
Similarly, [c] = [t] = [T¹] = [L⁰M⁰T¹]
∴ \(\frac{[\mathrm{b}]}{\left[\mathrm{T}^{1}\right]}\) = [v] = [L¹T^-1]
∴ [b] = [L¹T^-1] × [T¹] = [L¹] = [L¹M⁰T⁰]

Question 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Given: l = 4.234 m, b = 1.005 m,
t = 2.01 cm = 2.01 × 10^-2 m = 0.0201 m

To find:
i) Area of sheet to correct significant figures (A)
ii) Volume of sheet to correct significant figures (V)
Formulae: i. A = 2(lb + bt + tl)
iii) V = l × b × t

Calculation: From formula (i),
A = 2(4.234 × 1.005 + 1.005 × 0.0201 +0.0201 × 4.234)
= 2 |[antilog(log 4.234 + logl .005) + antiiog(log 1.005 + log0.0201) + antilog(log 0.0201 + log 4.234)]}
= 2{[antilog(0.6267 + 0.0021) + antilog(0.0021 + \(\overline{2}\) .3010) + antilog (\(\overline{2}\) .3010 + 0.6267)]}
= 2 {[antilog(0.6288) + antilog (\(\overline{2}[/latex .3031) +antilog([latex]\overline{2}\) .9277)]}
= 2 [4.254 + 0.02009 + 0.08467]
= 2 [4.35876]
= 8.71752m²

In correct significant figure,
A = 8.72 m2 From formula (ii),
V =4.234 × 1.005 × 0.0201
= antilog [log (4.234) + log (1.005) + log (0.0201)]
= antilog [0.6269 – 0.0021 – \(\overline{2}\).3032]
= antilog [0.6288 – \(\overline{2}\).3032]
= antilog [ 2 .9320]
= 8.551 × 10^-2
= 0.08551 m³
In correct significant figure (rounding off),
V = 0.086 m³

i.) Area of sheet to correct significant figures is 8.72 m².
ii) Volume of sheet to correct significant figures is 0.086 m³.
[Note: The given solution is arrived to by considering a rectangular sheet.]

Question 12.
If the length of a cylinder is l = (4.00 ± 0.001) cm, radius r = (0.0250 ± 0.001) cm and mass m = (6.25 ± 0.01) gm. Calculate the percentage error in the determination of density.
Answer:
Given: l = (4.00 ± 0.001) cm,
In order to have same precision, we use, (4.000 ± 0.001),
r = (0.0250 ± 0.00 1) cm,
In order to have same precision, we use, (0.025 ± 0.001)
m = (6.25 ± 0.01) g
To find: percentage error in density

Formulae:
i) Relative error in volume, \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}\)
….(∵ Volume of cylinder, V = πr²l)

ii) Releative error \(\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{\Delta V}{V}\)

iii) Percentage error= Relative error × 100%

Calculation.
From formulae (i) and (ii),
∴ \(\frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta l}{l}\)
= \(\frac{0.01}{6.25}+\frac{2(0.001)}{0.025}+\frac{0.001}{4.000}\)
= 0.00 16 + 0.08 + 0.00025
= 0.08 185
From formula (iii).
% error in density = \(\frac{\Delta \rho}{\rho}\) × 100
= 0.08185 × 100
= 8.185%
Percentage error in density is 8.185%.

Question 13.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of the Jupiter.
Answer:
Given: Angular diameter (a) = 35.72″
= 35.72″ × 4.847 × 10^-6 rad
1.73 × 10^-4 rad

Distance from Earth (D)
= 824.7 million km
= 824.7 × 10⁶ km
= 824.7 × 10⁹ m.

To find: Diameter of Jupiter (d)
Formula: d = α D
Calculation: From formula,
d = 1.73 × 10^-4 × 824.7 × 10⁹
= 1.428 × 10⁸ m
= 1.428 × 10⁵ km
Diameter of Jupiter is 1.428 × 10⁵ km.

Question 14.
If the formula for a physical quantity is X = \(\frac{a^{4} b^{3}}{c^{1 / 3} d^{1 / 2}}\) and if the percentage error in the measurements of a, b, c and d are 2%, 3%, 3% and 4% respectively. Calculate percentage error in X.
Answer:
Given X = \(\frac{a^{4} b^{3}}{c^{1 / 3} d^{1 / 2}}\)
Percentage error in a, b, c, d is respectively 2%, 3%, 3% and 4%.
Now, Percentage error in X

Question 15.
Write down the number of significant figures in the following: 0.003 m², 0.1250 gm cm^-2, 6.4 × 10⁶ m, 1.6 × 10^-19 C, 9.1 × 10^-31 kg.
Answer:

Question 16.
The diameter of a sphere is 2.14 cm. Calculate the volume of the sphere to the correct number of significant figures.
Answer:
Volume of sphere = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.142 × (\(\frac{2.14}{2}\))³ ………….. (∵ r = \(\frac{d}{2}\))
= \(\frac{4}{3}\) × 3.142 × (1.07)³
= 1.333 × 3.142 × (1.07)³
= {antilog [log (1.333) + log(3.142)+3 log(1.07)]}
= {antilog [0.1249 + 0.4972 + 3 (0.0294)])
= {antilog [0.6221 + 0.0882]}
= {antilog [0.7103]}
= 5.133cm³
In multiplication or division, the final result should retain as many significant figures as there are in the original number with the least significant figures.
Volume in correct significant figures
∴ 5.13 cm³

11th Physics Digest Chapter 1 Units and Measurements Intext Questions and Answers

Can you recall (Textbook Page No. 1)

Question 1.
i) What is a unit?
ii) Which units have you used in the laboratory for measuring
a. length
b. mass
c. time
d. temperature?
iii. Which system of units have you used?
Answer:

The standard measure of any quantity is called the unit of that quantity.
MKS or SI system is used mostly. At times. even CGS system is used.

Can you tell? (Textbook Page No. 8)

Question 1.
If ten students are asked to measure the length of a piece of cloth upto a mm, using a metre scale, do you think their answers will be identical? Give reasons.
Answer:
Answers of the students are likely to be different. Length of cloth needs to be measured up to a millimetre (mm) length. Hence, to obtain accurate and precise reading one must use measuring instrument having least count smaller than 1 mm.

But least count of metre scale is 1 mm. As a result, even smallest uncertainty in reading would vary reading significantly. Also, skill of students doing measurement may also introduce uncertainty in observation.
Hence, their answers are likely to be different.

Activity (Textbook Page No. 10)

Perform an experiment using a Vernier callipers of least count 0.01cm to measure the external diameter of a hollow cylinder. Take 3 readings at different positions on the cylinder and find (i) the mean diameter (ii) the absolute mean error and (iii) the percentage error in the measurement of diameter.
Answer:
Given: L.C. = 0.01 cm
To measure external diameter of hollow cylinder readings are taken as follows:

[Note: The above table is made assuming zero error in Vernier callipers. If calliper has positive or negative zero error, the zero error correction needs to be introduced into observed reading.]

Internet my friend (Textbook Page No. 12)

i. ideoiectures.net/mit801f99_lewin_lec0l/
ii. hyperphysicsphy-astr.gsu.ed u/libase/hfra me. html

[Students can use links given above as reference and collect information about units and measurements]

Class 11 Physics Questions And Answers

11th Std Physics Questions And Answers:

11th Physics Chapter 10 Exercise Electrostatics Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 10 Electrostatics Textbook Exercise Questions and Answers.

Electrostatics Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 10 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 10 Exercise Solutions

1. Choose the correct option.

Question 1.
A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be
(A) same as that on the glass rod and equal in quantity
(B) opposite to that on the glass of and equal in quantity
(C) same as that on the glass rod but lesser in quantity
(D) same as that on the glass rod but more in quantity
Answer:
(A) same as that on the glass rod and equal in quantity

Question 2.
An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will
(A) be attracted to the +ve plate
(B) be attracted to the -ve plate
(C) remain stationary
(D) will move parallel to the plates
Answer:
(A) be attracted to the +ve plate

Question 3.
A charge of + 7 µC is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be
(A) 1 : 4
(B) 1 : 2
(C) 1 : 1
(D) 1 : 16
Answer:
(C) 1 : 1

Question 4.
Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be
(A) 1.0 N
(B) 9 × 10⁹ N
(C) 9 × 10^-9 N
(D) 10 N
Answer:
(B) 9 × 10⁹ N

Question 5.
Two point charges of +5 µC are so placed that they experience a force of 80 × 10^-3 N. They are then moved apart, so that the force is now 2.0 × 10^-3 N. The distance between them is now
(A) 1/4 the previous distance
(B) double the previous distance
(C) four times the previous distance
(D) half the previous distance
Answer:
(B) double the previous distance

Question 6.
A metallic sphere A isolated from ground is charged to +50 µC. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two sphere will be now in the ratio
(A) 1 : 2
(B) 2 : 1
(C) 4 : 1
(D) 1 : 1
Answer:
(D) 1 : 1

Question 7.
Which of the following produces uniform electric field?
(A) point charge
(B) linear charge
(C) two parallel plates
(D) charge distributed an circular any
Answer:
(C) two parallel plates

Question 8.
Two point charges of A = +5.0 µC and B = -5.0 µC are separated by 5.0 cm. A point charge C = 1.0 µC is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards
(A) point A
(B) point B
(C) a direction parallel to line AB
(D) a direction along the perpendicular bisector
Answer:
(C) a direction parallel to line AB

2. Answer the following questions.

Question 1.
What is the magnitude of charge on an electron?
Answer:
The magnitude of charge on an electron is 1.6 × 10^-19 C

Question 2.
State the law of conservation of charge.
Answer:
In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant”
OR
For an isolated system, total charge cannot be created nor destroyed.

Question 3.
Define a unit charge.
Answer:
Unit charge (one coulomb) is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 × 10⁹ N.

Question 4.
Two parallel plates have a potential difference of 10V between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.
Answer:
V=10V
d = 0.5 mm = 0.5 × 10^-3 m
To find: The strength of electric field (E)
Formula: E = \(\frac {V}{d}\)
Calculation: From formula,
E = \(\frac {10}{0.5×10^{-3}}\)
20 × 10³ V/m

Question 5.
What is uniform electric field?
Answer:
A uniform electric field is a field whose magnitude and direction are same at all points. For example, field between two parallel plates as shown in the diagram.

Question 6.
If two lines of force intersect of one point. What does it mean?
Answer:
If two lines of force intersect of one point, it would mean that electric field has two directions at a single point.

Question 7.
State the units of linear charge density.
Answer:
SI unit of λ is (C / m).

Question 8.
What is the unit of dipole moment?
Answer:
i. Strength of a dipole is measured in terms of a quantity called the dipole moment.

ii. Let q be the magnitude of each charge and 2\(\vec{l}\) be the distance from negative charge to positive charge. Then, the product q(2\(\vec{l}\)) is called the dipole moment \(\vec{p}\).

iii. Dipole moment is defined as \(\vec{p}\) = q(2\(\vec{l}\))

iv. A dipole moment is a vector whose magnitude is q (2\(\vec{l}\)) and the direction is from the negative to the positive charge.

v. The unit of dipole moment is coulomb-metre (C m) or debye (D).

Question 9.
What is relative permittivity?
Answer:
i. Relative permittivity or dielectric constant is the ratio of absolute permittivity of a medium to the permittivity of free space.
It is denoted as K or ε_r.
i.e., K or ε_r = \(\frac {ε}{ε_0}\)

ii. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.
i.e., K or ε_r = \(\frac {F_{vacuum}}{F_{medium}}\)

iii. It is also called as specific inductive capacity or dielectric constant.

3. Solve numerical examples.

Question 1.
Two small spheres 18 cm apart have equal negative charges and repel each other with the force of 6 × 10^-8 N. Find the total charge on both spheres.
Solution:
Given: F = 6 × 10^-8 N, r = 18 cm = 18 × 10^-2 m
To find: Total charge (q₁ + q₂)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,

Taking square roots from log table,
∴ q = -4.648 × 10^-10 C
….(∵ the charges are negative)
Total charge = q₁ + q₂ = 2q
= 2 × (-4.648) × 10^-10
= -9.296 × 10^-10 C

Question 2.
A charge + q exerts a force of magnitude – 0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q.
Answer:
Given: q₁ = + q, q₂ = -2q, F = -0.2 N
r = 25 cm = 25 × 10^-2 m
To find: Charge (q)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,

[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 3.
Four charges of +6 × 10^-8 C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show in direction on a diagram drawn to scale.
Answer:
Given: q_A = q_B = q_C = q_D = 6 × 10^-8 C, a = 3 cm

∴ Resultant force on ‘A’
= F_AD cos 45 + F_AB cos 45 + F_AC
= (3.6 × 10^-2 × \(\frac {1}{√2}\)) + (3.6 × 10^-2 × \(\frac {1}{√2}\)) + 1.8 × 10^-2
= 6.89 × 10^-2 N directed along \(\vec{F_{AC}}\)

Question 4.
The electric field in a region is given by \(\vec{E}\) = 5.0 \(\hat{k}\) N/C Calculate the electric flux through a square of side 10.0 cm in the following cases
i. The square is along the XY plane
ii. The square is along XZ plane
iii. The normal to the square makes an angle of 45° with the Z axis.
Answer:
Given: \(\vec{E}\) = 5.0 \(\hat{k}\) N/C, |E| = 5 N/C
l = 10 cm = 10 × 10^-2 m = 10^-1 m
A = l² – 10^-2m²
To find: Electric flux in three cases.
(ø₁) (ø₂) (ø₃)
Formula: (ø₁) = EA cos θ
Calculation:
Case I: When square is along the XY plane,
∴ θ = 0
ø₁ = 5 × 10^-2 cos 0
= 5 × 10^-2 V m

Case II: When square is along XZ plane,
∴ θ = 90°
ø₁ = 5 × 10^-2 cos 90° = 0 V m

Case III: When normal to the square makes an angle of 45° with the Z axis.
∴ 0 = 45°
∴ ø₃ = 5 × 10^-2 × cos 45°
= 3.5 × 10^-2 V m

Question 5.
Three equal charges of 10 × 10^-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle.
Answer:
Given: q_A = q_B = q_C = 10 × 10^-8

Force on B due to A,

Question 6.
A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of 9.6 x 10-19 C falls between the plates. Find (i) electric field intensity between the plates and (ii) the force on the oil drop.
Answer:
Given: V = 5000 volt, d = 5 cm = 5 × 10^-2 m
q = 9.6 × 10^-19 C
To find:
i. Electric field intensity (E)
ii. Force (F)
Formula:
i. E = \(\frac {V}{d}\) \(\frac {q}{r}\)
ii. E = \(\frac {F}{q}\)
Calculation: From formula (i),
E = \(\frac {F}{q}\) = 10⁵ N/C
From formula (ii)
F = E x q
= 10⁵ × 9.6 × 10^-19
= 9.6 × 10^-14 N

Question 7.
Calculate the electric field due to a charge of -8.0 × 10^-8 C at a distance of 5.0 cm from it.
Answer:
Given: q = – 8 × 10^-8 C, r = 5 cm = 5 × 10^-2 m
To Find: Electric field (E)
Formula: E = \(\frac {1}{4πε_0}\) \(\frac {q}{r^2}\)
Calculation: From formula,
E = 9 × 10⁹ × \(\frac {(-8×10^{-8})}{(5×10^{-2})^2}\)
= -2.88 × 10⁵ N/C

11th Physics Digest Chapter 10 Electrostatics Intext Questions and Answers

Can you recall? (Textbookpage no. 188)

Question 1.
Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend?
Answer:
Yes, sometimes a shock while getting up from a plastic chair and shaking hand with friend is experienced.

Question 2.
Ever heard a crackling sound while taking out your sweater in winter?
Answer:
Yes, sometimes while removing our sweater in winter, some crackling sound is heard and the sweater appears to stick to body.

Question 3.
Have you seen the lightning striking during pre-monsoon weather?
Answer:
Yes, sometimes lightning striking during pre-pre-monsoon weather seen.

Can you tell? (Textbook page no. 189)

i. When a petrol or a diesel tanker is emptied in a tank, it is grounded.
ii. A thick chain hangs from a petrol or a diesel tanker and it is in contact with ground when the tanker is moving.
Answer:
i. When a petrol or a diesel tanker is emptied in a tank, it is grounded so that it has an electrically conductive connection from the petrol or diesel tank to ground (Earth) to allow leakage of static and electrical charges.

ii. Metallic bodies of cars, trucks or any other big vehicles get charged because of friction between them and the air rushing past them. Hence, a thick chain is hanged from a petrol or a diesel tanker to make a contact with ground so that charge produced can leak to the ground through chain.

Can you tell? (Textbook page no. 194)

Three charges, q each, are placed at the vertices of an equilateral triangle. What will be the resultant force on charge q placed at the centroid of the triangle?
Answer:

Since AD. BE and CF meets at O, as centroid of an equilateral triangle.
∴ OA = OB = OC
∴ Let, r = OA = OB = OC
Force acting on point O due to charge on point A,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OA}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{i}^{2}} \hat{\mathrm{r}}_{\mathrm{AO}}\)
Force acting on point O due to charge on point B,
\(\overrightarrow{\mathrm{F}}_{O \mathrm{~B}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{\mathrm{BO}}\)
Force acting on point O due to charge on point C,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OC}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \mathrm{co}\)
∴ Resultant force acting on point O,
F = \(\vec{F}\)_OA + \(\vec{F}\)_OB + \(\vec{F}\)_OC
On resolving \(\vec{F}\)_OB and \(\vec{F}\)_OC, we get –\(\vec{F}\)_OA
i.e., \(\vec{F}\)_OB + \(\vec{F}\)_OC = –\(\vec{F}\)_OA
∴ \(\vec{F}\) = \(\vec{F}\)_OA – \(\vec{F}\)_OA = 0
Hence, the resultant force on the charge placed at the centroid of the equilateral triangle is zero.

Can you tell? (Textbook page no. 197)

Why a small voltage can produce a reasonably large electric field?
Answer:

Electric field produced depends upon voltage as well as separation distance.
Electric field varies linearly with voltage and inversely with distance.
Hence, even if voltage is small, it can produce a reasonable large electric field when the gap between the electrode is reduced significantly.

Can you tell? (Textbook page no. 198)

Lines of force are imaginary; can they have any practical use?
Answer:
Yes, electric lines of force help us to visualise the nature of electric field in a region.

Can you tell? (Textbook page no. 204)

The surface charge density of Earth is σ = -1.33 nC/m². That is about 8.3 × 10⁹ electrons per square metre. If that is the case why don’t we feel it?
Answer:
The Earth along with its atmosphere acts as a neutral system. The atmosphere (ionosphere in particular) has nearly equal and opposite charge.

As a result, there exists a mechanism to replenish electric charges in the form of continual thunder storms and lightning that occurs in different parts of the globe. This makes average charge on surface of the Earth as zero at any given time instant. Hence, we do not feel it.

Internet my friend (Textbook page no. 205)

i. https://www.physicsclassroom.com/class
ii. https://courses.lumenleaming.com/physics/
iii. https://www,khanacademy.org/science
iv. https://www.toppr.com/guides/physics/
[Students are expected to visit the above mentioned websites and collect more information about electrostatics.]

11th Std Physics Questions And Answers:

11th Physics Chapter 5 Exercise Gravitation Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 5 Gravitation Textbook Exercise Questions and Answers.

Gravitation Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 5 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 5 Exercise Solutions

1. Choose the correct option.

Question 1.
The value of acceleration due to gravity is maximum at
(A) the equator of the Earth .
(B) the centre of the Earth.
(C) the pole of the Earth.
(D) slightly above the surface of the Earth.
Answer:
(C) the pole of the Earth.

Question 2.
The weight of a particle at the centre of the Earth is _________
(A) infinite.
(B) zero.
(C) same as that at other places.
(D) greater than at the poles.
Answer:
(B) zero.

Question 3.
The gravitational potential due to the Earth is minimum at
(A) the centre of the Earth.
(B) the surface of the Earth.
(C) a points inside the Earth but not at its centre.
(D) infinite distance.
Answer:
(A) the centre of the Earth.

Question 4.
The binding energy of a satellite revolving around planet in a circular orbit is 3 × 10⁹ J. Its kinetic energy is _________
(A) 6 × 10⁹ J
(B) -3 × 10⁹ J
(C) -6 × 10⁺⁹ J
(D) 3 × 10⁺⁹J
Answer:
(D) 3 × 10⁺⁹J

2. Answer the following questions.

Question 1.
State Kepler’s law equal of area.
Answer:
The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.

Question 2.
State Kepler’s law of period.
Answer:
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.

Question 3.
What are the dimensions of the universal gravitational constant?
Answer:
The dimensions of universal gravitational constant are: [L³M^-1T^-2].

Question 4.
Define binding energy of a satellite.
Answer:
The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.

Question 5.
What do you mean by geostationary satellite?
Answer:
Some satellites that revolve around the Earth in equatorial plane have same sense of rotation as that of the Earth. The also have the same period of rotation as that of the Earth i.e.. 24 hours. Due to this, these satellites appear stationary from the Earth’s surface and are known as geostationary satellites.

Question 6.
State Newton’s law of gravitation.
Answer:
Statement:
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 7.
Define escape velocity of a satellite.
Answer:
The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (v_e) of the body.

Question 8.
What is the variation in acceleration due to gravity with altitude?
Answer:
Variation in acceleration due to gravity due to altitude is given by, g_h = g\(\left(\frac{R}{R+h}\right)^{2}\)
where,
g_h = acceleration due to gravity of an object placed at h altitude
g = acceleration due to gravity on surface of the Earth
R = radius of the Earth
h = attitude height of the object from the surface of the Earth.
Hence, acceleration due to gravity decreases with increase in altitude.

Question 9.
On which factors does the escape speed of a body from the surface of Earth depend?
Answer:
The escape speed depends only on the mass and radius of the planet.
[Note: Escape velocity does not depend upon the mass of the body]

Question 10.
As we go from one planet to another planet, how will the mass and weight of a body change?
Answer:

As we go from one planet to another, mass of a body remains unaffected.
However, due to change in mass and radius of planet, acceleration due to gravity acting on the body changes as, g ∝ \(\frac{\mathrm{M}}{\mathrm{R}^{2}}\).
Hence, weight of the body also changes as, W ∝ \(\frac{\mathrm{M}}{\mathrm{R}^{2}}\)

Question 11.
What is periodic time of a geostationary satellite?
Answer:
The periodic time of a geostationary satellite is same as that of the Earth i.e., one day or 24 hours.

Question 12.
State Newton’s law of gravitation and express it in vector form.
Answer:

Statement:
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
In vector form, it can be expressed 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.

Question 13.
What do you mean by gravitational constant? State its SI units.
Answer:

From Newton’s law of gravitation,
F = G \(\frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
where, G = constant called universal gravitational constant Its value is 667 X 10^-11 N m²/kg².
G = \(\frac{\mathrm{Fr}^{2}}{\mathrm{~m}_{1} \mathrm{~m}_{2}}\)
If m₁ = m₂ = 1 kg, r = 1 m thenF = G.
Hence, the universal gravitational constant is the force of gravitation between two particles of unit mass separated by unit distance.
Unit: N m²/kg² in SI system.

Question 14.
Why is a minimum two stage rocket necessary for launching of a satellite?
Answer:

For the projection of an artificial satellite, it is necessary for the satellite to have a certain velocity.
In a single stage rocket, when the fuel in first stage of rocket is ignited on the surface of the Earth, it raises the satellite vertically.
The velocity of projection of satellite normal to the surface of the Earth is the vertical velocity.
If this vertical velocity is less than the escape velocity (v_e), the satellite returns to the Earth’s surface. While, if the vertical velocity is greater than or equal to the escape velocity, the satellite will escape from Earth’s gravitational influence and go to infinity.
Hence, minimum two stage rocket, one to raise the satellite to desired height and another to provide required hori7ontal velocity, is necessary for launching of a satellite.

Question 15.
State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection
Answer:
The path of the satellite depends upon the value of horizontal speed of projection v_h relative to critical velocity v_c and escape velocity v_e.
Case (I) v_h < v_c:
The orbit of satellite is an ellipse with point of projection as apogee and Earth at one of the foci. During this elliptical path, if the satellite passes through the Earth’s atmosphere. it experiences a nonconservative force of air resistance. As a result it loses energy and spirals down to the Earth.
Case (II) v_h = v_c:
The satellite moves in a stable circular orbit around the Earth.
Case (III) v_c < v_h < v_e:
The satellite moves in an elliptical orbit round the Earth with the point of projection as perigee.
Case (IV) v_h = v_e
The satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity.
Case (V) v_h > v_e:
The satellite escapes from gravitational influence of Earth traversing a hyperbolic path.

3. Answer the following questions in detail.

Question 1.
Derive an expression for critical velocity of a satellite.
Answer:
Expression for critical velocity:

Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity v_c.
The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
∴ Centripetal force = Gravitational force

This is the expression for critical speed at the orbit of radius (R + h).
The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.

Question 2.
State any four applications of a communication satellite.
Answer:
Applications of communication satellite:

For the transmission of television and radiowave signals over large areas of Earth’s surface.
For broadcasting telecommunication.
For military purposes.
For navigation surveillance.

Question 3.
Show that acceleration due to gravity at height h above the Earth’s surface is g_h = g(\(\frac{R}{R+h}\))²
Answer:
Variation of g due to altitude:

Let,
R = radius of the Earth,
M = mass of the Earth.
g = acceleration due to gravity at the surface of the Earth.
Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is given by,
The body is taken at height h above the surface of the Earth as shown in figure. The acceleration due to gravity now changes to,
g_h = \(\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}\) …………. (2)
Dividing equation (2) by equation (1), we get,

We can rewrite,

This expression can be used to calculate the value of g at height h above the surface of the Earth as long as h<< R.

Question 4.
Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
Answer:

v_h = horizontal speed of projection
v _c = critical velocity
v_e = escape velocity

Question 5.
Derive an expression for binding energy of a body at rest on the Earth’s surface.
Answer:

Let,
M = mass of the Earth
m = mass of the satellite
R = radius of the Earth.
Since the satellite is at rest on the Earth, v = 0
∴ Kinetic energy of satellite.
K.E = \(\frac{1}{2}\) mv² = 0
Gravitational potential at the Earth’s surface
= – \(\frac{\mathrm{GM}}{\mathrm{R}}\)
∴ Potential energy of satellite = Gravitational potential × mass of satellite
= –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Total energy of sitellite = T.E = P.E + K.E
∴ T.E. = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\) + 0 = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Negative sign in the energy indicates that the satellite is bound to the Earth, due to gravitational force of attraction.
For the satellite to be free form Earth’s gravitational influence, its total energy should become positive. That energy is the binding energy of the satellite at rest on the surface of the Earth.
∴ B.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\)

Question 6.
Why do astronauts in an orbiting satellite have a feeling of weightlessness?
Answer:

For an astronaut, in a satellite, the net force towards the centre of the Earth will always be, F = mg – N.
where, N is the normal reaction.
In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.
In this case, the downward acceleration, a_d = g, or the satellite (along with the astronaut) is in the state of free fall.
Thus, the net force acting on astronaut will be, F = mg – ma_d i.e., the apparent weight will be zero, giving the feeling of total weightlessness.

Question 7.
Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth’s surface.
Answer:

Question 8.
At which place on the Earth’s surface is the gravitational acceleration maximum? Why?
Answer:

Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
At poles, latitude θ = 90°.
∴ g’ = g
i.e., there is no reduction in acceleration due to gravity at poles.
Also, shape of the Earth is actually an ellipsoid, bulged at equator. The polar radius of the Earth is 6356 km which minimum. As g ∝ \(\frac{1}{\mathrm{R}^{2}}\), acceleration due to gravity is maximum at poles i.e., 9.8322 m/s².

Question 9.
At which place on the Earth surface the gravitational acceleration minimum? Why?
Answer:

Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
At equator, latitude θ = 0°.
∴ g’ = g – Rω²
i.e., the acceleration due to gravity ¡s reduced by amount Rω²(≈ 0.034 m/s²) at equator.
Also, shape of the Earth is actually an ellipsoid, bulged at equator. The equatorial radius of the Earth is 6378 km, which is maximum. As g ∝ \(\frac{1}{\mathrm{R}^{2}}\) acceleration due to gravity is minimum on equator i.e., 9.7804 m/s².

Question 10.
Define the binding energy of a satellite. Obtain an expression for binding energy of a satellite revolving around the Earth at certain attitude.
Answer:
The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.
Expression for binding energy of satellite revolving in circular orbit round the Earth:

Consider a satellite of mass m revolving at height h above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy.
Let M be the mass of the Earth, R be the Radius of the Earth, v_c be critical velocity of satellite, r = (R + h) be thc radius of the orbit.
Kinetic energy of satellite = \(\frac{1}{2} \mathrm{mv}_{\mathrm{c}}^{2}=\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\)
The gravitational potential at a distance r from the centre of the Earth is –\(\frac{\mathrm{GM}}{\mathrm{r}}\)
∴ Potential energy of satellite = Gravitational potential × mass of satellite
= –\(\frac{\mathrm{GMm}}{\mathrm{r}}\)
The total energy of satellite is given as T.E. = KF. + P.E.
= \(\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}-\frac{\mathrm{GMm}}{\mathrm{r}}=-\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\)
Total energy of a circularly orbiting satellite is negative. Negative sign indicates that the satellite is bound to the Earth, due to gravitational force of attraction. For the satellite to be free from the Earth’s gravitational influence its total energy should become zero or positive.
Hence the minimum energy to be supplied to unbind the satellite is +\(\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{r}}\). This is the binding energy of a satellite.

Question 11.
Obtain the formula for acceleration due to gravity at the depth ‘d’ below the Earth’s surface.
Answer:

The Earth can be considered to be a sphere made of large number of concentric uniform spherical shells.
When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.
The acceleration due to gravity on the surface of the Earth is, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Assuming that the density of the Earth is uniform, mass of the Earth is given by
M = volume x density = \(\frac{4}{3}\) πR³ρ
∴ g = \(\frac{\mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho}{\mathrm{R}^{2}}\) = \(\frac{4}{3}\) πRρG ………….. (1)
Consider a body at a point P at the depth d below the surface of the Earth as shown in figure.

Here the force on a body at P due to outer spherical shell shown by shaded region, cancel out due to symmetry.
The net force on P is only due to the inner sphere of radius OP = R – d.
Acceleration due to gravity because of this sphere is,

This equation gives acceleration due to gravity at depth d below the Earth’s surface.

Question 12.
State Kepler’s three laws of planetary motion.
Answer:

All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.
The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
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.

Question 13.
State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’ Hence show that their ratio is equal to \(\left(\frac{R-d}{R-2 h}\right)\) by assuming that the altitude is very small as compared to the radius of the Earth.
Answer:

For an object at depth d, acceleration due to gravity of the Earth is given by,
g_d = g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) ………………. (1)
Also, the acceleration due to gravity at smaller altitude h is given by,
g_h = g(1 – \(\frac{2 \mathrm{~h}}{\mathrm{R}}\)) ……………. (2)
Hence, dividing equation (1) by equation (2),
we get,

Question 14.
What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?
Answer:
The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbIt round the Earth is called the critical velocity or orbital velocity (v_c).
Expression for critical velocity:

Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity v_c.
The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
∴ Centripetal force = Gravitational force

This is the expression for critical speed at the orbit of radius (R + h).
The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.

Question 15.
Define escape speed. Derive an expression for the escape speed of an object from the surface of the earth.
Answer:

The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (v_e) of the body.
As the gravitational force due to Earth becomes zero at infinite distance, the object has to reach infinite distance in order to escape.
Let us consider the kinetic and potential energies of an object thrown vertically upwards with escape velocity v_e.
On the surface of the Earth,
K.E.= \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}\)
P.E. = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Total energy = P.E. + K.E.
∴ T.E. = \(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}\) ………………. (1)
The kinetic energy of the object will go on decreasing with time as it is pulled back by Earth’s gravitational force. It will become zero when it reaches infinity. Thus, at infinite distance from the Earth,
K.E. = 0
Also,
P.E. = –\(\frac{\mathrm{GMm}}{\infty}\) = 0
∴ Total energy = P.E. + K.E. = 0
As energy is conserved
\(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}=0\) ……[From(1)]
or, v_e = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)

Question 16.
Describe how an artificial satellite using two stage rocket is launched in an orbit around the Earth.
Answer:

Launching of a satellite in an orbit around the Earth cannot take place by use of single stage rocket. It requires minimum two stage rocket.
With the help of first stage of rocket, satellite can be taken to a desired height above the surface of the Earth.
Then the launcher is rotated in horizontal direction i.e.. through 900 using remote control and the first stage of the rocket is detached.
With the help of second stage of rocket, a specific horizontal velocity (v_h) is given to satellite so that it can revolve in a circular path around the Earth.
The satellite follows different paths depending upon the horizontal velocity provided to it.

4. Solve the following problems.

Question 1.
At what distance below the surface of the Earth, the acceleration due to gravity decreases by 10% of its value at the surface, given radius of Earth is 6400 km.
Solution:
Given: g_d = 90% of g i.e., \(\frac{\mathrm{g}_{\mathrm{d}}}{\mathrm{g}}\) = 0.9,
R = 6400km = 6.4 × 10⁶ m
To find: Distance below the Earth’s surface (d)

At distance 640 km below the surface of the Earth, value of acceleration due to gravity decreases by 10%.

Question 2.
If the Earth were made of wood, the mass of wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate escape speed from the surface of this Earth.
Solution:

As, we know that the escape speed from surface of the Earth is 11.2 km/s, Substituting value of v_e = 11.2 km/s
V_e_w = 11.2 × \(\frac{1}{\sqrt{10}}=\frac{11.2}{3.162}\)
= 11.2 × \(\frac{1}{3.162}\)
…………… [Taking square root value]
= antilog {log(1 1.2) – Log(3.162)}
= antilog {1.0492 – 0.5000}
= antilog {0.5492} = 3.542
∴ V_e_w = 3.54km/s
The escape velocity from the surface of wooden Earth is 3.54 km/s.

Question 3.
Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth.
Given:- G = 6.67 × 10^-11 Nm²/kg²
R = 6400 km
M = 6 × 10²⁴ kg
Solution:
Given:- m = 2000 kg, h = 3600 km = 3.6 × 10⁶ m,
G = 6.67 × 10^-11 Nm²/kg²
R = 6400 km
M = 6 × 10²⁴ kg

To find: i) Kninetic energy (K.E.)
ii) Potential Energy (P.E.)
iii) Total Energy (T.E.)
iv) Binding Energy (B.E.)

From formula (ii),
P.E. = -2 × 40.02 × 10⁹
= -80.04 × 10⁹ J
From formula (iii),
T.E. = (40.02 × 10⁹) + (-80.02 × 10⁹)
= -40.02 × 10⁹ J
From formula (iv),
B.E.= -(-40.02 × 10⁹)
= 40.02 × 10⁹ J
Kinetic energy of the satellite is 40.02 × 10⁹ J, potential energy is -80.04 × 10⁹ J, total energy is -40.02 × 10⁹ J and binding energy is 40.02 × 10⁹ J.
[Note: Total energy of orbiting satellite is negative.]

Question 4.
Two satellites A and B are revolving around a planet. Their periods of revolution are 1 hour and 8 hours respectively. The radius of orbit of satellite B is 4 × 10⁴ km. find radius of orbit of satellite A .
Solution:
Given: T_A = 1 hour, T_B = 8 hour,
r_B = 4 × 10⁴ km
To find: Radius of orbit of satellite A (r_A)

Radius of orbit of satellite A will be 1 × 10⁴ km.

Question 5.
Find the gravitational force between the Sun and the Earth.
Given Mass of the Sun = 1.99 × 10³⁰ kg
Mass of the Earth = 5.98 × 10²⁴ kg
The average distance between the Earth and the Sun = 1.5 × 10¹¹ m.
Solution:
Given: M_S = 1.99 × 10³⁰ kg
M_E = 5.98 × 10²⁴ kg, R = 1.5 × 10¹¹ m.
To find: Gravitational force between the Sun and the Earth (F)
Formula: F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)
Calculation:As, we know, G = 6.67 × 10^-11 N m²/kg²
From formula,

= antilog {(log(6.67) + log( 1.99) + log(5.98) – log(2.25)} × 10²¹
= antilog {(0.8241) + (0.2989) + (0.7767) – (0.3522)} × 10²¹
= antilog {1.5475} × 10²¹
= 35.28 × 10²¹
= 3.5 × 10²² N
The gravitational force between the Sun and the Earth is 3.5 × 10²² N.

Question 6.
Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 × 10²⁴ kg, R = 6400 km).
Solution:
Given: h = 300 km = 0.3 × 10⁶ m,
M = 5.98 × 10²⁴ kg,
R = 6400km = 6.4 × 10⁶ m
G = 6.67 × 10^-11 Nm²/kg²
To find: Acceleration due to gravity at height (g_h)
Formula: g_h = \(\frac{G M}{(R+h)^{2}}\)

Calculation: From formula,
g_h = \(\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{\left[\left(6.4 \times 10^{6}\right)+\left(0.3 \times 10^{6}\right)\right]^{2}}\)
= \(\frac{6.67 \times 5.98 \times 10^{13}}{(6.7)^{2} \times 10^{12}}\)
6.67 X 10” x 5.98 X iO
= antilog {log(6.67) + log(5.98) – 2log(6.7)} × 10
= antilog{0.8241 + 0.7767 – 2(0.8261)} × 10
= antilog {1.6008 – 1.6522} × 10
= antilog {\(\overline{1}\) .9486} × 10
= 0.8884 × 10 = 8.884 m/s²
Acceleration due to gravity at 300 km will be 8.884 m/s².

Question 7.
Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface. M_E = 5.98 × 10²⁴ kg, R = 6.4 × 10⁶ m.
Solution:
Given: h = 1000 km = 1 × 10⁶ m,
ME = 5.98 × 10²⁴ kg, R = 6.4 × 10⁶ m,
G = 6.67 × 10^-11 N m²/kg²
To find: Speed of satellite (v_c)

Speed of the satellite at height 1000 km is 7.34 × 10³ m/s.

Question 8.
Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars = 3.4 × 10³ km and its mass is 6.4 × 10²³ kg.
Solution:
Given:
M = 6.4 × 10²³ kg
R = 3.4 × 10³ = 3.4 × 10⁶ m,
To find: Acceleration due to gravity on the surface of the Mars (g_M)
Formula: g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: As, G = 6.67 × 10^-11 N m²/kg²
From formula,
g_M = \(\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{\left(3.4 \times 10^{6}\right)^{2}}=\frac{6.67 \times 6.4}{3.4 \times 3.4}\)
= antilog {log(6.67) + log(6.4) – log(3.4) – log(3.4)}
= antilog {(0.8241) + (0.8062) – (0.5315) – (0.53 15)}
= antilog {0.5673}
= 3.693 m/s²
Acceleration due to gravity on the surface of Mars is 3.693 m/s².

Question 9.
A planet has mass 6.4 × 10²⁴ kg and radius 3.4 × 10⁶ m. Calculate energy required to remove on object of mass 800 kg from the surface of the planet to infinity.
Solution:
Given: M = 6.4 × 10²⁴ kg, R = 3.4 × 10⁶ m, m = 800 kg
To find: Energy required to remove the object from surface of planet to infinity = B.E.
Formula: B.E. = \(\frac{\mathrm{GMm}}{\mathrm{R}}\)
Calculation: We know that,
G = 6.67 × 10^-11 N m²/kg²
From formula,

= antilog{log(6.67) + log(51.2) – log(3.4)} × 10⁹
= antilog{0.8241 + 1.7093 – 0.5315} × 10⁹
= antilog {2.0019} × 10⁹
= 1.004 × 10² × 10⁹
= 1.004 × 10¹¹ J
Energy required to remove the object from the surface of the planet is 1.004 × 10¹¹ J.
[Note: Answer calculated above ¡s in accordance with retual methods of calculation.]

Question 10.
Calculate the value of the universal gravitational constant from the given data. Mass of the Earth = 6 × 10²⁴ kg, Radius of the Earth = 6400 km and the acceleration due to gravity on the surface = 9.8 m/s²
Solution:
Given: M = 6 × 10²⁴ kg,
R = 6400km = 6.4 × 10⁶ m,
g = 9.8 m/s²
To find: Gravitational constant (G)
Formula. g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Calculation: From formula,
G = \(\frac{\mathrm{gR}^{2}}{\mathrm{M}}\)
G = \(\frac{9.8 \times\left(6.4 \times 10^{6}\right)^{2}}{6 \times 10^{24}}=\frac{401.4 \times 10^{12}}{6 \times 10^{24}}\)
∴ G = 6.69 × 10^-11 N m²/kg²
The value of gravitational constant is 6.69 × 10^-11 N m²/kg².

Question 11.
A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 1/7 times the mass of the Earth and radius twice that of the Earth’s radius.
Solution:
Given: W_E = 5.6 kg-wt.,
\(\frac{\mathrm{M}_{\mathrm{p}}}{\mathrm{M}_{\mathrm{E}}}=\frac{1}{7}, \frac{\mathrm{R}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{E}}}\) = 2
To find: Weight of the body on the surface of planet (W_p)
Formula: W = mg = \(\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\)
Calculation: From formula,

Weight of the body on the surface of a planet will be 0.2 kg-wt.
[Note: The answer given above is calculated in accordance with textual method considering the given data].

Question 12.
What is the gravitational potential due to the Earth at a point which is at a height of 2R_E above the surface of the Earth, Mass of the Earth is 6 × 10²⁴ kg, radius of the Earth = 6400 km and G = 6.67 × 10^-11 Nm² kg^-2.
Solution:
Given: M = 6 × 10²⁴ kg,
R_E = 6400km = 6.4 × 10⁶ m,
G = 6.67 × 10^-11 Nm²/kg²,
h = 2R_E
To find: Gravitational potential (V)
Formula: V = – \(\frac{\mathrm{GM}}{\mathrm{r}}\)
Calculation: From formula,

= -2.08 × 10⁷ J kg^-1
Negative sign indicates the attractive nature of gravitational potential.
Gravitational potential due to Earth will be 2.08 × 10⁷ J kg^-1 towards the centre of the Earth.
[Note: According lo definition of gravitational potential its SI unit is J/kg.]

11th Physics Digest Chapter 5 Gravitation Intext Questions and Answers

Can you recall? (Textbook Page No. 78)

Question 1.
i) What are Kepler’s laws?
ii)What is the shape of the orbits of planets?
Answer:

The Kepler’s laws are:
- Kepler’s first law: The orbit of a planet is an ellipse with the Sun at one of the foci.
- Kepler’s second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
- Kepler’s third law: The square of orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
The orbits of the planet are elliptical in shape.

Question 2.
When released from certain height why do objects tend to fall vertically downwards?
Answer:
When released from certain height, objects tend to fall vertically downwards because of the gravitational force exerted by the Earth.

11th Std Physics Questions And Answers:

11th Chemistry Chapter 13 Exercise Nuclear Chemistry and Radioactivity Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 13 Nuclear Chemistry and Radioactivity Textbook Exercise Questions and Answers.

Nuclear Chemistry and Radioactivity Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 13 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 13 Exercise Solutions

1. Choose correct option.

Question A.
Identify nuclear fusion reaction

Answer:
Among the given options, reactions (i) and (ii) represent nuclear fusion reaction wherein lighter nuclei combine to form a heavy nucleus.

Question B.
The missing particle from the nuclear reaction is

Answer:
(A) \({ }_{15}^{30} \mathrm{P}\)

Question C.
\({ }_{27}^{60} \mathrm{CO}\) decays with half-life of 5.27 years to produce \({ }_{28}^{60} \mathrm{Ni}\). What is the decay constant for such radioactive disintegration ?
a. 0.132 y^-1
b. 0.138
c. 29.6 y
d. 13.8%
Answer:
a. 0.132 y^-1

Question D.
The radioactive isotope used in the treatment of Leukemia is
a. ⁶⁰Co
b. ²²⁶Ra
c. ³²P
d. ¹³¹I
Answer:
c. ³²P

Question E.
The process by which nuclei having low masses are united to form nuclei with large masses is
a. chemical reaction
b. nuclear fission
c. nuclear fusion
d. chain reaction
Answer:
c. nuclear fusion

2. Explain

Question A.
On the basis of even-odd of protons and neutrons, what type of nuclides are most stable ?
Answer:

Nuclides with even number of protons (Z) and even number of neutrons (N) are most stable.
These nuclides tend to form proton-proton and neutron-neutron pairs.
This impart stability to the nucleus.

Question B.
Explain in brief, nuclear fission.
Answer:
i. Nuclear fission: It is a process which involves splitting of the heavy nucleus of an atom into two nearly equal fragments accompanied by release of the large amount of energy.
e.g. Nuclear fission of ²³⁵U

ii. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation. This can be given as,

iii. Characteristics of nuclear fission reactions:

The mass of the fission products is less than the parent nucleus. A large amount of energy corresponding to the mass loss is released in each fission.
When one uranium 235 nucleus undergoes fission, three neutrons are emitted, which subsequently disintegrate three more uranium nuclei and thereby produce nine neutrons. Such a chain continues by itself.
In a very short time enormous amount of energy is liberated, which can be utilized for destructive or peaceful purposes.
Energy released per fission is approximately 200 MeV.

Note:

Each fission may lead to different products.
There is no unique way for fission of ²³⁵U that produces Ba and Kr. There are 400 ways for fission of ²³⁵U leading to 800 fission products.
Many of these fission products are radioactive which undergo spontaneous disintegrations giving rise to new elements in the periodic table.

Question C.
The nuclides with odd number of both protons and neutrons are the least stable. Why ?
Answer:

The nuclides with odd number of both protons and neutrons are the least stable because, odd number of protons and neutrons results in the presence of two unpaired nucleons.
These unpaired nucleons result in instability. Hence, such nuclides are the least stable.

Question D.
Referring the stabilty belt of stable nuclides, which nuclides are β^– and β⁺ emitters ? Why ?
Answer:

Beta decay occurs when an unstable nucleus emits a beta particle and energy. A beta particle is either an electron or a positron. An electron is a negatively charged particle, and a positron is a positively charged electron (or anti-electron).
When the beta particle is an electron, the decay is called beta-minus (β^–) decay. In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus.
When the beta particle is a positron, the decay is called beta-plus (β⁺) decay. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Thus, beta-minus decay occurs when a nucleus has too many neutrons relative to protons (i.e., N/Z > 1) and beta-plus decay occurs when a nucleus has too few neutrons relative to protons (i.e., N/Z < 1).
By referring the stability belt of stable nuclides, nuclides with N/Z > 1 are to the left of the stability zone. Such nuclides are beta-minus emitters as they become stable when a neutron converts to a proton.
Nuclides with N/Z < 1 are to the right of the stability zone. Such nuclides are beta-plus emitters as they become stable when a proton converts to a neutron.

Question E.
Explain with an example each nuclear transmutation and artifiacial radioactivity. What is the difference between them ?
Answer:
i. Nuclear transmutation: It involves transformation of a stable nucleus into another nucleus takes place which can be either stable or unstable.
ii. Artificial (induced) radioactivity: It is nuclear transmutation where the product nucleus is radioactive. The product nucleus decays spontaneously with emission of radiation and particles.

Step-I can be considered as nuclear transmutation as it produces a new nuclide \({ }_{7}^{13} \mathrm{~N}\).
However, the new nuclide is unstable (radioactive). Hence, step-I involves artificial (induced) radioactivity. Thus, in artificial transmutation, a stable element is collided with high speed particles to form another radioactive element.

Question F.
What is binding energy per nucleon ? Explain with the help of diagram how binding energy per nucleon affects nuclear stability ?
Answer:

i. Binding energy per nucleon (\(\overline{\mathrm{B}}\)), for nucleus containing (A) nucleons with binding energy (B.E.) is given as,
\(\overline{\mathrm{B}}\) = B.E./A
ii. Mean binding energy per nucleon (\(\overline{\mathrm{B}}\)) for the most stable isotopes as a function of mass number is shown above. This plot leads to the following inferences:
a. Light nuclides: (A < 30)
The peaks with A values in multiples of 4. For example, \({ }_{2}^{4} \mathrm{He},{ }_{6}^{12} \mathrm{C},{ }_{8}^{16} \mathrm{O}\) are more stable.
b. Medium mass nuclides: (30 < A < 90)
\(\overline{\mathrm{B}}\) increases typically from 8 MeV for A = 16 to nearly 8.3 MeV for A between 28 and 32 and it remains nearly constant 8.5 MeV beyond this and shows a broad maximum. The nuclides falling on the maximum are most stable which turns possess high values. ⁵⁶Fe with \(\overline{\mathrm{B}}\) value of 8.79 MeV is the most stable.
c. Heavy nuclides (A > 90)
\(\overline{\mathrm{B}}\) decreases from maximum 8.79 MeV to 7.7 MeV for A ≅ 210, 209Bi is the stable nuclide. Beyond this, all nuclides are radioactive (α-emitters).

Question G.
Explain with example α-decay.
Answer:
i. The emission of α-particle from the nuclei of an radioelement is called α-decay.
ii. The charge on an α-particle is +2 with a mass of 4 u.
It is identical with helium nucleus and hence an α-particle is designated as \({ }_{2}^{4} \mathrm{He}\).
iii. In the α-decay process, the parent nucleus \({ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{X}\) emits an α-particle and produces daughter nucleus Y. The parent nucleus thus loses two protons (charge +2) and two neutrons. The total mass lost is 4 u. The daughter nucleus will therefore, have mass 4 units less and charge 2 units less than its parent.
iv. General equation for α-decay process can be given as:

In α-decay process of radium, radon (daughter nuclei) is formed with loses of two protons (charge +2) and two neutrons. The total mass lost is 4 u.
Thus, radon has a mass of 4 units less and charge 2 units less than its parent radium.

Question H.
Energy produced in nuclear fusion is much larger than that produced in nuclear fission. Why is it difficult to use fusion to produce energy ?
Answer:

Nuclear fusion involves the fusion of lighter nuclei to form a heavy nucleus which is accompanied by an enormous amount of energy (heat).
Fusion reaction requires extremely high temperature typically of the order of 108 K.

Question I.
How does N/Z ratio affect the nuclear stability ? Explain with a suitable diagram.
Answer:

When the graph of number of neutrons (N) against protons (Z) is drawn, and all the stable isotopes are plotted on it, there is quite a clear correlation between N and Z. This graph is shown in the adjacent figure.
A large number of elements have several stable isotopes and hence, the curve appears as a belt or zone called stability zone. All stable nuclides fall with this zone and the nuclei that are to the left or to the right of the stability zone are unstable and exhibit radioactivity. Below the belt, a straight line which represents the ratio N/Z to be nearly unity (i.e., N = Z) is shown.
For nuclei lighter than \({ }_{20}^{40} \mathrm{Ca}\), the straight line (N = Z) passes through the belt. The lighter nuclides are therefore stable (N/Z being 1).
The N/Z ratio for the stable nuclides heavier than calcium gives a curved appearance to the belt with gradual increase of N/Z (> 1). The heavier nuclides therefore, need more number of neutrons than protons to attain stability. The heavier nuclides with increasing number of protons render large coulombic repulsions. With increased number of neutrons, the protons within the nuclei get more separated, which renders them stable.
Thus, nuclear stability is linked to the number of nucleons (neutrons and protons). In general, the lighter stable nuclei have equal numbers of protons and neutrons while heavier stable nuclei have increasingly more neutrons than protons.

[Note: Atoms with unstable nuclei are radioactive (exhibit radioactivity). To become more stable, the nuclei undergo radioactive decay.]

Question J.
You are given a very old sample of wood. How will you determine its age ?
Answer:
The age of the wood sample can be determined by radiocarbon dating as ¹⁴C becomes a part of a plant due to the photosynthesis reaction (i.e., absorption of [¹⁴CO₂ + ¹²CO₂]).
i. The activity (N) of given wood sample and that of fresh sample of live plant (N₀) is measured, where, N₀ denotes the activity of the given sample at the time of death.
ii. The age of the given wood sample. can be determined by applying following Formulae:

Note: The oldest rock found so far in Northern Canada is 3.96 billion years old.

3. Answer the following question

Question A.
Give example of mirror nuclei.
Answer:
Example of mirror nuclei: \({ }_{1}^{3} \mathrm{H}\) and \({ }_{2}^{3} \mathrm{He}\)

Question B.
Balance the nuclear reaction:

Answer:

Question C.
Name the most stable nuclide known. Write two factors responsible for its stability.
Answer:
The most stable nuclide known is lead (\({ }_{82}^{208} \mathrm{~Pb}\)).
Two factors responsible for its stability are as follows:

It is a nuclide with even number of both protons (Z) and neutrons (N).
It has two magic numbers i.e., 82 (for protons) and 126 (for neutrons).

Question D.
Write relation between decay constant of a radioelement and its half life.
Answer:
Relation between decay constant of a radioelement and its half-life is given as, λ = \(\frac{0.693}{\mathrm{t}_{1 / 2}}\)
Where, λ = Decay constant, t_1/2 = Half-life of a radioelement

Question E.
What is the difference between an α-particle and helium atom ?
Answer:

Helium atom is composed of 2 protons and 2 neutrons (or 1 neutron) along with 2 electrons in the outer shell.
On the other hand, α-particle constitutes 2 protons and 2 neutrons bound together to form a particle which is similar to helium (except presence of electrons).
Helium is one of the inert gas which is stable (duplet complete) whereas α-particle is unstable and highly reactive.

Question F.
Write one point that differentiates nuclear reations from chemical reactions.
Answer:
Chemical reactions:

Rearrangement of atoms by breaking and forming of chemical bonds.
Different isotopes of an element have same behaviour.

Nuclear reactions:

Elements or isotopes of one element are converted into another element in a nuclear reaction.
Isotopes of an element behave differently.

Question G.
Write pairs of isotones and one pair of mirror nuclei from the following :

Answer:
Isotones: i. \({ }_{5}^{10} \mathrm{~B} \text { and }{ }_{6}^{11} \mathrm{C}\)
ii. \({ }_{13}^{27} \mathrm{Al} \text { and }{ }_{14}^{28} \mathrm{~S}\)
Mirror nuclei: Since there are no isobars the given set of nuclides does not contain a pair of mirror nuclei.

Question H.
Derive the relationship between half life and decay constant of a radioelement.
Answer:
Equation for the decay constant is given as,
λ = \(\frac{2.303}{t} \log _{10} \frac{\mathrm{N}_{0}}{\mathrm{~N}}\) …(i)
Where, λ = Decay constant
N = Number of nuclei (atoms) present at time t
At t = 0, N = N₀.
Hence, at t = t_1/2, N = N₀/2
Substitution of these values of N and t in equation (i) gives,

Question I.
Represent graphically log₁₀ (activity /dps) versus t/s. What is its slope ?
Answer:
Equation for a decay constant (λ) is given as,

Hence, instead if log₁₀N versus t, log₁₀ \(\left(\frac{-\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right)\) which is log₁₀ (activity) is plotted.
The graph of log₁₀ (activity/dps) versus t/s gives a straight line which can be represented as follows:

Question J.
Write two units of radioactivity. How are they interrelated ?
Answer:
The unit of radioactivity is curie (Ci).
1 Ci = 3.7 × 10¹⁰ dps
ii. Other unit of radioactivity is Becquerel (Bq).
1 Bq = 1 dps
Thus, 1 Ci = 3.7 × 10¹⁰ dps = 3.7 × 10¹⁰ Bq

Question K.
Half life of ²⁴Na is 900 minutes. What is its decay constant?
Answer:

Question L.
Decay constant of ¹⁹⁷Hg is 0.017 h^-1. What is its half life ?
Answer:

Question M.
The total binding energy of ⁵⁸Ni is 508 MeV. What is its binding energy per nucleon ?
Answer:
Given: B.E. of ⁵⁸Ni = 508 MeV,
A = 58
To find: Binding energy per nucleon \(\bar{B}\)

Question N.
Atomic mass of \({ }_{16}^{32} \mathrm{~S}\) is 31.97 u. If masses of neutron and H atom are 1.0087 u and 1.0078 u respectively. What is the mass defect ?
Answer:
Given: m = 31.97 u, Z = 16, A = 32
m_n = 1.0087 u
m_H = 1.0078 u
To find: Δm
Formula: Δm = Zm_H + (A – Z)m_n – m
Calculation: Δm = Zm_H + (A – Z)m_n – m
= 16 × 1.0078 + (16 × 1.0087) – 31.97
= [16.1248 + 16.1392] – 31.97
= 0.294 u
Ans: The mass defect is 0.294 u.

Question O.
Write the fusion reactions occuring in the Sun and stars.
Answer:
Fusion reactions occurring in the Sun and stars are can be represented as,

Question P.
How many α and β – particles are emitted in the trasmutation
\({ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}\)
Answer:
\({ }_{90}^{232} \mathrm{Th} \longrightarrow{ }_{82}^{208} \mathrm{~Pb}\)
The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 232 – 208 = 24.
This decrease is only due to α-particles. Hence, number of α-particles emitted = \(\frac {24}{4}\) = 6
Now, the emission of one α-particle decrease the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 90 – 82 = 8
The emission of 6 α-particles causes decrease in atomic number by 12. However, the actual decrease is only 8. Thus, atomic number increases by 4. This increase is due to emission of 4 β-particles.
Thus, 6 α and 4 β-particles are emitted.

Question Q.
A produces B by α- emission. If B is in the group 16 of periodic table, what is the group of A ?
Answer:

When α-emission occurs, atomic number decreases by 2 and atomic mass number by 4.
Thus, if ‘B’ belongs to group 16 of periodic table, that means outermost orbit will contain 6 electrons.
Thus, ‘A’ will have 8 electrons in its valence shell and it will belong to group 18 of the periodic table.

Question R.
Find the number of α and β- particles emitted in the process
\({ }_{86}^{222} \mathrm{Rn} \longrightarrow{ }_{84}^{214} \mathrm{PO}\)
Answer:
The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 222 – 214 = 8. This decrease is only due to α-particle. Hence, number of α-particle emitted = 8/4 = 2
Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 86 – 84 = 2
The emission of 2 α-particles causes decrease in atomic number by 4. However, the actual decrease is only 2. It means atomic number increases by 2. This increase is due to emission of 2 β-particles.
Thus, 2 α and 2 β-particles are emitted.

[Note: The above question is modified to include the final decay product so as to determine the number of α-particles and β-particles emitted in the process. Here, the final decay product is assumed to be Po-214.]

4. Solve the problems

Question A.
Half life of ¹⁸F is 110 minutes. What fraction of ¹⁸F sample decays in 20 minutes ?
Answer:
Given: t_1/2 = 110 min
t = 20 min
To find: Fraction of ¹⁸F simple that decays

∴ Fraction of ¹⁸F sample that decays = 1 – 0.882 = 0.118
Ans: Fraction of ¹⁸F sample that decays in 20 minutes is 0.118.

Question B.
Half life of ³⁵S is 87.8 d. What percentage of ³⁵S sample remains after 180 d ?
Answer:
Given: t_1/2 = 87.8 d,
N₀ = 100,
t = 180 d
To find: % of ³⁵S that remains after 180 days

Question C.
Half life ⁶⁷Ga is 78 h. How long will it take to decay 12% of sample of Ga ?
Answer:
Given: t_1/2 = 78 h,
N₀ = 100,
N = 100 – 12 = 88
To find: t

Question D.
0.5 g Sample of ²⁰¹Tl decays to 0.0788 g in 8 days. What is its half life ?
Answer:
Given: N₀ = 0.5 g,
N = 0.0788 g,
t = 8 days
To find: t_1/2

Question E.
65% of ¹¹¹In sample decays in 4.2 d. What is its half life ?
Answer:
Given: N₀ = 100,
N = 100 – 65 = 35,
t = 4.2d
To find: t_1/2

Question F.
Calculate the binding energy per nucleon of \({ }_{36}^{84} \mathrm{Kr}\) whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u).
Answer:
Given: A = 84, Z = 36,
m = 83.913 u
m_n = 1.0087 u
m_H = 1.0078 u
To find: Binding energy per nucleon \((\bar{B})\)

Question G.
Calculate the energy in Mev released in the nuclear reaction
\({ }_{77}^{174} \mathrm{Ir} \longrightarrow{ }_{75}^{170} \mathrm{Re}+{ }_{2}^{4} \mathrm{He}\)
Atomic masses : Ir = 173.97 u,
Re = 169.96 u and
He = 4.0026 u
Answer:
Given: m_Ir= 173.97 u
m_Re = 169.96 u
m_He = 4.0026 u
To find: Energy released
Formulae: i. Δm = (mass of ¹⁷⁴Ir) – (mass of ¹⁷⁰Re + mass of ⁴He)
ii. E = Δm × 931.4 MeV
Calculation:i. Δm = (mass of ¹⁷⁴Ir) – (mass of ¹⁷⁰Re + mass of ⁴He)
= 173.97 – (169.96 + 4.0026)
= 7.4 × 10^-3 u
ii. E = Δm × 931.4
= 7.4 × 10^-3 × 931.4
= 6.89236 MeV ≈ 6.892 MeV
Ans: The energy released in given nuclear reaction is 6.892 MeV.

Question H.
A 3/4 of the original amount of radioisotope decays in 60 minutes. What is its half life ?
Answer:

Question I.
How many – particles are emitted by 0.1 g of ²²⁶Ra in one year?
Answer:
Given: t = 1 y,
Amount of sample = 0.1 g
To find: Number of particles emitted

Activity = \(\frac{-\mathrm{d} \mathrm{N}}{\mathrm{dt}}\) = λN
= 4.28 × 10^-4 × 2.665 × 10²⁰ atoms
= 1.141 × 10¹⁷ particles/year
Ans: Particles emitted by 0.1 g of ²²⁶Ra in one year = 1.141 × 10¹⁷ particles/year.
[Note: The half-life of radium is 1620 years. In order to apply appropriate textual concept, we have used this value in calculation.]

Question J.
A sample of ³²P initially shows activity of one Curie. After 303 days the activity falls to 1.5× 10⁴ dps. What is the half life of ³²P ?
Answer:

Question K.
Half life of radon is 3.82 d. By what time would 99.9 % of radon will be decayed.
Answer:
Given: t_1/2 = 3.82 d,
N₀ = 100
N = 100 – 99.9 = 0.1
To find: t

Question L.
It has been found that the Sun’s mass loss is 4.34 × 10⁹ kg per second. How much energy per second would be radiated into space by the Sun ?
Answer:
Given: Sun’s mass loss = 4.34 × 10⁹ kg per second
To find: Energy radiated per second into space by Sun
calculation: Δm = 4.34 × 10⁹ kg per second
Now, 1.66 × 10^-27 kg = 1u
∴ Δm = \(\frac{4.34 \times 10^{9}}{1.66 \times 10^{-27}}\) u per second
= 2.614 × 10³⁶ u per second
Now, 1 u = 931.4 MeV
2.614 × 10³⁶ u per second = 2.614 × 10³⁶ × 931.4
= 2.435 × 10³⁹ MeV/s
Now, 1 MeV = 1.6022 × 10^-19 J and 1 eV = 1 × 10^-6 MeV
1 MeV = 1.6022 × 10^-13 J
= 1.6022 × 10^-16 LJ
E = 2.435 × 10³⁹ MeV/s × 1.6022 × 10^-16 kJ/MeV
= 3.901 × 10²³ kJ/s
Ans: Energy radiated per second into space by Sun is 3.901 × 10²³ kJ/s.

Question M.
A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, How old is the given sample of wood ? Half life of ¹⁴C 5730 y.
Answer:

Activity :

1. Discuss five applications of radioactivity for peaceful purpose.
Answer:

Development in earth sciences: Like to understand various geographical changes occurring on earth.
Development in space technology: To study nuclear reactions in stars which may lead to new discoveries.
Development in medical sciences: Diagnosis and treatment of various diseases.
Development in industries: As a potent source of electricity or a power generator.
Development in agriculture: To study or monitor changes in soil like uptake of nutrients from the soil etc.

[Note: Students can use above points are reference to discuss topic in class].

2. Organize a trip to Bhabha Atomic Reasearch Centre, Mumbai to learn about nuclear reactor. This will have to be organized through your college.
Answer:
Students are expected to visit the place to understand more about nuclear reactors.

11th Chemistry Digest Chapter 13 Nuclear Chemistry and Radioactivity Intext Questions and Answers

Do you know? (Textbook Page no. 190)

Question 1.
How small is the nucleus in comparison to the rest of the atom?
Answer:
The radius of nucleus is of the order of 10^-15 m whereas that of the outer sphere is of the order of 10^-10 m. The size of outer sphere, is 10⁵ times larger than the nucleus i.e., if we consider the atom of size of football stadium then its nucleus will be the size of a pea.

(Textbook Page no. 191)

Question 1.
Identify the following nuclides as: isotopes, isobars and isotones.

Answer:

(Textbook Page No. 194)

Question 1.
i. What do you understand by the term rate of decay and give its mathematical expression.
ii. Why is minus sign required in the expression of decay rate?
Answer:
i. Rate of decay of a radioelement denotes the number of nuclei of its atoms which decay in unit time. It is also called activity of radioelement.
Rate of decay at any time t can be expressed as follows:
Rate of decay (activity) = \(-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\)
where, dN is the number of nuclei that decay within time interval dt.
ii. Minus sign in the expression indicates that the number of nuclei decreases with time. Therefore, dN is a negative quantity. But, the rate of decay is a positive quantity. The negative sign is introduced in the rate expression to make the rate positive.

Try this. (Textbook Page No. 194)

Question 1.
Prepare a chart of comparative properties of the above three types of radiations.
Answer:

Just think (Textbook Page No. 195)

Question 1.
Does half-life increase, decrease or remain constant? Explain.
Answer:
Half-life of a particular radioelement remains constant at a given instant. A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay. It is related to decay constant by the expression: t_1/2 = 0.693 / λ

From the expression, it is evident that half-life of a radio isotope is dependent only on the decay constant and is independent of the initial amount of the radio isotope. Each successive half-life in which the amount of radio isotope decreases to its half value is the same.

Thus, half-life remains constant.

Try this (Textbook Page No. 198)

Question 1.
²⁴Mg and ²⁷Al, both undergo (α, n) reactions and the products are radioactive. These emit β particles having positive charge (called positrons). Write balanced nuclear reactions in both.
Answer:

Do you know? (Textbook Page No. 198)

Question 1.
What is the critical mass of ²³⁵U?
Answer:
i. The critical mass is the minimum mass of uranium-235 required to achieve a self-sustaining fission chain reaction under stated conditions.
ii. The chain reaction in fission of U-235 becomes self-sustaining when the critical mass of uranium-235 is about 50 kilograms.

Activity (Textbook Page No. 200)

Question 1.
You have learnt in Std. 9th, medical, industrial and agricultural applications of radioisotopes. Write at least two applications each.
Answer:
i. The uses of radioactive isotopes in the field of medicine:
a. Polycythaemia: The red blood cell count increases in the disease polycythaemia. Phosphorus-32 is used in its treatment.
b. Bone cancer: Strontium-89, strontium-90, samarium-153 and radium-223 are used in the treatment of bone cancer.

ii. The uses of radioactive isotopes in the industrial field:
a. Luminescent paint and radioluminescence: The radioactive substances radium, promethium, tritium with some phosphorus are used to make certain objects visible in the dark.
e.g. Hands of a clock, krypton-85 is used in HID (High Intensity Discharge) lamps.
b. Use in ceramic articles:
1. Luminous colours are used to decorate ceramic tiles, utensils, plates, etc.
2. Uranium oxide was earlier used to colour ceramics.

iii. The uses of radioactive isotopes in the agriculture field:
a. The genes and chromosomes that give seeds its properties like fast growth, higher productivity, etc., can be modified by means of radiation.
b. Onions and potatoes are irradiated with gamma rays from cobalt-60 to prevent their sprouting.

11th Std Chemistry Questions And Answers:

11th Physics Chapter 6 Exercise Mechanical Properties of Solids Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

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

Mechanical Properties of Solids Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 6 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 6 Exercise Solutions

1. Choose the correct answer:

Question 1.
Change in dimensions is known as …………..
(A) deformation
(B) formation
(C) contraction
(D) strain.
Answer:
(A) deformation

Question 2.
The point on stress-strain curve at which strain begins to increase even without increase in stress is called…………
(A) elastic point
(B) yield point
(C) breaking point
(D) neck point
Answer:
(B) yield point

Question 3.
Strain energy of a stretched wire is 18 × 10^-3 J and strain energy per unit volume of the same wire and same cross section is 6 × 10^-3 J/m³. Its volume will be………….
(A) 3cm³
(B) 3 m³
(C) 6 m³
(D) 6 cm³
Answer:
(B) 3 m³

Question 4.
……………. is the property of a material which enables it to resist plastic deformation.
(A) elasticity
(B) plasticity
(C) hardness
(D) ductility
Answer:
(C) hardness

Question 5.
The ability of a material to resist fracturing when a force is applied to it, is called……………
(A) toughness
(B) hardness
(C) elasticity
(D) plasticity.
Answer:
(A) toughness

2. Answer in one sentence:

Question 1.
Define elasticity.
Answer:
If a body regains its original shape and size after removal of the deforming force, it is called an elastic body and the property is called elasticity.

Question 2.
What do you mean by deformation?
Answer:
The change in shape or size or both of u body due to an external force is called deformation.

Question 3.
State the SI unit and dimensions of stress.
Answer:

SI unit: N m^-2 or pascal (Pa)
Dimensions: [L^-1M¹T^-2]

Question 4.
Define strain.
Answer:
Strain:

Strain is defined as the ratio of change in dimensions of the body to its original dimensions.
Strain = \(\frac{\text { change in dimensions }}{\text { original dimensions }}\)
Types of strain:
- Longitudinal strain,
- Volume strain,
- Shearing strain.

Question 5.
What is Young’s modulus of a rigid body?
Answer:
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.

Question 6.
Why bridges are unsafe after a very long use?
Answer:
A bridge during its use undergoes recurring stress depending upon the movement of vehicles on it. When bridge is used for long time, it loses its elastic strength and ultimately may collapse. Hence, the bridges are declared unsafe after long use.

Question 7.
How should be a force applied on a body to produce shearing stress?
Answer:
A tangential force which is parallel to the top and the bottom surface of the body should be applied to produce shearing stress.

Question 8.
State the conditions under which Hooke’s law holds good.
Answer:
Hooke’s Taw holds good only when a wire/body is loaded within its elastic limit.

Question 9.
Define Poisson’s ratio.
Answer:
Within elastic limit, the ratio of lateral strain to the linear strain is called the Poisson‘s ratio.

Question 10.
What is an elastomer?
Answer:
A material that can be elastically stretched to a larger value of strain is called an elastomer.

Question 11.
What do you mean by elastic hysteresis?
Answer:

In case of some materials like vulcanized rubber, when the stress applied on a body decreases to zero, the strain does not return to zero immediately. The strain lags behind the stress. This lagging of strain behind the stress is called elastic hysteresis.
Below figure shows the stress-strain curve for increasing and decreasing load. It encloses a loop. Area of loop gives the energy dissipated during deformation of a material.

Question 12.
State the names of the hardest material and the softest material.
Answer:
Hardest material: Diamond
Softest material: Aluminium
[Note: Material with highest strength is steel whereas material with lowest strength is plasticine clay.]

Question 13.
Define friction.
Answer:
The property which resists the relative motion between two surfaces in contact is called friction.

Question 14.
Why force of static friction is known as ‘self-adjusting force?
Answer:
The force of static friction varies in accordance with applied force. Hence, it is called as self adjusting force.

Question 15.
Name two factors on which the coefficient of friction depends.
Answer:
Coefficient of friction depends upon:

the materials of the surfaces in contact.
the nature of the surfaces.

3. Answer in short:

Question 1.
Distinguish between elasticity and plasticity.
Answer:

No.	Elasticity	Plasticity
i.	Body regains its original shape or size after removal of deforming force.	Body does not regain its original shape or size after removal of deforming force.
ii.	Restoring forces are strong enough to bring the displaced molecules to their original positions.	Restoring forces are not strong enough to bring the molecules back to their original positions.
	Examples of elastic materials: metals, rubber, quartz, etc	Examples of plastic materials: clay, putty, plasticine, thick mud, etc

Question 2.
State any four methods to reduce friction.
Answer:
Friction can be reduced by using polished surfaces, using lubricants, using grease and using ball bearings.

Question 3.
What is rolling friction? How does it arise?
Answer:

Friction between two bodies in contact when one body is rolling over the other, is called rolling friction.
Rolling friction arises as the point of contact of the body with the surface keep changing continuously.

Question 4.
Explain how lubricants help in reducing friction?
Answer:

The friction between lubricant to surface is much less than the friction between two same surfaces. Hence using lubricants reduces the friction between the two surfaces.
When lubricant is applied to machine parts, it fills the depression present on the surface in contact. Thus, less friction is occurred between machine parts.
Application of lubricants also reduces wear and tear of machine parts which in turn reduces friction.
Advantage: Reduction in function reduces dissipation of energy in machines due to which efficiency of machines increases.

Question 5.
State the laws of static friction.
Answer:
Laws of static friction:

First law: The limiting force of static friction (F_L) is directly proportional to the normal reaction (N) between the two surfaces in contact.
F_L ∝ N
∴ F_L = µ_s N
where, µ_s = constant called coefficient of static friction.
Second law: The limiting force of friction is
independent of the apparent area between the surfaces in contact, so long as the normal reaction remains the same.
Third law: The limiting force of friction depends upon materials in contact and the nature of their surfaces.

Question 6.
State the laws of kinetic friction.
Answer:
Laws of kinetic friction:

First law: The force of kinetic friction (Fk) is directly proportional to the normal reaction (N) between two surfaces in contact.
F_k ∝ N
∴ F_k = µ_kN
where, µ_k = constant called coefficient of kinetic friction.
Second law: Force of kinetic friction is independent of shape and apparent area of the surfaces in contact.
Third law: Force of kinetic friction depends upon the nature and material of the surfaces in contact.
Fourth law: The magnitude of the force of kinetic friction is independent of the relative velocity between the object and the surface provided that the relative velocity is neither too large nor too small.

Question 7.
State advantages of friction.
Answer:
Advantages of friction:

We can walk due to friction between ground and feet.
We can hold object in hand due to static friction.
Brakes of vehicles work due to friction; hence we can reduce speed or stop vehicles.
Climbing on a tree is possible due to friction.

Question 8.
State disadvantages of friction.
Answer:
Disadvantages of friction:

Friction opposes motion.
Friction produces heat in different parts of machines. It also produces noise.
Automobile engines consume more fuel due to friction.

Question 9.
What do you mean by a brittle substance? Give any two examples.
Answer:

Substances which breaks within the elastic limit are called brittle substances.
Examples: Glass, ceramics.

4. Long answer type questions:

Question 1.
Distinguish between Young’s modulus, bulk modulus and modulus of rigidity.
Answer:

No	Young’s modulus	Bulk modulus	Modulus of rigidity
i.	It is the ratio of longitudinal stress to longitudinal strain.	It is the ratio of volume stress to volume strain.	It is the ratio of shearing stress to shearing strain.
ii.	It is given by, Y = \(\frac{\mathrm{MgL}}{\pi \mathrm{r}^{2} l}\)	It is given by, K = \(\frac{V d P}{d V}\)	It is given by, \(\eta=\frac{F}{A \theta}\)
iii.	It exists in solids.	It exists in solid, liquid and gases.	It exists in solids.
iv.	It relates to change in length of a body.	It relates to change in volume of a body.	It relates to change in shape of a body.

Question 2.
Define stress and strain. What are their different types?
Answer:
i) Stress:

The internal restoring force per unit area of a both is called stress.
Stress = \(\frac{\text { deforming force }}{\text { area }}=\frac{|\vec{F}|}{\mathrm{A}}\)
where \(\vec{F}\) is internal restoring force or external applied deforming force.
Types of stress:
- Longitudinal stress,
- Volume stress,
- Shearing stress.

ii. Strain:

Strain is defined as the ratio of change in dimensions of the body to its original dimensions.
Strain = \(\frac{\text { change in dimensions }}{\text { original dimensions }}\)
Types of strain:
- Longitudinal strain,
- Volume strain,
- Shearing strain.

Question 3.
What is Young’s modulus? Describe an experiment to find out Young’s modulus of material in the form of a long straight wire.
Answer:
Definition: Young ‘s modulus is the ratio of longitudinal stress to longitudinal strain.
It is denoted by Y.
Unit: N/m² or Pa in SI system.
Dimensions: [L^-1M¹T^-2]

Experimental description to find Young’s modulus:

i. Consider a metal wire suspended from a rigid support. A load is attached to the free end of the wire. Due to this, deforming force gets applied to the free end of wire in downward direction and it produces a change in length.
Let,
L = original length of wire,
Mg = weight suspended to wire,
l = extension or elongation,
(L + l) = new length of wire.
r = radius of the cross section of wire

ii. In its equilibrium position,

Question 4.
Derive an expression for strain energy per unit volume of the material of a wire.
Answer:
Expression for strain energy per unit volume;

i. Consider a wire of original length L and cross sectional area A stretched by a force F acting along its length. The wire gets stretched and elongation l is produced in it

ii. If the wire is perfectly elastic then,
Longitudinal stress = \(\frac{F}{A}\)
Longitudinal strain = \(\frac{l}{L}\)

iii. The magnitude of stretching force increases from zero to F during elongation of wire.
Let ‘f’ be the restoring force and ‘x’ be its corresponding extension at certain instant during the process of extension.
∴ f = \(\frac{\text { YAx }}{\mathrm{L}}\) ……………. (2)

iv. Let ‘dW’ be the work done for the further small extension ‘dx’.
Work = force × displacement
∴ dW = fdx
∴ dW= \(\frac{\text { YAx }}{L}\) dx …………..(3) [From (2)]

v. The total amount of work done in stretching the wire from x = 0 to x = l can be found out by integrating equation (3).

∴ Work done in stretching a wire,
W = \(\frac{1}{2}\) × load × extension

vi. Work done by stretching force is equal to strain energy gained by the wire.
∴ Strain energy = \(\frac{1}{2}\) × load × extension

vii. Work done per unit volume

∴ Strain energy per unit volume = \(\frac{1}{2}\) × stress × strain

viii. Other forms:

Question 5.
What is friction? Define coefficient of static friction and coefficient of kinetic friction. Give the necessary formula for each.
Answer:

The property which resists the relative motion between two surfaces in contact is called friction.
The coefficient of static friction is defined as the ratio of limiting force of friction to the normal reaction.
Formula: \(\mu_{\mathrm{S}}=\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}\)
The coefficient of kinetic friction is defined as the ratio of force of kinetic friction to the normal reaction between the two surfaces in contact.
Formula: \(\mu_{\mathrm{k}}=\frac{\mathrm{F}_{\mathrm{K}}}{\mathrm{N}}\)

Question 6.
State Hooke’s law. Draw a labelled graph of tensile stress against tensile strain for a metal wire up to the breaking point. In this graph show the region in which Hooke’s law is obeyed.
Answer:
i) Statement: Within elastic limit, stress is directly proportional to strain.
Explanation;

According to Hooke’s law,
Stress ∝ Strain

This constant of proportionality is called modulus of elasticity.
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.
The graph of strain (on X-axis) and stress (on Y-axis) within elastic limit is shown in the figure.

ii)

iii) Hooke’s law is completely obeyed in the region OA.

5. Answer the following

Question 1.
Calculate the coefficient of static friction for an object of mass 50 kg placed on horizontal table pulled by attaching a spring balance. The force is increased gradually it is observed that the object just moves when spring balance shows 50N.
[Answer: µs = 0.102]
Solution:
Given: m = 50 kg, F_L = 50 N, g = 9.8 m/s²
To find: Coefficient of static friction (µ_s)
Formula: µ_s = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}=\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{mg}}\)
µ_s = \(\frac{50}{50 \times 9.8}\) = 0.102
Answer:
The coefficient of static friction is 0.102.

Question 2.
A block of mass 37 kg rests on a rough horizontal plane having coefficient of static friction 0.3. Find out the least force required to just move the block horizontally.
[Answer: F= 108.8N]
Solution:
Given: m = 37 kg, µ_s = 0.3, g = 9.8 m /s²
To find: Limiting force (F_L)
Formula: F_L = µ_SN = µ_S mg
Calculation: From formula,
F_L = 0.3 × 37 × 9.8 = 108.8 N
Answer:
The force required to move the block is 108.8 N.

Question 3.
A body of mass 37 kg rests on a rough horizontal surface. The minimum horizontal force required to just start the motion is 68.5 N. In order to keep the body moving with constant velocity, a force of 43 N is needed. What is the value of
a) coefficient of static friction? and
b) coefficient of kinetic friction?
Asw:
a) µ_s = 0.188
b) µ_k = 0.118]
Solution:
Given:
F_L = 68.5 N, F_k = 43 N,
m = 37 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 = \(\frac{F_{L}}{N}\) = \(\frac{F_{L}}{m g}\)
ii. µ_k = \(\frac{F_{k}}{N}\) = \(\frac{\mathrm{F}_{\mathrm{k}}}{\mathrm{mg}}\)

Calculation:
From formula (i),
∴ µ_s = \(\frac{F_{S}}{N}=\frac{68.5}{37 \times 9.8}\) = 0.1889
From formula (ii),
∴ µ_k = \(\frac{F_{k}}{N}=\frac{43}{37 \times 9.8}\) = 0.1186
Answer:

The coefficient of static friction is 0.1889.
The coefficient of kinetic friction is 0.1186.

[Note: Answers calculated above are in accordance with textual methods of calculation.]

Question 4.
A wire gets stretched by 4mm due to a certain load. If the same load is applied to a wire of same material with half the length and double the diameter of the first wire. What will be the change in its length?
Solution:
Given. l₁ = 4mm = 4 × 10^-3 m
L₂ = \(\frac{\mathrm{L}_{1}}{2}\), D₂ = 2D, r₂ = 2r₁
To find: Change in length (l₂)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{FL}}{\pi \mathrm{r}^{2} l}\)
Calculation: From formula,

= 0.5 × 10^-3 m
= 0.5 mm
The new change in length of the wire is 0.5 mm.

Question 5.
Calculate the work done in stretching a steel wire of length 2m and cross sectional area 0.0225mm² when a load of 100 N is slowly applied to its free end. [Young’s modulus of steel= 2 × 10¹¹ N/m²]
Solution:
Given. L = 2m, F = 100 N,
A = 0.0225 mm² = 2.25 × 10^-8 m²,
Y = 2 × 10^-11 N/m²,
To find: Work (W)
Formula: W = \(\frac{1}{2}\) × F × l
Claculation:

= antilog [log 10 – log 4.5]
= antilog [1.0000 – 0.6532 ]
= antilog [0.3468]
∴ W = 2.222 J
Answer:
The work done in stretching the steel wire is 2.222 J.

Question 6.
A solid metal sphere of volume 0.31m³ is dropped in an ocean where water pressure is 2 × 10⁷ N/m². Calculate change in volume of the sphere if bulk modulus of the metal is 6.1 × 10¹⁰ N/m²
Solution:
Given: V= 0.31 m³, dP = 2 × 10⁷ N/m²,
K = 6.1 × 10¹⁰ N/m²
To find: Change in volume (dV)
Formula: K = V × \(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,
dV = \(\frac{\mathrm{V} \times \mathrm{dP}}{\mathrm{K}}\)
∴ dV = \(\frac{0.31 \times 2 \times 10^{7}}{6.1 \times 10^{10}}\) ≈ 10^-4 m³
The change in volume of the sphere is 10^-4 m³.

Question 7.
A wire of mild steel has initial length 1.5 m and diameter 0.60 mm is extended by 6.3 mm when a certain force is applied to it. If Young’s modulus of mild steel is 2.1 × 10¹¹ N/m², calculate the force applied.
Solution:
Given:
L = 1.5m, d = 0.60 mm,
r = \(\frac{d}{2}\) = 0.30 mm = 3 × 10^-4 m,
Y = 2.1 × 10¹¹ N/m²,
l = 6.3 mm = 6.3 × 10^-3 m
To find: Force (F)
Calculation:
From formula,

= 2.1 × 3.142 × 6 × 6.3
= antilog [log 2.1 + log 3.142 + log 6 + log 6.3]
= antilog [0.3222 + 0.4972 + 0.7782 + 0.7993]
= antilog [2.3969]
= 2.494 × 10²
≈ 250 N
The force applied on wire is 250 N.

Question 8.
A composite wire is prepared by joining a tungsten wire and steel wire end to end. Both the wires are of the same length and the same area of cross section. If this composite wire is suspended to a rigid support and a force is applied to its free end, it gets extended by 3.25mm. Calculate the increase in length of tungsten wire and steel wire separately.
[Given: Y_steel = 2 × 10¹¹ Pa, Y_Tungsten = 4.11 × 10¹¹ Pa]
Solution:
Given: l_s + l_T = 3.25 mm,
Y_T = 4.11 × 10¹¹ Pa
Y_s = 2 × 10¹¹ Pa
To find: Extension in tungsten wire (l_T)
Extension in steel wire (l_s)

But l_s + l_T = 3.25
l_s + 0.487 l_s = 3.25
l_s(1 + 0.487) = 3.25
l_s = 2.186 mm
∴ l_T = 3.25 – 2.186
= 1.064 mm
The extension in tungsten wire is 1.064 mm and the extension in steel wire is 2.186 mm.

[Note: Values of Young’s modulus of tungsten and steel considered above are standard values. Using them, calculation is carried out ¡n accordance with textual method.]

Question 9.
A steel wire having cross sectional area 1.2 mm² is stretched by a force of 120 N. If a lateral strain of 1.455 mm is produced in the wire, calculate the Poisson’s ratio.
Solution:
Given: A = 1.2 mm² = 1.2 × 10^-6 m²,
F = 120 N, Y_steel = 2 × 10¹¹ N/m²,
Lateral strain = 1.455 × 10^-4
To find: Poisson’s ratio (σ)

The Poisson’s ratio of steel is 0.291.
[Note: Lateral strain being ratio of two same physical quantities, is unitless. hence, value given in question ¡s modified to 1.455 × 10^-4 to reach the answer given in textbook.]

Question 10.
A telephone wire 125m long and 1mm in radius is stretched to a length 125.25m when a force of 800N is applied. What is the value of Young’s modulus for material of wire?
Solution:
Given: L = 125m,
r = 1 mm= 1 × 10^-3 m
l = 125.25 – 125 = 0.25 m,
F = 800N
To find: Young’s modulus (Y)
Formula: Y \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{FL}}{\pi \mathrm{r}^{2} l}\)
Calculation: From formula,
Y = \(\frac{800 \times 125}{3.142 \times 10^{-6} \times 0.25}\)
= {antilog [log 800 + log 125 – log 3.142 – log 0.25 ]} × 10⁶
= {antilog [2.9031 + 2.0969 – 0.4972 – \(\overline{1}\) .3979]} × 10⁶
= {antilog[5.1049]} × 10⁶
= 1.274 × 10⁵
= 1.274 × 10¹¹ N/m²
The Young’s modulus of telephone wire is 1.274 × 10¹¹ N/m².

Question 11.
A rubber band originally 30cm long is stretched to a length of 32cm by certain load. What is the strain produced?
Solution:
Given: L = 30 cm = 30 × 10 ^-2 m,
∆l = 32 cm – 30 cm = 2cm = 2 × 10 ^-2 m
To find. Strain
Formula: Strain = \(\frac{\Delta l}{\mathrm{~L}}\)
Calculation: From formula,
Strain = \(\frac{2 \times 10^{-2}}{30 \times 10^{-2}}\) = 6.667 × 10 ^-2
The strain produced in the wire is 6.667 × 10 ^-2.

Question 12.
What is the stress in a wire which is 50m long and 0.01cm² in cross section, if the wire bears a load of 100kg?
Solution:
Given: M = 100 kg, L 50 m, A = 0.01 × 10^-4 m
To find: Stress
Formula: Stress = \(\frac{\mathrm{F}}{\mathrm{A}}=\frac{\mathrm{Mg}}{\mathrm{A}}\)
Calculation: From formula,
Stress = \(\frac{100 \times 9.8}{0.01 \times 10^{-4}}\) = 9.8 × 10⁸ N/m²
The stress in the wire is 9.8 × 10⁸ N/m².

Question 13.
What is the strain in a cable of original length 50m whose length increases by 2.5cm when a load is lifted?
Solution:
Given: L = 50m, ∆l = 2.5cm = 2.5 × 10 ^-2 m
To find: Strain
Formula: Strain = \(\frac{\Delta l}{\mathrm{~L}}\)
Calculation: From formula,
Strain = \(\frac{2.5 \times 10^{-2}}{50}\) = 5 × 10^-4
The Strain produced in wire is 5 × 10^-4 .

11th Physics Digest Chapter 6 Mechanical Properties of Solids Intext Questions and Answers

Can you recall? (Textbook Page No. 100)

Question 1.

Can you name a few objects which change their shape and size on application of a force and regain their original shape and size when the force is removed?
Can you name objects which do not regain their original shape and size when the external force is removed?

Answer:

Objects such as rubber, metals, quartz, etc. change their shape and size on application of a force (within specific limit) and regain their original shape and size when the force is removed.
Objects such as putty, clay, thick mud. etc. do not regain their original shape and size when the external force is removed.

Can you tell? (Textbook Page No. 107)

Question 1.
Why does a rubber band become loose after repeated use?
Answer:

After repeated use of rubber band, its stress-strain curve does not remain linear.
In such case, since rubber crosses its elastic limit, there is a permanent set formed on the rubber due to which it becomes loose.

Can you tell? (Textbook Page No.111)

Question 1.
i. It is difficult to run fast on sand.
ii. It is easy to roll than pull a barrel along a road.
iii. An inflated tyre rolls easily than a flat tyre.
iv. Friction is a necessary evil.
Answer:
i.

The intermolecular space between crystals of sand is very large as compared to that in a rigid surface.
Thus, there are number of depressions at the points of contact of feet and sand surface.
Projections and depressions between sand and feet are not completely interlocked.
Thus, action and reaction force become unbalanced. The horizontal component of force helps to move forward and vertical component of the force resist to move.
Hence, it becomes difficult to run fast on sand.

ii.

When a barrel is pulled along a road, the friction between the tyres and road is kinetic friction, but when its rolls along the road it undergoes rolling friction.
The force of kinetic friction is greater than force of rolling friction.
Hence, it is easy to roll than pull a barrel along a road.

iii.

When the tyre is inflated, the pressure inside the tyre is reducing the normal force between tyre and the ground, and thus reducing the friction between the tyre and the road.
When the tyre gets deflated, it gets deformed during rolling, the supplied energy is used up in changing the shape and not overcoming the friction, and thus due to deformation, friction increases.
Hence, an inflated tyre rolls easily than a flat tyre.

iv.

Friction helps us to walk, hold objects in hand, lift objects and without friction we cannot walk, we cannot grip or hold objects with our hands,
Friction is responsible for wear and tear of various part of machines, it produces heat in different parts of machine and also produces noise but it also helps in ball bearing or connecting screws.
Hence, friction is said to be a necessary evil because it is useful as well as harmful.

Internet my friend (Textbook Page No. 111)

Question 1.
i. https ://opentextbc. ca/physicstestbook2/ chapter/friction/
ii. https://www.livescience.com/
iii. https://www.khanacademy.org/science/physics
iv. https://courses.lumenleaming.com/physics/ chapter/5-3-elasticity-stress-and-strain/
v. https://www.toppr.com/guides/physics/
Answer:
[Students are expected to visit the above mentioned websites and collect more information about mechanical properties of solid.]

11th Std Physics Questions And Answers:

11th Chemistry Chapter 6 Exercise Redox Reactions Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 6 Redox Reactions Textbook Exercise Questions and Answers.

Redox Reactions Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 6 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 6 Exercise Solutions

1. Choose the most correct option

Question A.
Oxidction numbers of Cl atoms marked as Cl^a and Cl^b in CaOCl₂ (bleaching powder) are

a. zero in each
b. -1 in Cl^a and +1 in Cl^b
c. +1 in Cl^a and -1 in Cl^b
d. 1 in each
Answer:
b. -1 in Cl^a and +1 in Cl^b

Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H₂ → Cu + H₂O
b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂
c. 2K + F₂ → 2KF
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Answer:
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A₂(BC₃)₂
b. A₃(BC₄)₂
c. A₃(B₄C)₂
d. ABC₂
Answer:
b. A₃(BC₄)₂

Question D.
The coefficients p, q, r, s in the reaction
\(\mathrm{pCr}_{2} \mathrm{O}_{7}^{2-}\) + q Fe^2⊕ → r Cr^3⊕ + s Fe^3⊕ + H₂O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6

Question E.
For the following redox reactions, find the correct statement.
Sn^2⊕ + 2Fe^3⊕ → Sn^4⊕ + 2Fe^2⊕
a. Sn^2⊕ is undergoing oxidation
b. Fe^3⊕ is undergoing oxidation
c. It is not a redox reaction
d. Both Sn^2⊕ and Fe^3⊕ are oxidised
Answer:
a. Sn^2⊕ is undergoing oxidation

Question F.
Oxidation number of carbon in H₂CO₃ is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4

Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O₂
b. Mn(II)O₂
c. Mn(III)O₂
d. Mn(IV)O₂
Answer:
d. Mn(IV)O₂

Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –\(\frac {1}{2}\)
d. 0
Answer:
c. –\(\frac {1}{2}\)

Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F

Question K.
The process SO₂ → S₂Cl₂ is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction

2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl₂
ii. Tl₂SO₄
iii. SnO₂
iv. Cr₂O₃

3. Answer the following questions

Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
In CH₄, the oxidation state of carbon is -4 while in CO₂, the oxidation state of carbon is +4.

Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:

C. Calculate the oxidation number of underlined atoms.
a. H₂SO₄
b. HNO₃
c. H₃PO₃
d. K₂C₂O₄
e. H₂S₄O₆
f. Cr₂O₇^2-
g. NaH₂PO₄
Answer:
i. H₂SO₄
Oxidation number of H = +1
Oxidation number of O = -2
H₂SO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H₂SO₄ = +6

ii. HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
HNO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO₃ = +5

iii. H₃PO₃
Oxidation number of O = -2
Oxidation number of H = +1
H₃PO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H₃PO₃ = +3

iv. K₂C₂O₄
Oxidation number of K = +1
Oxidation number of O = -2
K₂C₂O₄ is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +\(\frac {6}{2}\)
∴ Oxidation number of C in K₂C₂O₄ = +3

v. H₂S₄O₆
Oxidation number of H = +1
Oxidation number of O = -2
H₂S₄O₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +\(\frac {10}{4}\)
∴ Oxidation number of S in H₂S₄O₆ = +2.5

vi. Cr₂O₇^2-
Oxidation of O = -2
Cr₂O₇^2- is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +\(\frac {12}{2}\)
∴ Oxidation number of Cr in Cr₂O₇^2- = +6

vii. NaH₂PO₄
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH₂PO₄ is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH₂PO₄ = +5

Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
b. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
c. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
Answer:
i. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.

Result:

The given reaction is a redox reaction.
Oxidant/oxidising agents (Reduced species): Cu₂O/ Cu₂S
Reductant/reducing agent (Oxidised species): Cu₂S

[Note: Cu in both Cu₂O and Cu₂S undergoes reduction. Hence, both Cu₂O and Cu₂S can be termed as oxidising agents in the given reaction.]

ii. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.

iii. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

The given reaction is a redox reaction.
Oxidant/oxidising agent (Reduced species): I₂
Reductant/reducing agent (Oxidised species): S₂O₃^2-

Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–
Answer:
a. Mg / Mg²⁺
Here, Mg loses two electrons to form Mg²⁺ ion.
\(\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Mg / Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺
Here, Cu loses two electrons to form Cu²⁺ ion.
\(\mathrm{Cu}_{(\mathrm{s})} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O^2-
Here, each O gains two electrons to form O^2- ion.
\(\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}_{(\mathrm{aq})}^{2-}\)
Hence, O₂ / O^2- is in a reduced state.

d. Cl₂ / Cl^–
Here, each Cl gains one electron to form Cl^– ion.
\(\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence, Cl₂ / Cl^– is in a reduced state.

Question F.
Justify the following reaction as redox reaction.
2 Na_2(s) + S_(s) → Na₂S_(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + S_(s) → 2Na⁺ + S^2-
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S^2-. This can be represented as follows:

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electrons to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S^2- in Na₂S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

Question G.
Provide the stock notation for the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, MnO and CuO.
Answer:

Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)₄^–
b. Na₂S₂O₃
c. H₃BO₃
Answer:
i. Cr(OH)₄^–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)₄^– is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)₄^– = +3

ii. Na₂S₂O₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂S₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +\(\frac {4}{2}\)
∴ Oxidation number of S in Na₂S₂O₃ = +2

iii. H₃BO₃
Oxidation number of H = +1
Oxidation number of O = -2
H₃BO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H₃BO₃ = +3

Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)
b. \(\mathrm{MnO}_{4}^{\Theta}\) (E0 = 1.51 V) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\) (E⁰ = 1.33 V)
Answer:
a. Cl₂ has a larger positive value of E⁰ than Br₂. Thus, Cl₂ is a stronger oxidizing agent than Br₂.
b. \(\mathrm{MnO}_{4}^{\Theta}\) has larger positive value of E⁰ than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\). Thus, \(\mathrm{MnO}_{4}^{\Theta}\) is stronger oxidizing agent than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\)

Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)
b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)
Answer:
a. Li has a larger negative value of E⁰ than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E⁰ than Fe. Thus, Zn is a stronger reducing agent than Fe.

4. Balance the reactions/equations :

Question A.
Balance the following reactions by oxidation number method

Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \quad(\text { acidic })\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(a q)}^{2-}+\mathrm{SO}_{3(a)}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{(a q)} \quad \text { (basic) }\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{ }_{(\mathrm{aq})}\)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Mn.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}\)
Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
\(2 \mathrm{MnO}_{4(a q)}^{-}+\mathrm{Br}_{(2 q)}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3 \text { (aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

iii. H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of S.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l) + H₂O_(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H₂SO_4(aq) + C_(s) → CO₂ + 2SO_2(g) + H₂O_(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)

iv. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.

Question B.
Balance the following redox equation by half reaction method

Answer:
i. H₂C₂O_4(aq) + \(\mathrm{MnO}_{4(a q)}^{-}\) → CO_2(g) + \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

ii. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{SnO}_{2(\mathrm{aq})}^{2-} \longrightarrow \mathrm{SnO}_{3(\mathrm{aq})}^{2-}+\mathrm{Bi}_{(\mathrm{s})}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1H₂O to left side of oxidation half equation and 3H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 3H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH^– ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:

5. Complete the following table :

Assign oxidation number to the underlined species and write Stock notation of compound

Compound	Oxidation number	Stock notation
AuCl₃	……………..	……………..
SnCl₂	……………..	……………..
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	……………..	……………..
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	……………..	……………..
H₃AsO₃	……………..	……………..

Answer:

Compound	Oxidation number	Stock notation
AuCl₃	+3	Au(III)Cl₃
SnCl₂	+2	Sn(II)Cl₂
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	+5	V₂(V)\(\mathrm{O}_{7}^{4-}\)
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	+4	Pt(IV)\(\mathrm{Cl}_{6}^{2-}\)
H₃AsO₃	+3	H₃As(III)O₃

11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers

Can you tell? (Textbook Page No. 81)

Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.

Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.

Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.

Can you recall? (Textbook Page No. 81)

Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.

Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.

Try this. (Textbook Page No. 82)

Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

Reactants	Products
Zn_(s) + ————_(aq)	————-_(aq) + Cu_(s)
Cu_(s) + 2Ag⁺_(aq)	—————– + ————–
———– + ————-	\( \mathrm{Co}_{(\mathrm{aq})}^{2+}\) + Ni_(s)

Answer:

Try this (Textbook Page No. 88)

Question 1.
Classify the following unbalanced half equations as oxidation and reduction.

Answer:

11th Std Chemistry Questions And Answers:

11th Biology Chapter 15 Exercise Excretion and Osmoregulation Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 15 Excretion and Osmoregulation Textbook Exercise Questions and Answers.

Excretion and Osmoregulation Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 15 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 15 Exercise Solutions

1. Choose correct option

Question (A).
Which one of the following organisms would spend maximum energy in production of nitrogenous waste?
a. Polar bear
b. Flamingo
c. Frog
d. Shark
Answer:
b. Flamingo

Question (B).
In human beings, uric acid is formed due to metabolism of __________.
a. amino acids
b. fatty acids
c. creatinine
d. nucleic acids
Answer:
d. nucleic acids

Question (C).
Visceral layer : Podocytes :: PCT : _______
a. Cilliated cells
b. Squamous cells
c. Columnar cells
d. Cells with brush border
Answer:
d. Cells with brush border

Question (D).
Deproteinised plasma is found in __________.
a. Bowman’s capsule
b. Descending limb
c. Glomerular capillaries
d. Ascending limb
Answer:
a. Bowman’s capsule, b. Descending limb, d. Ascending limb

Question (E).
Specific gravity of urine would _______ if level of ADH increases.
a. remain unaffected
b. increases
c. decreases
d. stabilise
Answer:
b. increases

Question (F).
What is micturition?
a. Urination
b. Urine formation
c. Uremia
d. Urolithiasis
Answer:
a. Urination

Question (G).
Which one of the following organisms excrete waste through nephridia?
a. Cockroach
b. Earthworm
c. Crab
d. Liver Fluke
Answer:
c. Crab

Question (H).
Person suffering from kidney stone is advised not to have tomatoes as it has _______.
a. seeds
b. lycopene
c. oxalic acid
d. sour taste
Answer:
c. oxalic acid

Question (I).
Tubular secretion does not take place in ________.
a. DCT
b. PCT
c. collecting duct
d. Henle’s loop
Answer:
b. PCT

Question (J).
The minor calyx ____________.
a. collects urine
b. connects pelvis to ureter
c. is present in the cortex
d. receives column of Bertini
Answer:
a. collects urine

Question (K).
Which one of the followings is not a part of human kidney?
a. Malpighian body
b. Malpighian tubule
c. Glomerulus
d. Loop of Henle
Answer:
b. Malpighian tubule

Question (L).
The yellow colour of the urine is due to presence of ___________
a. uric acid
b. cholesterol
c. urochrome
d. urea
Answer:
c. urochrome

Question (M).
Hypotonic filtrate is formed in _______
a. PCT
b. DCT
c. LoH
d. CT
Answer:
a. PCT

Question (N).
In reptiles, uric acid is stored in _____
a. cloaca
b. fat bodies
c. liver
d. anus
Answer:
a. cloaca

Question (O).
The part of nephron which absorbs glucose and amino acid is______
a. collecting tubule
b. proximal tubule
c. Henle’s loop
d. DCT
Answer:
b. proximal tubule

Question (P).
Bowman’s capsule is located in kidney in the ________
a. cortex
b. medulla
c. pelvis
d. pyramids
Answer:
a. cortex

Question (Q).
The snakes living in desert are mainly__________
a. aminotelic
b. ureotelic
c. ammonotelic
d. uricotelic
Answer:
d. uricotelic

Question (R).
Urea is a product of breakdown of ___________
a. fatty acids
b. amino acids
c. glucose
d. fats
Answer:
b. amino acids

Question (S).
Volume of the urine is regulated by__________
a. aldosterone
b. ADH
c. both a and b
d. none
Answer:

Question 2.
Answer the following questions

Question (A).
Doctors say Mr. Shaikh is suffering from urolithiasis. How it could be explained in simple words?
Answer:
Urolithiasis is the condition of having calculi in the urinary tract (which also includes the kidneys), which may pass into urinary bladder.

Question (B).
Anitaji needs to micturate several times and feels very thirsty. This is an indication of change in permeability of certain part of nephron. Which is this part?
Answer:

Need to micturate several times (polyuria) and feeling very thirsty (polydipsia) is a symptom of diabetes insipidus (imbalance of fluids in the body).
ADH prevents diuresis and due to absence of ADH, large amount of dilute urine is excreted.
ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
If the permeability of these cells changes, it will result in increase in urine volume (frequent micturition) and increase in the osmolarity of blood. An imbalance in volume and osmolarity of body fluids increases thirst.

[Note: Water is reabsorbed by osmosis in PCT, DCT and descending limb of loop of Henle)

Question (C).
Effective filtration pressure was calculated to be 20 mm Hg; where glomerular hydrostatic pressure was 70 mm of Hg. Which other pressure is affecting the filtration process? How much is it?
Answer:
The other pressure affecting the filtration process is osmotic pressure of blood and filtrate hydrostatic pressure. Commonly effective filtration pressure (EFP) is represented as;
EFP = Glomerular Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
If EFP = 20 mmHg and Glomerular Hydrostatic pressure = 70 mmHg
20 = 70 – (Osmotic pressure of blood + Filtrate hydrostatic pressure)
∴ (Osmotic pressure of blood + Filtrate hydrostatic pressure) = 70-20
Then (Osmotic pressure of blood + Filtrate Hydrostatic pressure) = 50 mmHg .

[Note: Given values are insufficient to calculate the exact osmotic pressure of blood and filtrate hydrostatic pressure. The sum of the two values can be calculated to be 50 mmHg ]

Question (D).
Name any one guanotelic organism.
Answer:
Spiders, scorpions and penguins are guanotelic organisms as they excrete guanine.

Question (E).
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question (F).
State role of liver in urea production.
Answer:

Ammonia formed during the breakdown of amino acids is converted into urea in the liver of ureotelic animals.
This conversion takes place by the help of the ornithine / urea cycle.
3 ATP molecules are used to produce one molecule of urea using the ornithine/ urea cycle. Since, the liver contains carrier molecules and enzymes necessary for urea cycle, it plays a major role in urea production.

Question (G).
Why do we get bad breath after eating garlic or raw onion?
Answer:

Raw onion and garlic contain volatile sulphur-containing compounds.
Sulphur-containing compounds have a distinctive odour which remain in the mouth after consumption of onion and garlic.
Also, volatile compounds (like certain sulphur containing compounds) in foodstuffs are generally excreted through the lungs and may result in bad breath.

3. Answer the following questions

Question (A).
John has two options as treatment for his renal problem : Dialysis or kidney transplants. Which option should he choose? Why?
Answer:

If John has two options of dialysis and kidney transplant, readily available he must opt for kidney transplant.
A kidney transplant, if successful, can improve the quality of life of a patient and reduce the risk of death.
The patient would not have to endure frequent dialysis procedures. Repeated visits for dialysis takes time and may not allow the patient to perform normal activities or go to office regularly.
Dialysis is regarded as a holding measure until kidney transplant can be performed or a supportive measure in those for whom a transplant would be inappropriate. However, dialysis cannot replace all the functions of a normal kidney such as production of hormones like erythropoietin, calcitriol and renin. Hence, if John has an option of kidney transplant, he must opt for it.

Question (B).
Amphibian tadpole can afford to be ammonotelic. Justify.
Answer:

Tadpole (larval stage of life cycle of amphibian) is aquatic. They are ammonotelic as they excrete nitrogenous waste in the form of ammonia.
Ammonia is very toxic and requires large amount of water for its elimination.
It is readily soluble in water and diffuses across the body surface and into the surrounding water.
Also, the water lost during excretion can be made up through the surrounding water in ammonotelic organisms.

Hence, amphibian tadpole can afford to be ammonotelic.

Question (C).
Birds are uricotelic in nature. Give reason.
Answer:

Birds are capable of converting ammonia into uric acid by ‘inosinic acid pathway’ in their liver.
Uric acid is least toxic and hence, it can be retained in the body for some time.
It is least soluble water hence, negligible amount of water is required for its elimination.
This mode of excretion can also help reduce body weight (an adaptation for flight) and those animals which
need to conserve more water follow uricotelism.

Hence, in order to conserve water as an adaptation for flight, birds are uricotelic in nature.

Question 4.
Write the explanation in your word

Question (A).
Nitya has been admitted to hospital after heavy blood loss. Till proper treatment could be given; how did Nitya’s body must have tackled the situation?
Answer:

Heavy blood loss is called haemorrhage. In case of haemorrhage or severe dehydration, the osmoreceptors stimulate Antidiuretic hormone (ADH) secretion.
ADH is important in regulating water balance through the kidneys. For detailed mechanism of reabsorption by ADH:

Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

Another regulatory mechanism that must have been activated is RAAS. For detailed mechanism of electrolyte reabsorption:

Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

Question 5.
Complete the diagram / chart with correct labels / information. Write the conceptual details regarding it

Question (A).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (B).

Answer:

Nephrons are structural and functional units of kidney.
Each nephron consists of a 4 – 6 cm long, thin-walled tube called the renal tubule and a bunch of capillaries known as the glomerulus.
The wall of the renal tubule is made up of a single layer of epithelial cells.
Its proximal end is wide, blind, cup-like and is called as Bowman’s capsule, whereas the distal end is open.
The nephron is divisible into Ilowman’s capsule, neck, proximal convoluted tubule (PCT), Loop of Henle (LoH), distal convoluted tubule (DCT) and collecting tubule (CT).
The glomerulus is present in the cup-like cavity of Bowman’s capsule and both are collectively known as renal corpuscle or Malpighian body.

Question (C)

Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Question (D).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

Question (E).

Answer:

When renal function of a person falls below 5 – 7 %, accumulation of harmful substances in blood begins. In such a condition the person has to go for artificial means of filtration of blood i.e. haemodialysis.
In haemodialysis, a dialysis machine is used to filter blood. The blood is filtered outside the body using a dialysis unit.
In this procedure, the patients’ blood is removed; generally from the radial artery and passed through a cellophane tube that acts as a semipermeable membrane.
The tube is immersed in a fluid called dialysate which is isosmotic to normal blood plasma. Hence, only excess salts if present in plasma pass through the cellophane tube into the dialysate.
Waste substances being absent in the dialysate, move from blood into the dialyzing fluid.
Filtered blood is returned to vein.
In this process it is essential that anticoagulant like heparin is added to the blood while it passing through the tube and before resending it into the circulation, adequate amount of anti-heparin is mixed.
Also, the blood has to move slowly through the tube and hence the process is slow.

Question 6.
Prove that mammalian urine contains urea.
Answer:

Urea is a nitrogenous waste formed by breakdown of protein (deamination of amino acids).
During this process, amino groups are removed from the amino acids present in the proteins and converted to highly toxic ammonia. The ammonia is finally converted to area through ornithine cycle. Thus, the urea formed is passed to kidneys and excreted out of the body through urine.
Reabsorption of urea (proximal tubule, collecting ducts) and active secretion of urea (Henle loop) leads to a urea circulation (urea recycling) between the lumen of the nephron and renal medulla, which is an important element of the renal urine concentration.
About 54 g of urea is filtered per day in the glomerular capsule, of which approximately 30 g is excreted in the urine and 24 g is reabsorbed into blood (assuming GFR is 180 litres/day).
Urinalysis can help detect the amount of urea in urine (Urine urea nitrogen test, urease test, etc.).

Practical / Project :

Visit to a nearby hospital or pathological laboratory and collect detailed information about different blood and urine tests.
Answer:
Testing the urine is known as urinalysis. It generally has three parts:

Visual examination: Check sample colour and clearness.
Dipstick examination: Checks for abnormal amounts of glucose, protein, etc.
Microscopic examination: Check for presence of RBCs. WBCs, bacteria, crystals, etc.
Apart from routine urine examination, specific tests may also be done. They are as follows:
- BUN (Blood Urea Nitrogen) Test: It measures the amount of nitrogen in blood and evaluates kidney function.
- Urease Test/ Urea Nitrogen Test: It is done to check the amount of urea in urine sample.
- Urine albumin to creatine ratio (UACR) test: Estimates the amount of albumin in urine.

[Students are expected to collect more information and perform the given activity on their own]

12th Biology Digest Chapter 15 Excretion and Osmoregulation Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Question 1.
Why are various waste products produced in the body of an organism?
Answer:
Metabolism produces a variety of by-products, some of which need to be eliminated. Such by-produçts are called metabolic waste products.

Question 2.
How are these waste eliminated?
Answer:
Depending on the type of waste product, they are eliminated through various organs of the body:

The various excretory products produced by the human body are as follows:

Fluids such as water; gaseous wastes like CO₂ nitrogenous wastes like ammonia, urea and uric acid, creatinine; minerals; salts of sodium, potassium. calcium, etc. if present in body in excess are excreted through urine, faeces and sweat.
Pigments formed due to breakdown of haemoglobin like bilirubin (excreted through faeces) and urochrome (eliminated through urine).
The pigments present in consumed foodstuffs like beet root or excess of vitamins, hormones and drugs.
Volatile substances present in spices (eliminated through lungs).

Have you ever observed? (Textbook Page No. 174)

Question 1.
When does urine appear deeply coloured?
Answer:
Urine can appear deeply coloured due to various reasons:

Severe dehydration resulting in production of concentrated urine.
Consumption of foodstuff like beet root, which contain coloured pigments.
Some medications can also cause the urine to appear deeply coloured.

Think about it. (Textbook Page No. 174)

Question 1.
Do organisms differ in type of metabolic wastes they produce?
Answer:
Yes, organisms differ in the type of metabolic wastes they produce. Some organisms excrete ammonia while some excrete urea or uric acid as metabolic wastes.

Question 2.
Do environment or evolution have any effect on type of waste produced by an organism?
Answer:

The theory of evolution proposes that life started in an aquatic environment.
Aquatic organisms are generally ammonotelic. It is believed that the urea cycle evolved to adapt to a changing environment when terrestrial life forms evolved.
Arid conditions probably led to the evolution of the uric acid pathway as a means of conserving water.
However, the correlation between evolution and type of waste production is uncertain.

Question 3.
How do thermoregulation and food habits affect saste production?
Answer:

To generate heat. endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more tì.od in order to meet their energy requirements.
Also, carnivorous diet contains more proteins than herbivores.
Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste
These animals would also require more energy to eliminate the high levels oF nitrogenous wastes which build up when animal protein is digested.

Use your brain power. (Textbookpage No. 175)

Question 1.
Why ammonia is highly toxic?
Answer:

Ammonia is basic in nature and its retention in the body would disturb the pH of the body.
An increase in pH would disturb all enzyme catalysed reactions in the body and also make the plasma membrane unstable.

Hence, ammonia is highly toxic to the body.

Find out. (Textbook page No. 175)

You will study about a type of arthritis called gouty arthritis caused due to accumulation of uric acid in joints. Where does uric acid come from in case of ureotelic human beings?
Answer:

Uric acid produced as a waste product during the normal breakdown of nucleic acids (purines) and certain naturally occurring substances found in foods such as mushrooms. Mackerel, dried beans. etc.
This uric acid is generally excreted out along with urine.
If uric acid is produced in excess or not excreted, it accumulates in joints causing gouty arthritis.

Think about it. (Textbook Page No.175)

Endotherms consume more food in order to meet energy requirements. Also, carnivorous diet contains more proteins than herbivorous. Does it affect excretion of nitrogenous waste?
Answer:

To generate heat, endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more food in order to meet their energy requirements.
Also, carnivorous diet contains more proteins than herbivores.
Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste.
These animals would also require more energy to eliminate the high levels of nitrogenous wastes which build up when animal protein is digested.

Observe and Discuss (Textbook Page No. 176)

Question 1.
These are blood reports of patients undergoing investigations for kidney function. What is creatinine? What is your observation and opinion about the findings? Why is it used as an index of kidney function?

Answer:
i. Creatinine:

Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
It provides a ready source of high energy phosphate.

ii. Observations and Opinion:
Report A indicates a value of creatinine which is higher than the normal range. This would indicate impaired kidney function.
Report B indicates high fasting blood sugar and detection of sugar in the blood is known as glucosuria. High fasting blood sugar (>126 mg/dL) is usually indicative of diabetes.

iii. Creatinine used as an index of kidney function:

Normally blood creatinine levels remain steady because the rate of production matches his excretion in urine.
Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

Think about it. (Textbook Page No. 176)

Question 1.
During summer, we tend to produce less urine, is it so?
Answer:

Generally, excess water containing wastes is lost from the body in the time of urine. sweat and faeces.
During summer when the surrounding temperature is high. we also lose a considerable amount of water in the form of sweat.

Thus, the kidneys retain water for maintaining the concentration of body fluids and reduce the amount of water lost through urine.

Question 2.
Marine birds like Albatross spend their life on the sea. That means the water the, drink is salty. how do they manage osnioregulation then?
Answer:

Marine birds like Albatross have special glands called salt glands near their nostrils.
These glands are capable of secreting salts by active transport and help to manage osmotic balance,

[Note: The salt glands in Albatross are located in or on the skull in the area of eyes.]

Question 3.
like ectothermic and endothermic animals, do organisms differ in the way they maintain salt balance?
Answer:
Yes, organisms differ in the way they maintain salt balance.

Animals can be either isosmotic to the surrounding (osmoconformers or control the internal environment independent of external environment (osmoregulators).
Marine organisms are mostly osmoconfòrmers because their body fluids and external environment are isosmotic in nature while fresh water forms and terrestrial organisms are osmoregulators,
Generally, most organisms can tolerate only a narrow range of salt concentrations. Such organism are known as stenohaline organisms.
Organisms which are capable of handling a wide change in salinity are called euryhaline organisms.e.g. Barnacles, clams etc.

Find out. (Textbook Page No 176)

Question 1.
How do freshwater fishes and marine fishes carry out osmoregulation?
Answer:
Osmoregulation is the process of maintaining an internal concentration of salt and water in the body of fishes.

i. Freshwater fishes:
The salt concentration inside the body of freshwater fishes is higher than their surrounding water. Due to this, water enters the body due to osmosis.
If the flow of water into the body is not regulated. fishes would swell and get bigger.
To compensate this, the kidneys produce a large amount of urine,
Excretion of large amounts of urine regulates the level of water in the body hut results in the loss of salts.
Thus, in order to maintain a sufficient salt level, special cells in the gills (chloride cells) take tip ions from
the water, which are then directly transported into the blood.

ii. Marine fishes:
Since the salt content in blood of marine fishes is much lower than that of seawater, they constantly tend to lose water and build up salt.
To replace the water loss, they continually need to drink seawater.
Since their small kidney can only excrete a relatively small amount of urine, salt is additionally excreted through gills, where chloride cells work in reverse as in freshwater fishes.

Make a table. (Textbook Page No. 178)

Question 1.
The details of modes of excretion of nitrogenous wastes.
Answer:
The three main modes of excretion in animals are as follows:

i. Ammonotelism
ii. Ureotelism
iii. Uricotelism

i. Ammonotelism:

Elimination of nitrogenous wastes in the form of ammonia is called as ammonotelism.
Ammonia is basic in nature and hence it can disturb the pH of the body, if not eliminated immediately.
Any change in pH would disturb all enzyme catalyzed reactions in the body and would also make the plasma membrane unstable.
Ammonia is readily soluble in water and needs large quantity of water to dilute and reduce its toxicity.
This is however an energy saving mechanism of excretion and hence all animals that have plenty of water available for dilution of ammonia, excrete nitrogenous wastes in the form of ammonia.
Animals that follow this mode of excretion are known as ammonotelic animals.
1 gm ammonia needs about 300 – 500 ml of water for elimination.
Ammonotelic animals excrete ammonia through general body surface (skin), gills and kidneys.
e.g. Ammonotelism is found in aquatic invertebrates, bony fishes, and aquatic / larval amphibians. Animals without excretory system (Protozoa) are also ammonotelic.

ii. Ureotelism:

Elimination of nitrogenous wastes in the form of urea is called as ureotelism.
Urea is comparatively less toxic and less water-soluble than ammonia. Hence, it can be concentrated to some extent in body.
The body requires less water for elimination.
Since it is less toxic and less water soluble, ureotelism is suitable for animals that need to conserve water to some extent. Hence, ureotelism is common in terrestrial animals, as they have to conserve water.
It takes about 50 ml H₂O for removal of 1 gm NH₂ in form of urea.
Ureotelic animals generally convert ammonia to urea in the liver by operating ornithine / urea cycle in which 3 ATP molecules are used to produce one molecule of urea.
e.g. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most of the adult amphibians, etc. are ureotelic.

iii. Uricotelism:

Elimination of nitrogenous wastes in the form of uric acid is called as uricotelism.
Uric acid is least toxic and hence, it can be retained in the body for some time in concentrated form.
It is least soluble in water. Hence there is minimum (about 5 – 10 ml for 1 gm) or no need of water for its elimination.
Those animals which need to conserve more water follow uricotelism. However, these animals need to spend more energy.
Ammonia is converted into uric acid by ‘inosinic acid pathway’ in the liver of birds, e.g. Birds, some insects, many reptiles, land snails, are uricotelic.

No.	Ureotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of urea.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of urea requires less (moderate ) amount of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of urea requires 50 ml of water.	Removal of 1 gm of uric acid requires 5 – 10 ml of
iv.	rea is less toxic.	Uric acid is least toxic.
e.g.	It is generally seen in terrestrial animals. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most adult amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

No.	Ammonotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of ammonia.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of ammonia requires plenty of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of ammonia requires 300 – 500 ml of water.	Removal of 1 gm of uric acid requires 10ml of water.
iv.	Ammonia is very toxic.	Uric acid is less toxic.
e.g.	It is found in aquatic invertebrates, bony fishes and aquatic/ larval amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

[Students can Refer these and make a chart on their own.]

Use your brain power. (Textbook Page No. 178)

Question 1.
Creatinine is considered as index of kidney function. Give reason.
Answer:

Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
It provides a ready source of high energy phosphate.
Normally blood creatinine levels remain steady because the rate of production matches its excretion in urine.
Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

[Note: Plasma creatinine is a waste product produced by muscles from the breakdown of a compound called ‘creatine phosphate ’.]

Make a table. (Textbook Page No. 178)

Question 1.
The excretory organs found in various animal phyla.
Answer:

Sr. No.	Animal Phyla	Excretory organs
i.	Porifera	Lack excretory organ instead rely on water transport system/ Canal system
ii.	Coelenterata	Lack specialised excretory organs. Excretion takes place through simple diffusion or through the mouth.
iii.	Ctenophora	Lack specialised excretory organs
iv.	Platyhelminthes	Protonephridia or Flame cells
v.	Aschelminthes	Excretory tube and pore
vi.	Annelida	Nephridia
vii.	Arthropoda	Malpighian tubules
viii.	Mollusca	Organ of Bojanus
ix.	Echinodermata	Lack specialized excretory organs, waste materials directly diffuse into water or are excreted through tube feet
x.	Hemichordata	Proboscis gland
xi.	Chordata	Kidney

Observe and complete. (Textbook Page No. 178)

Question 1.
Label the diagram and complete following paragraphs.

i. Kidney: A pair of ____ shaped kidneys are present on either side of the ____ from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind ___. Hence are called retroperitoneal. Dimensions of
each kidney are 10 × ____ × ____ cms. Average weight is ___ g in males and 135 g in ____. Outer surface is ___ and inner is concave. Notch on the inner concave surface is called ___. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called ___.

ii. Ureters: A pair of ureters arise from ___ of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into ___ by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of ___ of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median ___ sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the ___ there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional ___. It helps bladder to stretch.

iv. Urethra: It is a ___ structure arising from urinary bladder and opening to the exterior of the body.
There are ___ urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of ___ muscles, involuntary in nature.
b. External sphincter: Made up of ___ muscles, voluntary in nature.
Answer:

i. Kidney: A pair of bean shaped kidneys are present on either side of the backbone from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind peritoneum. Hence are called retroperitoneal. Dimensions of each kidney are 10 × 5 × 4 cms. Average weight is 150 g in males and 135 g in females. Outer surface is convex and inner is concave. Notch on the inner concave surface is called hilum. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called nephron.

ii. Ureters: A pair of ureters arise from hilum of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into urinary bladder by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of backward flow of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median pear-shaped sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the urinary bladder there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional epithelial tissue. It helps bladder to stretch.

iv. Urethra: It is a fibromuscular tube-like structure arising from urinary bladder and opening to the exterior of the body. There are two urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of detrusor muscles, involuntary in nature.
b. External sphincter: Made up of striated muscles, voluntary in nature.
If this valve is not functioning properly during inflammation of bladder, it can lead to kidney infection.

Internet is my friend. (Textbook Page No. 179)

Question 1.
Find out what is floating kidney.
Answer:

Floating kidney or nephroptosis, is an inferior displacement or dropping of the kidney.
This condition occurs when the kidney slips from its normal position because it is not held securely in place by the adjacent organs or its fat covering.
It generally develops in extremely thin people whose adipose capsule or renal fascia is deficient.
It may result in twisting of the ureter and cause blockage of urine flow. The resulting backup of urine would put pressure on the kidney and damage the tissues.
Twisting of the ureter may also cause pain and discomfort.
This condition is more common in females than males and happens commonly among one in four people.
Weakening of the fibrous bands that hold the kidney in place can predispose to floating kidney.

Can you recall? (Textbook Page No. 179)

Question 1.
Observe the figure carefully and label various regions of L.S. of kidney.

Answer:

Can you tell? (Textbook Page No.182)

Question 1.
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question 2.
Why urinary tract infections are more common in females than males?
Answer:

The urethra in women (4 cm) is much shorter than that of males (20 cm).
This allows easy passage of bacteria into the urinary bladder.

Hence, urinary tract infections are more common in females than males.

Question 3.
What is nephron? Which are its main parts? Why are they important?
Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

ii. Renal tubule:

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Think about ¡t. (Textbook Page No. 182)

Question 1.
How much blood ¡s supplied to kidney?
Answer:
Around 600 ml of blood passes through each kidney per minute.

Do this. (Textbook Page No. 183)

Question 1.
Check blood reports of patients and comment about possibility of glucosuria.
Answer:
Glucosuria is the presence of glucose sugar in urine. High glucose in urine is usually indicative of diabetes mellitus.

Condition	Glucose range in urine
Normal	0 to 15 mg/dL (0 to 0.8 mmol/L)
Prediabetes	100 to 125 mg/dL (5.6 to 6.9 mmol/L)
Diabetes	126 mg/dL (7 mmol/L)

[Students can get access to sample reports on the internet and refer the above table to comment on blood reports of patients on their own.]

Use your brain power. (Textbook Page No. 185)

Question 1.
In which regions of nephron the filtrate will he isotonic to blood?
Answer:
Filtrate leasing the proximal convoluted tubule (PCT) is isotonic to the blood plasma.

Can you tell? (Textbook Page No. 185)

Question 1.
Explain the process of urine formation in details.
Answer:
Process of urine formation is completed in three steps, namely;

i. Ultrafiltration/ Glomerular filtration,
ii. Selective reabsorption,
iii. Tubular secretion / Augmentation

i. Ultrafiltration / Glomerular filtration :
Diameter of afferent arteriole is greater than the efferent arteriole. The diameter of capillaries is still smaller than both arterioles. Due to the difference in diameter, blood flows with greater pressure through the glomerulus. This is called as glomerular hydrostatic pressure (GHP) and normally, it is about 55 mmHg. GIIP is opposed by osmotic pressure of blood (normally, about 30 mm Hg) and capsular pressure (normally, about 15 mm Hg).

Hence net / effective filtration pressure (EFP) is 10 mm Hg.
EFP = Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
= 55 – (30 + 15)
= 10 mm Hg

Under the effect of high pressure, the thin walls of the capillary become permeable to major components of blood (except blood cells and macromolecules like protein).
Thus, plasma except proteins oozes out through wall of capillaries.
About 600 ml blood passes through each kidney per minute.

The blood (plasma) flowing through kidney (glomeruli) is filtered as glomerular filtrate, at a rate of 125 ml / min. (180 L/d).
Glomerular filtrate / deproteinized plasma / primary urine is alkaline, contains urea, amino acids, glucose, pigments, and inorganic ions.
Glomerular filtrate passes through filtration slits into capsular space and then reaches the proximal convoluted tubule.

ii. Selective reabsorption :
Selective reabsorption occurs in proximal convoluted tubule (PCT). It is highly coiled so that glomerular filtrate passes through it very slowly. Columnar cells of PCT are provided with microvilli due to which absorptive area increases enormously.
This makes the process of reabsorption very effective.
These cells perform active (ATP mediated) and passive (simple diffusion) reabsorption.

Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca⁺⁺, K⁺, Na⁺, Cl^– are absorbed actively, against the concentration gradient. Low threshold substances like water, sulphates, nitrates, etc., are absorbed passively.
In this way, about 99% of glomerular filtrate is reabsorbed in PCT and DCT.

iii. Tubular secretion / Augmentation :
Finally filtrate reaches the distal convoluted tubule via loop of Henle. Peritubular capillaries surround DCT. Cells of distal convoluted tubule and collecting tubule actively absorb the wastes like creatinine and ions like K⁺, H⁺ from peritubular capillaries and secrete them into the lumen of DCT and CT, thereby augmenting the concentration of urine and changing its pH from alkaline to acidic.
Secretion of H⁺ ions in DCT and CT is an important homeostatic mechanism for pH regulation of blood. Tubular secretion is the only process of excretion in marine bony fishes and desert amphibians.

Question 2.
How does counter current mechanism help concentration of urine?
Answer:
Under the conditions like low water intake or high water loss due to sweating, humans can produce concentrated urine. This urine can be concentrated around four times i.e. 1200 mOsm/L than the blood (300 mOsm/L). Hence, a mechanism called countercurrent mechanism is operated in the human kidneys. The countercurrent mechanism operating in the Limbs of Henle’s loop of juxtamedullary nephrons and vasa recta is as follows:

It involves the passage of fluid from descending to ascending limb of Henle’s loop.
This mechanism is called countercurrent mechanism, since the flow of tubular fluid is in opposite direction through both limbs.
In case of the vasa recta, blood flows from ascending to descending parts of itself.
Wall of descending limb is thin and permeable to water, hence, water diffuses from tubular fluid into tissue fluid due to which, tubular fluid becomes concentrated.
The ascending limb is thick and impermeable to water. Its cells can reabsorb Na⁺ and Cl^– from tubular fluid and release into tissue fluid.
Due to this, tissue fluid around descending limb becomes concentrated. This makes more water to move out from descending limb into tissue fluid by osmosis.
Thus, as tubular fluid passes down through descending limb, its osmolarity (concentration) increases gradually due to water loss and on the other hand, progressively decreases due to Na⁺ and Cl^– secretion as it flows up through ascending limb.
Whenever retention of water is necessary, the pituitary secretes ADH. ADH makes the cells in the wall of collecting ducts permeable to water.
Due to this, water moves from tubular fluid into tissue fluid, making the urine concentrated.
Cells in the wall of deep medullar part of collecting ducts are permeable to urea. As concentrated urine flows through it, urea diffuses from urine into tissue fluid and from tissue fluid into the tubular fluid flowing through thin ascending limb of Henle’s loop.
This urea cannot pass out from tubular fluid while flowing through thick segment of ascending limb, DCT and cortical portion of collecting duct due to impermeability for it in these regions.
However, while flowing through collecting duct, water reabsorption is operated under the influence of ADII. Due to this, urea concentration increases in the tubular fluid and same urea again diffuses into tissue fluid in deep medullar region.
Thus, same urea is transferred between segments of renal tubule and tissue fluid of inner medulla. This is called urea recycling; operated for more and more water reabsorption from tubular fluid and thereby excreting small volumes of concentrated urine.
Osmotic gradient is essential in the renal medulla for water reabsorption by counter current multiplier system.
This osmotic gradient is maintained by vasa recta by operating counter current exchange system.
Vasa recta also have descending and ascending limbs. Blood that enters the descending limb of the vasa recta has normal osmolarity of about 300 mOsm/L.
As it flows down in the region of renal medulla where tissue fluid becomes increasingly concentrated, Na⁺, Cl^– and urea molecules diffuse from tissue fluid into blood and water diffuse from blood into tissue fluid.
Due to this, blood becomes more concentrated which now flows through ascending part of vasa recta. This part runs through such region of medulla where tissue fluid is less concentrated.
Due to this, Na⁺, Cl^– and urea molecules diffuse from blood to tissue fluid and water from tissue fluid to blood. This mechanism helps to maintain the osmotic gradient.

Try this. (Textbook Page No. 185)

Question 1.
Read the given urine report and prepare a note on composition of normal urine.

Answer:
The composition of normal urine is as follows:

A volume of 1 – 2 litres of urine in 24 hours is normal. This volume can however vary considerably as it depends on fluid intake, physical activity, temperature, etc.
The colour of normal urine is generally pale yellow due to urochrome (pigment produced by breakdown of bile). The colour of urine may vary slightly due to urochrome concentration and diet.
The appearance of urine is generally clear and transparent.
Any form of deposits (sediments/ crystals) is generally absent in normal urine.
The pH of normal urine is acidic and is generally around 6.0 (Range: 4.6 to 8.0). The pH varies considerably with the diet of a person.
The specific gravity of urine is an average of 1.02 ( Range : 1.001 to 1.035).
Albumin, sugar, bile salts bile pigments, ketone bodies and casts are absent in normal urine.
Occult blood is generally not seen in normal urine.

Think (Textbook Page No. 185)

Question 1.
What would happen if ADH secretion decreases due to any reason?
Answer:
In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus. Frequent excretion of large amount of dilute urine may cause a person to feel thirsty.

Think and appreciate. (Textbook Page No. 185)

Question 1.
How do kidneys bring about homeostasis? Is there any role of neuro endocrine system in it?
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.
If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

No. Both ADH and RAAS are essential for homeostasis.

Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
Action of ADH and RAAS leads to increase in blood volume and osmolarity.
For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

ADH is produced by the hypothalamus and is stored and released by the posterior pituitary or the neurohypophysis in response to appropriate trigger. Hence, there is a role of the neuroendocrine system in homeostasis.

Use your brain power. (Textbook Page No. 186)

Question 1.
Can we use this knowledge in treatment of high blood pressure? Why high BP medicines are many a times diuretics?
Answer:

Yes, the knowledge of homoeostasis is used in the treatment of high blood pressure.
Some commonly used theories for treatment of high blood pressure are as follows:
- Angiotensin II receptor blockers (ARBs) are used as medications to treat high blood pressure. These medications block the action of angiotensin II by preventing angiotensin II from binding to angiotensin II receptors on the muscles surrounding blood vessels. As a result, blood vessels enlarge (dilate), and blood pressure is reduced.
- Another method is the use of ‘Angiotensin converting enzyme’ ACE blockers. These inhibitors inhibit activity of ACE and therefore decrease the production of angiotensin II. As a result, these medications cause the blood vessels to enlarge or dilate, and this reduces blood pressure.
Vasodilation reduces arterial pressure. Reduced angiotensin II leads to natriuresis (increased excretion of Na⁺ in urine) and diuresis, thereby reducing blood pressure.
Too much salt can cause extra fluid to build up in the blood vessels, raising blood pressure. Diuretics are substances that slow renal absorption of water and thereby cause diuresis (elevated urine flow rate) which in turn reduces blood volume and blood pressure by flushing out salt and extra fluid. Hence, high BP medicines are many a times diuretics.

Can you tell? (Textbook page no. 186)
How do skin and lungs help in excretion?
OR
Can you tell? (Textbook page no. 187)
Explain role of lungs and skin in excretion.
Answer:
Yes, various organs other than the kidney participate in excretion. They are as follows:

i. Skin:

Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.

Sweat glands are distributed all over the skin. They are abundant in the palm and facial regions.
These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCl, lactic acid and urea.
Sebaceous glands are present at the neck of hair follicles. They secrete oily substance called sebum.
It forms a lubricating layer on skin making it softer. It protects skin from infection and injury.

ii. Lungs:

Lungs are the accessory excretory organs. They help in excretion of volatile substances like CO₂ and water vapour produced during cellular respiration. Along with CO₂, lungs also remove excess of H₂O in the form of vapours during expiration. They also excrete volatile substances present in spices and other food stuff.

Can you tell? (Textbook Page No. 187)

Question 1.
When does kidney produce renin? Where is it produced in kidney?
Answer:
Kidney produces renin whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration).
The juxtaglomerular Apparatus (JGA) cells secrete renin.

Question 2.
Explain how electrolyte balance of blood plasma maintained.
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

No. Both ADH and RAAS are essential for homeostasis.

Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
Action of ADH and RAAS leads to increase in blood volume and osmolarity.
For mechanism of Atrial natriuretic peptide:

Can you tell? (Textbook Page No. 187)

Question 1.
What is the composition of sweat?
Answer:
Sweat is composed of water, NaCl, lactic acid and urea.

Internet my friend. (Textbook Page No. 189)

Question 1.
Treatments other than surgical removal of kidney stone like Lithotripsy. (Breaking down of kidney stones using shock waves).
Answer:
a. Cystoscopy and ureteroscopy:
During cystoscopy, the doctor uses a cystoscope to look inside the urethra and bladder to find a stone in the urethra or bladder.
During ureteroscopy, the doctor uses a ureteroscope, which is longer and thinner than a cystoscope, to see detailed images of the lining of the ureters and kidneys.

The doctor inserts the cystoscope or ureteroscope through the urethra to see the rest of the urinary tract. Once the stone is found, the doctor can remove it or break it into smaller pieces.
The doctor performs these procedures in the hospital with anesthesia.

b. Percutaneous nephrolithotomy:
The doctor uses a thin viewing tool, called a nephroscope, to locate and remove the kidney stone.
The doctor inserts the tool directly into your kidney through a small cut made in your back.
For larger kidney stones, the doctor also may use a laser to break the kidney stones into small pieces. The doctor performs percutaneous nephrolithotomy in a hospital with anesthesia.

c. Generally for smaller stones doctors recommend drinking lots of water, consuming pain relievers and consuming medicines like alpha blocker to relax the ureter muscles, and help pass the kidney stones more quickly and with less pain

[Students are expected to find more information using the internet.]

Question 2.
Dietary restrictions suggested for kidney patients.
Dietary restrictions for kidney patients include the following:

Drinking large amounts of water.
Reduce consumption of oxalate rich food like rhubarb, beets, okra, spinach, Swiss chard, sweet potatoes, nuts, tea, chocolate and soy products.
Follow a diet low in salt and animal protein.
Reduce consumption of calcium supplements (if any) but consume appropriate amount of calcium in food.

[Students are expected to find more information using the internet.]

11th Std Biology Questions And Answers:

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

September 20, 2024September 18, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Miscellaneous Exercise 9 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

(I) Select the appropriate option from the given alternatives.

Question 1.
If y = \(\frac{x-4}{\sqrt{x+2}}\), then \(\frac{d y}{d x}\) is
(A) \(\frac{1}{x+4}\)
(B) \(\frac{\sqrt{x}}{\left(\sqrt{x+2)^{2}}\right.}\)
(C) \(\frac{1}{2 \sqrt{x}}\)
(D) \(\frac{x}{(\sqrt{x}+2)^{2}}\)
Answer:
(C) \(\frac{1}{2 \sqrt{x}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q1

Question 2.
If y = \(\frac{a x+b}{c x+d}\),then \(\frac{d y}{d x}\) =
(A) \(\frac{a b-c d}{(c x+d)^{2}}\)
(B) \(\frac{a x-c}{(c x+d)^{2}}\)
(C) \(\frac{a c-b d}{(c x+d)^{2}}\)
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Answer:
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q2

Question 3.
If y = \(\frac{3 x+5}{4 x+5}\), then \(\frac{d y}{d x}\) =
(A) \(-\frac{15}{(3 x+5)^{2}}\)
(B) \(-\frac{15}{(4 x+5)^{2}}\)
(C) \(-\frac{5}{(4 x+5)^{2}}\)
(D) \(-\frac{13}{(4 x+5)^{2}}\)
Answer:
(C) \(-\frac{5}{(4 x+5)^{2}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q3

Question 4.
If y = \(\frac{5 \sin x-2}{4 \sin x+3}\), then \(\frac{d y}{d x}\) =
(A) \(\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
(C) \(-\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(D) \(-\frac{15 \cos x}{(4 \sin x+3)^{2}}\)
Answer:
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q4

Question 5.
Suppose f(x) is the derivative of g(x) and g(x) is the derivative of h(x).
If h(x) = a sin x + b cos x + c, then f(x) + h(x) =
(A) 0
(B) c
(C) -c
(D) -2(a sin x + b cos x)
Answer:
(B) c
Hint:
h(x) = a sin x + b cos x + c
Differentiating w.r.t. x, we get
h'(x) = a cos x – b sin x = g(x) …..[given]
Differentiating w.r.t. x, we get
g'(x) = -a sin x – b cos x = f(x) …..[given]
∴ f(x) + h(x) = -a sin x – b cos x + a sin x + b cos x + c
∴ f(x) + h(x) = c

Question 6.
If f(x) = 2x + 6, for 0 ≤ x ≤ 2
= ax² + bx, for 2 < x ≤ 4
is differentiable at x = 2, then the values of a and b are
(A) a = \(-\frac{3}{2}\), b = 3
(B) a = \(\frac{3}{2}\), b = 8
(C) a = \(\frac{1}{2}\), b = 8
(D) a = \(-\frac{3}{2}\), b = 8
Answer:
(D) a = \(-\frac{3}{2}\), b = 8
Hint:
f(x) = 2x + 6, 0 ≤ x ≤ 2
= ax² + bx, 2 < x ≤ 4
Lf'(2) = 2, Rf'(2) = 4a + b
Since f is differentiable at x = 2,
Lf'(2) = Rf'(2)
∴ 2 = 4a + b …..(i)
f is continuous at x = 2.
∴ \(\lim _{x \rightarrow 2^{+}} f(x)=f(2)=\lim _{x \rightarrow 2^{-}} f(x)\)
∴ 4a + 2b = 2(2) + 6
∴ 4a + 2b = 10
∴ 2a + b = 5 …..(ii)
Solving (i) and (ii), we get
a = \(-\frac{3}{2}\), b = 8

Question 7.
If f(x) = x² + sin x + 1, for x ≤ 0
= x² – 2x + 1, for x ≤ 0, then
(A) f is continuous at x = 0, but not differentiable at x = 0
(B) f is neither continuous nor differentiable at x = 0
(C) f is not continuous at x = 0, but differentiable at x = 0
(D) f is both continuous and differentiable at x = 0
Answer:
(A) f is continuous at x = 0, but not differentiable at x = 0
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q7

Question 8.
If f(x) = \(\frac{x^{50}}{50}+\frac{x^{49}}{49}+\frac{x^{48}}{48}+\ldots .+\frac{x^{2}}{2}+x+1\), then f'(1) =
(A) 48
(B) 49
(C) 50
(D) 51
Answer:
(C) 50
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q8

(II).

Question 1.
Determine whether the following function is differentiable at x = 3 where,
f(x) = x² + 2, for x ≥ 3
= 6x – 7, for x < 3.
Solution:
f(x) = x² + 2, x ≥ 3
= 6x – 7, x < 3
Differentiability at x = 3
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q1

Here, Lf'(3) = Rf'(3)
∴ f is differentiable at x = 3.

Question 2.
Find the values of p and q that make function f(x) differentiable everywhere on R.
f(x) = 3 – x, for x < 1
= px² + qx, for x ≥ 1.
Solution:
f(x) is differentiable everywhere on R.
∴ f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q2
f(x) is differentiable at x = 1.
∴ Lf'(1) = Rf'(1)
∴ -1 = 2p + q …..(ii)
Subtracting (i) from (ii), we get
p = -3
Substituting p = -3 in (i), we get
p + q = 2
∴ -3 + q = 2
∴ q = 5

Question 3.
Determine the values of p and q that make the function f(x) differentiable on R where
f(x) = px³, for x < 2
= x² + q, for x ≥ 2
Solution:
f(x) is differentiable on R.
∴ f(x) is differentiable at x = 2.
∴ f(x) is continuous at x = 2.
Continuity at x = 2:
f(x) is continuous at x = 2.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3

f(x) is differentiable at x = 2.
∴ Lf'(2) = Rf'(2)
∴ 12p = 4
∴ p = \(\frac{1}{3}\)
Substituting p = \(\frac{1}{3}\) in (i), we get
8(\(\frac{1}{3}\) – q = 4
∴ q = \(\frac{8}{3}\) – 4 = \(\frac{-4}{3}\)

Question 4.
Determine all real values of p and q that ensure the function
f(x) = px + q, for x ≤ 1
= tan(\(\frac{\pi x}{4}\)), for 1 < x < 2
is differentiable at x = 1.
Solution:
f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Continuity at x= 1:
f(x) is continuous at x = 1.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q4

Question 5.
Discuss whether the function f(x) = |x + 1| + |x – 1| is differentiable ∀ x ∈ R.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q5

Here, Lf'(1) ≠ Rf'(1)
∴ f is not differentiable at x = 1.
∴ f is not differentiable at x = -1 and x = 1.
∴ f is not differentiable ∀ x ∈ R.

Question 6.
Test whether the function
f(x) = 2x – 3, for x ≥ 2
= x – 1, for x < 2
is differentiable at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q6

Question 7.
Test whether the function
f(x) = x² + 1, for x ≥ 2
= 2x + 1, for x < 2
is differentiable at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7

Question 8.
Test whether the function
f(x) = 5x – 3x², for x ≥ 1
= 3 – x, for x < 1
is differentiable at x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q8
Here, Lf'(1) = Rf'(1)
∴ f(x) is differentiable at x = 1.

Question 9.
If f(2) = 4, f'(2) = 1, then find \(\lim _{x \rightarrow 2}\left[\frac{x f(2)-2 f(x)}{x-2}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q9

Question 10.
If y = \(\frac{\mathbf{e}^{x}}{\sqrt{x}}\), find \(\frac{d y}{d x}\) when x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q10

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

September 20, 2024September 18, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

(I) Differentiate the following w.r.t. x

Question 1.
y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q1

Question 2.
y = √x + tan x – x³
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q2

Question 3.
y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q3

Question 4.
y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4

Question 5.
y = 7^x + x⁷ – \(\frac{2}{3}\) x√x – log x + 7⁷
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q5

Question 6.
y = 3 cot x – 5e^x + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q6

(II) Diffrentiate the following w.r.t. x

Question 1.
y = x⁵ tan x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q1

Question 2.
y = x³ log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q2

Question 3.
y = (x² + 2)² sin x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q3

Question 4.
y = e^x log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q4

Question 5.
y = \(x^{\frac{3}{2}} e^{x} \log x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q5

Question 6.
y = \(\log e^{x^{3}} \log x^{3}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q6

(III) Diffrentiate the following w.r.t. x

Question 1.
y = x²√x + x⁴ log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q1

Question 2.
y = e^x sec x – \(x^{\frac{5}{3}}\) log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q2

Question 3.
y = x⁴ + x√x cos x – x² e^x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q3

Question 4.
y = (x³ – 2) tan x – x cos x + 7^x . x⁷
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q4

Question 5.
y = sin x log x + e^x cos x – e^x √x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q5

Question 6.
y = e^x tan x + cos x log x – √x 5^x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q6

(IV) Diffrentiate the following w.r.t.x.

Question 1.
y = \(\frac{x^{2}+3}{x^{2}-5}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q1

Question 2.
y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q2

Question 3.
y = \(\frac{x e^{x}}{x+e^{x}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q3

Question 4.
y = \(\frac{x \log x}{x+\log x}\)
Solution:
y = \(\frac{x \log x}{x+\log x}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q4

Question 5.
y = \(\frac{x^{2} \sin x}{x+\cos x}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q5

Question 6.
y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q6

(V).

Question 1.
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution:
Let f(x) = ax² + bx + c …..(i)
∴ f(0) = a(0)² + b(0) + c
∴ f(0) = c
But, f(0) = 3 …..(given)
∴ c = 3 …..(ii)
Differentiating (i) w.r.t. x, we get
f'(x) = 2ax + b
∴ f'(2) = 2a(2) + b
∴ f'(2) = 4a + b
But, f'(2) = 2 …..(given)
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x² – 18x + 3

Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x² – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)² – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.

Question 2.
If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).
Solution:
f(x) = a sin x – b cos x
Differentiating w.r.t. x, we get
f'(x) = a cos x – b (- sin x)
∴ f'(x) = a cos x + b sin x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 V Q2

Now, f(x) = a sin x – b cos x
∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.
y = e^x . tan x
Diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q1
Solution:

Question 2.
y = \(\frac{\sin x}{x^{2}+2}\)
diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q2
Solution:

Question 3.
y = (3x² + 5) cos x
Diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q3
Solution:

Question 4.
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q4

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

September 20, 2024September 18, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x² + 3x – 1
Solution:
Let f(x) = x² + 3x – 1
∴ f(x + h) = (x + h)² + 3(x + h) – 1
= x² + 2xh + h² + 3x + 3h – 1
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (i)

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (ii)

(d) 3^x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (iv)

(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (v)

(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vi)

(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vii)

(h) x√x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (viii)

Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) \(\sqrt{2 x+5}\) at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (i)

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (ii)

(iii) 2^3x+1 at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (iii)

(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (iv)

(v) e^3x-4 at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (v)

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi)

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x² + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q3

∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.

Question 4.
Show that f(x) = x² is continuous and differentiable at x = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q4

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (i)

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii)

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q6
f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x² if x ≤ 2 at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q7

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8

Question 9.
Examine the function
f(x) = x² cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q9