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Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 11 Electric Current Through Conductors Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 1.
Define current. State its formula and SI unit.
Answer:

1. Current is defined as the rate of flow of electric charge.
2. Formula: I = \(\frac {q}{t}\)
3. SI unit: ampere (A)

Question 2.
Derive an expression for a current generated due to flow of charged particles
Answer:
i. Consider an imaginary gas of both negatively and positively charged particles moving randomly in various directions across a plane P.

ii. In a time interval t, let the amount of positive charge flowing in the forward direction be q⁺ and the amount of negative charge flowing in the forward direction be q^–. Thus, the net charge flowing in the forward direction is q = q⁺ – q^–

iii. Let I be the current varying with time. Let ∆q be the amount of net charge flowing across the plane P from time t to t + At, i.e. during the time interval ∆t.

iv. Then the current is given by
I(t) = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{q}}{\Delta \mathrm{t}}\)
Flere, the current is expressed as the limit of the ratio (∆q/∆t) as ∆t tends to zero.

Question 3.
Match the amount of current generated A given in column – II with the sources given in column -I.

Column I	Column II
1. Lightening	a. Few amperes
2. House hold circuits	b. 10000 A
	c. Order of µA

Answer:

Column I	Column II
1. Lightening	b. 10000 A
2. House hold circuits	a. Few amperes

Question 4.
Which are the most common units of current used in semiconductor devices?
Answer:

1. milliampere (mA)
2. microampere (µA)
3. nanoampere (nA)

Question 5.
Six ampere current flows through a bulb. Find the number of electrons that should flow through the bulb in a time of 4 hrs.
Answer:
Given: I = 6 A, t = 4 hrs = 4 × 60 × 60 s
To find: Number of electrons (N)
Formula: I = \(\frac {q}{t}\) = \(\frac {Ne}{t}\)
Calculation: As we know, e = 1.6 × 10^-19 C
From formula,
N = \(\frac {It}{e}\) = \(\frac {6×4×60×60}{1.6×10^{-19}}\) 6x4x60x60 = 5.4 × 10²³

Question 6.
Explain flow of current in different conductor.
Answer:

1. A current can be generated by positively or negatively charged particles.
2. In an electrolyte, both positively and negatively charged particles take part in the conduction.
3.  In a metal, the free electrons are responsible for conduction. These electrons flow and generate a net current under the action of an applied electric field.
4. As long as a steady field exists, the electrons continue to flow in the form of a steady current.
5. Such steady electric fields are generated by cells and batteries.

Question 7.
State the sign convention used to show the flow of electric current in a circuit.
Answer:
The direction of the current in a circuit is drawn in the direction in which positively charged particles would move, even if the current is constituted by the negatively charged particles, (electrons), which move in the direction opposite to that the electric field.

Question 8.
Explain the concept of drift velocity with neat diagrams.
Answer:
i. When no current flows through a copper rod, the free electrons move in random motion. Therefore, there is no net motion of these electrons in any direction.

ii. If an electric field is applied along the length of the copper rod, a current is set up in the rod. The electrons inside rod still move randomly, but tend to ‘drift’ in a particular direction.

iii. Their direction is opposite to that of the applied electric field.

iv. The electrons under the action of the applied electric field drift with a drift speed v_d.

Question 9.
What is current density? State its SI unit.
Answer:
i. Current density at a point in a conductor is the amount of current flowing per unit area of the conductor.
Current density, J = \(\frac {I}{A}\)
where, I = Current
A = Area of cross-section

ii. SI unit: A/m²

Question 10.
A metallic wire of diameter 0.02 m contains 10 free electrons per cubic metre. Find the drift velocity for free electrons, having an electric current of 100 amperes flowing through the wire.
(Given: charge on electron = 1.6 × 10^-19C)
Answer:
Given: e = 1.6 × 10^-19 C, n = 10²⁸ electrons/m³,
D = 0.02 m, r = D/2 = 0.01 m,
I = 100 A
To find: Drift velocity (v_d)
Formula: v_d = \(\frac {I}{nAe}\)
Calculation: From formula,
v_d = \(\frac {I}{nπr^2e}\)
∴ v_d = \(\frac {100}{10^{28}×3.142×10^{-4}×1.6×10^{-19}}\)
= \(\frac {10^{-3}}{3.142×1.6}\)
= 1.989 × 10^-4 m/s

Question 11.
A copper wire of radius 0.6 mm carries a current of 1 A. Assuming the current to be uniformly distributed over a cross sectional area, find the magnitude of current density. Answer:
Given: r = 0.6 mm = 0.6 × 10^-3 m, I = 1 A
To find: Current density (J)
Formula: J = \(\frac {I}{A}\)
Calculation: From formula,
J = \(\frac {1}{3.142×(0.6)^2×10^{-6}}\)
= 0.884 × 10⁶ A/m²

Question 12.
A metal wire of radius 0.4 mm carries a current of 2 A. Find the magnitude of current density if the current is assumed to be uniformly distributed over a cross sectional area.
Answer:
Given: r = 0.4 mm = 0.4 × 10^-3 m, I = 2 A
To find: Current density (J)
Formula: J = \(\frac {I}{A}\)
Calculation: From formula,
J = \(\frac {2}{3.142×(0.4)^2×10^{-6}}\)
= 3.978 × 10⁶ A/m²

Question 13.
State and explain ohm’s law.
Answer:
Statement: The current I through a conductor is directly proportional to the potential difference V applied across its two ends provided the physical state of the conductor is unchanged.
Explanation:
According to ohm’s law,
I ∝ V
∴ V = IR or R = \(\frac {V}{I}\)
where, R is proportionality constant and is called the resistance of the conductor.

Question 14.
Draw a graph showing the I-V curve for a good conductor and ideal conductor.
Answer:

Question 15.
Define one ohm.
Answer:
If potential difference of 1 volt across a conductor produces a current of 1 ampere through it, then the resistance of the conductor is one ohm.

Question 16.
Define conductance. State its SI unit.
Answer:

1. Reciprocal of resistance is called conductance.
   C = \(\frac {I}{R}\)
2. S.I unit statement or Ω^-1

Question 17.
Explain the concept of electrical conduction in a conductor.
Answer:

1. Electrical conduction in a conductor is due to mobile charge carriers (electrons).
2. These conduction electrons are free to move inside the volume of the conductor.
3. During their random motion, electrons collide with the ion cores within the conductor. Assuming that electrons do not collide with each other these random motions average out to zero.
4. On application of an electric field E, the motion of the electron is a combination of the random motion of electrons due to collisions and that due to the electric field \(\vec{E}\).
5. The electrons drift under the action of the field \(\vec{E}\) and move in a direction opposite to the direction of the field \(\vec{E}\). In this way electrons in a conductor conduct electricity.

Question 18.
Derive expression for electric field when an electron of mass m is subjected to an electric field (E).
Answer:
i. Consider an electron of mass m subjected to an electric field E. The force experienced by the electron will be \(\vec{F}\) = e\(\vec{E}\).

ii. The acceleration experienced by the electron will then be
\(\vec{a}\) = \(\frac {e\vec{E}}{m}\) …………. (1)

iii. The drift velocities attained by electrons before and after collisions are not related to each other.

iv. After the collision, the electron will move in random direction, but will still drift in the direction opposite to \(\vec{E}\).

v. Let τ be the average time between two successive collisions.

vi. Thus, at any given instant of time, the average drift speed of the electron will be,
v_d = a τ = \(\frac {eEτ}{m}\) ………………(From 1)
v_d = \(\frac {eEτ}{m}\) = \(\frac {J}{ne}\) ……………(2) [∵ v_d = \(\frac {J}{ne}\)]

vii. Electric field is given by,
E = (\(\frac {m}{e^2nτ}\))J ………… (from 2)
= ρJ = [∵ ρ = \(\frac {m}{ne^2τ}\)]
where, ρ is resistivity of the material.

Question 19.
A Flashlight uses two 1.5 V batteries to provide a steady current of 0.5 A in the filament. Determine the resistance of the glowing filament.
Answer:
Given: For each battery, V₁ = V₂ = 1.5 volt,
I = 0.5 A
To find: Resistance (R)
Formula: V = IR
Calculation: Total voltage, V = V₁ + V₂ = 3 volt
From formula,
R = \(\frac {V}{I}\) = \(\frac {3}{0.5}\) = 6.0 Ω

Question 20.
State an expression for resistance of non-ohmic devices and draw I-V curve for such devices.
Answer:
i. Resistance (R) of a non-ohmic device at a particular value of the potential difference V is given by,
R = \(\lim _{\Delta I \rightarrow 0} \frac{\Delta V}{\Delta I}=\frac{d V}{d I}\)
where, ∆V = potential difference between the
two values of potential V – \(\frac {∆V}{2}\) to V + \(\frac {∆V}{2}\),
and ∆I = corresponding change in the current.

Question 21.
Derive an expression for decrease in potential energy when a charge flows through an external resistance in a circuit.
Answer:
i. Consider a resistor AB connected to a cell in a circuit with current flowing from A to B.

ii. The cell maintains a potential difference V between the two terminals of the resistor, higher potential at A and lower at B.

iii. Let Q be the charge flowing in time ∆t through the resistor from A to B.

iv. The potential difference V between the two points A and B, is equal to the amount of work (W) done to carry a unit positive charge from A to B.
∴ V = \(\frac {W}{Q}\)

v. The cell provides this energy through the charge Q, to the resistor AB where the work is performed.

vi. When the charge Q flows from the higher potential point A to the lower potential point B, there is decrease in its potential energy by an amount
∆U = QV = I∆tV
where I is current due to the charge Q flowing in time ∆t.

Question 22.
Prove that power dissipated across a resistor is responsible for heating up the resistor. Give an example for it.
OR
Derive an expression for the power dissipated across a resistor in terms of its resistance R.
Answer:
i. When a charge Q flows from the higher potential point to the lower potential point, its potential energy decreases by an amount,
∆U = QV = I∆tV
where I is current due to the charge Q flowing in time ∆t.

ii. By the principle of conservation of energy, this energy is converted into some other form of energy.

iii. In the limit as ∆t → 0, \(\frac {dU}{dt}\) = IV
Here, \(\frac {dU}{dt}\) is power, the rate of transfer of energy ans is given by p = \(\frac {dU}{dt}\) = IV
Hence, power is transferred by the cell to the resistor or any other device in place of the resistor, such as a motor, a rechargeable battery etc.

iv. Due to the presence of an electric field, the free electrons move across a resistor and their kinetic energy increases as they move.

v. When these electrons collide with the ion cores, the energy gained by them is shared among the ion cores. Consequently, vibrations of the ions increase, resulting in heating up of the resistor.

vi. Thus, some amount of energy is dissipated in the form of heat in a resistor.

vii. The energy dissipated per unit time is actually the power dissipated which is given by,
P = \(\frac {V^2}{R}\) = I²R
Hence, it is the power dissipation across a resistor which is responsible for heating it up.

viii. For example, the filament of an electric bulb heats upto incandescence, radiating out heat and light.

Question 23.
Calculate the current flowing through a heater rated at 2 kW when connected to a 300 V d. c. supply.
Answer:
Given: P = 2 kW = 2000 W, V = 300 V
To find: Current (I)
Formula: P = IV
Calculation: From formula,
I = \(\frac {P}{V}\) = \(\frac {2000}{300}\) = 6.67 A

Question 24.
An electric heater takes 6 A current from a 230 V supply line, calculate the power of the heater and electric energy consumed by it in 5 hours.
Answer:
Given: I = 6 A, V = 230 V, t = 5 hours
To find: Power (P), Energy consumed
Formulae: i. P = IV
ii. Energy consumed = power × time
Calculation: From formula (i),
P = 6 × 230
= 1380 W = 1.38 kW
From formula (ii),
Energy consumed = 1.38 × 5 = 6.9 kWh
= 6.9 units

Question 25.
When supplied a voltage of 220 V, an electric heater takes 6 A current. Calculate the power of heater and electric energy consumed by its in 2 hours?
Answer:
Given: I = 6 A, V = 220 volt, t = 2 hour
To find: i. Power of heater (P)
ii. Electric energy consumed (E)
Formulae: i. P = IV
ii. Electric energy consumed
= Power × time
Calculation: From formula (i),
P = 6 × 220 = 1320 W = 1.32 kW
From formula (ii),
Electric energy consumed
= 1.32 × 2 = 2.64 kWh = 2.64 units

Question 26.
Explain the colour code system for resistors with an example.
Answer:
i. In colour code system, resistors has 4 bands on it.

ii. In the four band resistor, the colour code of the first two bands indicate two numbers and third band often called decimal multiplier.

iii. The fourth band separated by a space from the three value bands, indicates tolerance of the resistor.

iv. Following table represents the colour code of carbon resistor.

v. Example:
Let the colours of the rings of a resistor starting from one end be brown, red and orange and gold at the other end. To determine resistance of resistor we have,
x = 1, y = 2, z = 3 (From colour code table)
∴ Resistance = xy × 10^z Ω ± tolerance
= 12 × 10³ Ω ± 5%
= 12 kΩ ± 5%
[Note: To remember the colours in order learn the Mnemonics: B.B. ROY of Great Britain had Very Good Wife]

Question 27.
Explain the concept of rheostat.
Answer:

1. A rheostat is an adjustable resistor used in applications that require adjustment of current or resistance in an electric circuit.
2. The rheostat can be used to adjust potential difference between two points in a circuit, change the intensity of lights and control the speed of motors, etc.
3. Its resistive element can be a metal wire or a ribbon, carbon films or a conducting liquid, depending upon the application.
4. In hi-fi equipment, rheostats are used for volume control.

Question 28.
Explain series combination of resistors.
Answer:
i. In series combination, resistors are connected in single electrical path. Hence, the same electric current flows through each resistor in a series combination.

ii. Whereas, in series combination, the supply voltage between two resistors R₁ and R₂ is divided into V₁ and V₂ respectively.

iii. According to Ohm’s law,
R₁ = \(\frac {V_1}{I}\), R₂ = \(\frac {v_2}{I}\)
Total Voltage, V = V₁ + V₂
= I(R₁ + R₂)
∴ V = I R_s
Thus, the equivalent resistance of the series circuit is, R_s = R₁ + R₂

iv. When a number of resistors are connected in series, the equivalent resistance is equal to the sum of individual resistances.
For ‘n’ number of resistors,
R_s = R₁ + R₂ + R₂ + ………….. + R_n = \(\sum_{i=1}^{i=n} R_{i}\)

Question 29.
Explain parallel combination of resistors.
Answer:
i. In parallel combination, the resistors are connected in such a way that the same voltage is applied across each resistor.

ii. A number of resistors are said to be connected in parallel if all of them are connected between the same two electrical points each having individual path.

iii. In parallel combination, the total current I is divided into I, and I2 as shown in the circuit diagram.

iv. Since voltage V across them remains the same,
I = I₁ + I₂
where I₁ is current flowing through R₁ and I₂ is current flowing through R₂.

v. When Ohm’s law is applied to R₁,
V = I₁R₁
i.e. I₁ = \(\frac {V}{R_1}\) ………(1)
When Ohm’s law applied to R2,
V = I₂R₂
i.e., I₂ = \(\frac {V}{R_2}\) …………(2)

vi. Total current is given by,
I = I₁ + I₂
∴ I = \(\frac {V}{R_1}\) + \(\frac {V}{R_2}\) ………[From (1) and (2)]
Since, I = \(\frac {V}{R_p}\)
∴ \(\frac {V}{R_p}\) = \(\frac {V}{R_1}\) + \(\frac {V}{R_2}\)
∴ \(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\)
Where, R_p is the equivalent resistance in parallel combination.

vii. If ‘n’ number of resistors R₁, R₂, R₃, ………….. R_n are connected in parallel, the equivalent resistance of the combination is given by
\(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\) + \(\frac {1}{R_3}\) ……….. + \(\frac {1}{R_n}\) = \(\sum_{i=1}^{\mathrm{i}=\mathrm{n}} \frac{1}{\mathrm{R}}\)
Thus, when a number of resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of individual resistances.

Question 30.
Colour code of resistor is Yellow-Violet- Orange-Gold. Find its value.
Answer:

	Yellow (x)	Violet (y)	Orange (z)	Gold (T%)
Value	4	7	3	± 5

Value of resistance: xy × 10^z Ω ± tolerance
∴ Value of resistance = 47 × 10³ Ω ± 5%
= 47 kΩ ± 5%

Question 31.
From the given value of resistor, find the colour bands of this resistor.
Value of resistor: 330 Ω
Answer:
Value = 330 Ω = 33 × 10¹ Ω = xy × 10^z Ω

Value	3	3	1
Colour	Orange (x)	Orange (y)	Broen(z)

ii. Given: Green – Blue – Red – Gold

Question 32.
Evaluate resistance for the following colour-coded resistors:
i. Yellow – Violet – Black – Silver
ii. Green – Blue – Red – Gold
ill. Brown – Black – Orange – Gold
Answer:
i. Given: Yellow – Violet – Black – Silver
To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± T%)Ω

Colour	Yellow (x)	Violet (y)	Black (z)	Sliver (T%)
Code	4	7	0	±10

Hence x = 4, y = 7, z = 0, T = 10%
Value of resistance = (xy ×10^z ± T%) Ω
= (47 × 10° ± 10%) Ω
Value of resistance = 47 Ω ± 10%

To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± T%) Ω
Calculation:

Colour	Green (x)	Blue (y)	Red (z)	Gold (T%)
Code	5	6	2	±5

Hence x = 5, y = 6, z = 2, T = 5%
Value of resistance = (xy × 10^z ± T%) Q
= 56 × 102 Ω ± 5%
= 5.6 k Ω ± 5%

iii. Given: Brown – Black – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%) Ω
Calculation:

Colour	Brown (x)	Black (y)	Orange (z)	Gold (T%)
Code	1	0	3	±5

Hence x = 1, y = 0, z = 3, T = 5%
Value of resistance = (xy × 10^z ± T%) Ω
= 10 × 10³ Ω ± 5%
= 10 kΩ ± 5%

Question 33.
Calculate
i. total resistance and
ii. total current in the following circuit.
R₁ = 3 Ω, R₂ = 6 Ω, R₃ = 5 Ω, V = 14 V

Answer:
i. R₁ and R₂ are connected in parallel. This combination (R_p) is connected in series with R₃.
∴ Total resistance, R_T = R_p + R₃
R_p = \(\frac {R_1R_2}{R_1+R_2}\) = \(\frac {3×6}{3+6}\) = 2 Ω
∴ R_T = 2+ 5 = 7 Ω

ii. Total current: I = \(\frac {V}{R_T}\) = \(\frac {14}{7}\) = 2 A

Question 34.
State the factors affecting resistance of a conductor.
Answer:
Factors affecting resistance of a conductor:

1. Length of conductor
2. Area of cross-section
3. Nature of material

Question 35.
Derive expression for specific resistance of a material.
Answer:
At a particular temperature, the resistance (R) of a conductor of uniform cross section is
i. directly proportional to its length (l),
i.e., R ∝ l ……….. (1)

ii. inversely proportional to its area of cross section (A),
R ∝ \(\frac {1}{A}\) ……….. (1)
From equations (1) and (2),
R = ρ\(\frac {l}{A}\)
where ρ is a constant of proportionality and it is called specific resistance or resistivity of the material of the conductor at a given temperature.

iii. Thus, resistivity is given by,
ρ = \(\frac {RA}{l}\)

Question 36.
State SI unit of resistivity.
Answer:
SI unit of resistivity is ohm-metre (Ω m).

Question 37.
What is conductivity? State its SI unit.
Answer:
i. Reciprocal of resistivity is called as conductivity of a material.
Formula: σ = \(\frac {1}{ρ}\)
ii. SI unit: (\(\frac {1}{ohm m}\)) or siemens/metre

Question 38.
Explain the similarities between R = \(\frac {V}{I}\) and ρ = \(\frac {E}{J}\)
Answer:

1. Resistivity (ρ) is a property of a material, while the resistance (R) refers to a particular object.
2. The electric field \(\vec{E}\) at a point is specified in a material with the potential difference across the resistance and the current density \(\vec{J}\) in a material is specified instead of current I in the resistor.
3. For an isotropic material, resistivity is given by ρ = \(\frac {E}{J}\)
   For a particular resistor, the resistance R given by, R = \(\frac {V}{I}\)

Question 39.
State expression for current density in terms of conductivity.
Answer:
Current density, \(\vec{J}\) = \(\frac {1}{ρ}\) \(\vec{E}\) = σ \(\vec{E}\)
where, ρ = resistivity of the material
E = electric field intensity
σ = conductivity of the material

Question 40.
Calculate the resistance per metre, at room temperature, of a constantan (alloy) wire of diameter 1.25 mm. The resistivity of constantan at room temperature is 5.0 × 10^-7 Ωm.
Answer:
Given: ρ = 5.0 × 10^-7 Ω m, d = 1.25 × 10^-3 m,
∴ r = 0.625 × 10^-3 m
To find: Resistance per metre (\(\frac {R}{l}\))
Formula: ρ = \(\frac {RA}{l}\)
Calculation:
From formula,
\(\frac{\mathrm{R}}{l}=\frac{\rho}{\mathrm{A}}=\frac{\rho}{\pi \mathrm{r}^{2}}\)
= \(\frac{5 \times 10^{-7}}{3.142 \times\left(0.625 \times 10^{-3}\right)^{2}}\)
= \(\frac{5}{3.142 \times 0.625^{2}} \times 10^{-1}\)
= { antilog [log 5 – log 3.142 -2 log 0.625]} × 10^-1
= {antilog [ 0.6990 – 0.4972 -2(1.7959)]} × 10^-1
= {antilog [0.2018- 1.5918]} × 10^-1
= {antilog [0.6100]} × 10^-1
= 4.074 × 10^-1
∴ \(\frac {R}{l}\) ≈ 0.41 Ω m^-1

Question 41.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6 × 10^-7 m², and its resistance is measured to be 5 Ω. What is the resistivity of the material at the temperature of the experiment?
Answer:
Given: l = 15 m, A = 6.0 × 10^-7 m², R = 5 Ω
To find: Resistivity (ρ)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: From formula,
ρ = \(\frac {5×6×10^{-7}}{15}\)
∴ ρ = 2 × 10^-7 Ω m

Question 42.
A constantan wire of length 50 cm and 0.4 mm diameter is used in making a resistor. If the resistivity of constantan is 5 × 10^-7m, calculate the value of the resistor.
Answer:
Given: l = 50 cm = 0.5 m,
d = 0.4 mm = 0.4 × 10^-3 m,
r = 0.2 × 10^-3 m, p = 5 × 10^-7 Ωm
To Find: Value of resistor (R)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: from formula,

Question 43.
The resistivity of nichrome is 10^-6 Ωm. What length of a uniform wire of this material and of 0.2 mm diameter will have a resistance of 200 ohm?
Answer:
Given: ρ = 10^-6 Ω m, d = 0.2 mm,
∴ r = 0.1 mm = 0.1 × 10^-3 m, R = 200 Ω
To find: Length (l)
Formula: R = \(\frac {ρl}{A}\) = \(\frac {ρl}{πr^2}\)
Calculation: From formula,
l = \(\frac {πr^2}{ρ}\)
∴ l = \(\frac{200 \times 3.142 \times\left(0.1 \times 10^{-3}\right)^{2}}{10^{-6}}\) = 6 284 m

Question 44.
A wire of circular cross-section and 30 ohm resistance is uniformly stretched until its new length is three times its original length. Find its resistance.
Answer:
Given: R₁ = 30 ohm,
l₁ = original length, A₁ = original area,
l₂ = new length, A₂ = new area
l₂= 3l₁
To find: Resistance (R₂)
Formula: R= ρ\(\frac {l}{A}\)
Calculation: From formula,

The volume of wire remains the same in two cases, we have

Question 45.
Define temperature coefficient of resistivity. State its SI unit.
Answer:
i. The temperature coefficient of resistivity is defined as the increase in resistance per unit original resistance at the chosen reference temperature, per degree rise in temperature.
α = \(\frac{\rho-\rho_{0}}{\rho_{0}\left(T-T_{0}\right)}\)
= \(\frac{\mathrm{R}-\mathrm{R}_{0}}{\mathrm{R}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
For small difference in temperatures,
α = \(\frac {1}{R_0}\) \(\frac {dR}{dT}\)

ii. SI unit: °C^-1 (per degree Celsius) or K^-1 (per kelvin).

Question 46.
Give expressions for variation of resistivity and resistance with temperature. Represent graphically the temperature dependence of resistivity of copper.
Answer:
i. Resistivity is given by,
ρ = ρ₀ [1 + α (T – T₀)] where,
T₀ = chosen reference temperature
ρ₀ = resistivity at the chosen temperature
α = temperature coefficient of resistivity
T = final temperature

ii. Resistance is given by,
R = R₀ [1+ α (T – T₀)]
Where,
T₀ = chosen reference temperature
R₀ = resistance at the chosen temperature
α = temperature coefficient of resistance
T = final temperature

iii. For example, for copper, the temperature dependence of resistivity can be plotted as shown:

Question 47.
What is super conductivity?
Answer:

1. The resistivity of a metal decreases as the temperature decreases.
2. In case of some metals and metal alloys, the resistivity suddenly drops to zero at a particular temperature (T_c), this temperature is called critical temperature.
3. Super conductivity is the phenomenon where resistivity of a material becomes zero at particular temperature.
4. For example, mercury loses its resistance completely to zero at 4.2 K.

Question 48.
A piece of platinum wire has resistance of 2.5 Ω at 0 °C. If its temperature coefficient of resistance is 4 × 10^-3/°C. Find the resistance of the wire at 80 °C.
Answer:
Given: R₀ = 2.5 Ω
α = 4 × 10^-3/°C = 0.004/°C
T = 80 °C
To find: Resistance at 80 °C (R_T)
Formula: R_T = R₀(l + α T)
Calculation: From formula,
R_T = 2.5 [1+ (0.004 × 80)]
= 2.5(1 + 0.32)
R_T = 2.5 × 1.32
R_T = 3.3 Ω

Question 49.
The resistance of a tungsten filament at 150 °C is 133 ohm. What will be its resistance at 500 °C? The temperature coefficient of resistance of tungsten is 0.0045 per °C.
Answer:
Given: Let resistance at 150 °C be R₁ and resistance at 500 °C be R₂
Thus,
R₁= 133 Ω, α = 0.0045 °C^-1
To find: Resistance (R₂)
Formula: R_T = R₀ (1 + α∆T)
Calculation:
From formula,
R₁ = R₀ (1 + α × 150)
∴ 133 = R₀(1 + 0.0045 × 150) ……….(i)
R₂ = R₀ (1 + α × 500)
∴ R₂ = R₀(1 + 0.0045 × 500) ………(ii)
Dividing equation (ii) by (i), we get
\(\frac{\mathrm{R}_{2}}{133}=\frac{1+(0.0045 \times 500)}{1+(0.0045 \times 150)}=\frac{3.25}{1.675}\)
∴ R₂ = \(\frac {3.25}{1.675}\) × 133 = 258 Ω

Question 50.
A silver wire has resistance of 2.1 Ω at 27.5 °C. If temperature coefficient of silver is 3.94 × 10^-3/°C, find the silver wire resistance at 100 °C.
Answer:
Given: R₁ = 2.1 Ω, T₁ = 27.5 °C,
α = 3.94 × 10^-3/°C, T₂ = 100 °C
To find: Resistance (R₂)
Formula: R_T = R_o (1 + αT)
Calculation:
From the formula,
R₁ = R₀(1 + α × 27.5) ……….. (i)
R₂ = R₀(l + α × 100) ………….. (ii)
Dividing equation (i) by (ii), we get,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1+\left(3.94 \times 10^{-3} \times 27.5\right)}{1+\left(3.94 \times 10^{-3} \times 100\right)}\)
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1.10835}{1.394}\) = 0.795
∴ R₂ = \(\frac{\mathrm{R}_{1}}{0.795}=\frac{2.1}{0.795}\) = 2.641 Ω

Question 51.
At what temperature would the resistance of a copper conductor be double its resistance at 0 °C?
(a for copper = 3.9 × 10^-3/°C)
Answer:
Given: Let the resistance of the conductor at 0°C be R₀
R₁ = R₀ at T₁ = 0°C
R₂ = 2R₀ at T₂ = T
To find: Final temperature (T)
Formula: α = \(\frac {R_2-R_1}{R_1(T_2-T_1)}\)
Calculation: From formula,
α = \(\frac {2R_0-R_0}{R_1(T_2-T_1)}\) = \(\frac {1}{T}\)
∴ T = \(\frac {1}{α}\) = \(\frac {1}{3.9×10^{-3}}\) ≈ 256 °C

Question 52.
A conductor has resistance of 15 Ω at 10 °C and 18 Ω at 400 °C. Find the temperature coefficient of resistance of the material.
Answer:
Given: R₁ = 15 Ω, T₁ = 10 °C, R₂ = 18 Ω,
T₂ = 400 °C
To find: Temperature coefficient of resistance (α)
Formula: R_T = R₀ (1 + αT)
Calculation:
From formula,
R₁ = R₀ (1 + α × 10) ……..(i)
R₂ = R₀ (1 + α × 400) …….(ii)
Dividing equation (i) by (ii), we get,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1+(\alpha \times 10)}{1+(\alpha \times 400)}\)
∴ \(\frac{15}{18}=\frac{1+10 \alpha}{1+400 \alpha}\)
∴ 18 + 180 α = 15 + 6000 α
∴ 5820 α = 3
∴ α = \(\frac {3}{5820}\) = 5.155 × 10^-4/°C

Question 53.
Write short note on e.m.f. devices.
Answer:

1. When charges flow through a conductor, a potential difference get established between the two ends of the conductor.
2. For a steady flow of charges, this potential difference is required to be maintained across the two ends of the conductor.
3. There is a device that does so by doing work on the charges, thereby maintaining the potential difference. Such a device is called an emf device and it provides the emf E.
4. The charges move in the conductor due to the energy provided by the emf device. This energy is supplied by the e.m.f. device on account of its work done.
5. Power cells, batteries, Solar cells, fuel cells, and even generators, are some examples of emf devices.

Question 54.
Explain working of a circuit when connected to emf device.
Answer:
i. A circuit is formed with connecting an emf device and a resistor R. Flere, the emf device keeps the positive terminal (+) at a higher electric potential than the negative terminal (-)

ii. The emf is represented by an arrow from the negative terminal to the positive terminal.

iii. When the circuit is open, there is no net flow of charge carriers within the device.

iv. When connected in a circuit, the positive charge carriers move towards the positive terminal which acts as cathode inside the emf device.

v. Thus, the positive charge carriers move from the region of lower potential energy, to the region of higher potential energy.

vi. Consider a charge dq flowing through the cross section of the circuit in time dt.

vii. Since, same amount of charge dq flows throughout the circuit, including the emf device. Hence, the device must do work dW on the charge dq, so that the charge enters the negative terminal (low potential terminal) and leaves the positive terminal (higher potential terminal).

viii. Therefore, e.m.f. of the emf device is,
E = \(\frac {dW}{dq}\)
The SI unit of emf is joule/coulomb (J/C).

Question 55.
What is an ideal e.m.f. device?
Answer:

1. In an ideal e.m.f. device, there is no internal resistance to the motion of charge carriers.
2. The emf of the device is then equal to the potential difference across the two terminals of the device.

Question 56.
What is a real e.m.f. device?
Answer:

1. In a real emf device, there is an internal resistance to the motion of charge carriers.
2. If such a device is not connected in a circuit, there is no current through it.

Question 57.
Derive an expression for current flowing through a circuit when an external resistance is connected to a real e.m.f. device.
Answer:

i. If a current (I) flows through an emf device, there is an internal resistance (r) and the emf (E) differs from the potential difference across its two terminals (V).
V = E – Ir ……… (1)

ii. The negative sign is due to the fact that the current I flows through the emf device from the negative terminal to the positive terminal.

iii. By the application of Ohm’s law,
V = IR …….(2)
From equations (1) and (2),
IR = E – Ir
∴ \(\frac {E}{R+r}\)

Question 58.
Explain the conditions for maximum current.
Answer:

1. Current in a circuit is given by, I = \(\frac {E}{R+r}\)
2. Maximum current can be drawn from the emf device, only when R = 0, i.e.
   I_max = \(\frac {E}{R}\)
3. I_max is the maximum allowed current from an emf device (or a cell) which decides the maximum current rating of a cell or a battery.

Question 59.
A network of resistors is connected to a 14 V battery with internal resistance 1 Q as shown in the circuit diagram.
i. Calculate the equivalent resistance,
ii. Current in each resistor,
iii. Voltage drops V_AB, V_BC and V_DC.

Answer:

For equivalent resistance (R_eq):
R_AB is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\)
∴ R_AB = 2 Ω
R_BC = R₃ = 1 Ω
Also, R_CD is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{CD}}}=\frac{1}{\mathrm{R}_{4}}+\frac{1}{\mathrm{R}_{5}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}\)
∴ R_CD = 3 Ω
∴ R_eq = R_AB + R_BC + R_CD
= 2 + 1 + 3 = 6Ω

ii. Current through each resistor:
Total current, I = \(\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{eq}}+\mathrm{r}}\) = \(\frac {14}{6+1}\) = 2 A
Across AB, as, R₁ = R₂
V₁ = V₂
∴I₁ × 4 = I₂ × 4
∴ I₁ = I₂
But, I₁ + I₂ = I
∴ 2I₁ = I
∴ I₁ = I₂ =1 A ….(∵I = 2 A)
Similarly, as R₄ = R₅
I₃ = I₄ = 1 A
Current through resistor BC is same as I.
∴ I_BC = 2 A

iii. Voltage drops across AB, BC and CD:
V_AB = IR_AB = 2 × 2 = 4 V
V_BC = IR_BC = 2 × 1 = 2 V
V_CD = IR_CD = 2 × 3 = 6 V

Question 60.
i. Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
ii. If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.
Answer:
Given: R₁ = 2Ω, R₂ = 4 Ω, R₃ = 5 Ω,
V = 20 V
To Find: i. Total resistance (R)
ii. Current through each resistor (I₁, I₂, I₃ respectively)
iii. Total current (I)
Formulae:
i. \(\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}\)
ii. V = IR
iii. Total current, I = I₁ +I₂ + I₃
Calculation
From formula (i):
\(\frac{1}{R}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}\)
∴ R = \(\frac {20}{19}\) Ω
From formula (ii):

From formula (iii):
I = 10 + 5 + 4
∴ I = 19 A

Question 61.
i. Three resistors 1 Ω, 2 Ω and 3 Ω are combined in series. What is the total resistance of the combination?
ii. If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer:
Given: R₁ = 1Ω, R₂ = 2 Ω, R₃ = 3 Ω,
V = 12 V
To Find: i. Total resistance (R)
ii. P.D Across R₁, R₂, R₃ (V₁, V₂, V₃ respectively)
Formulae:
i. R_s = R₁ + R₂ + R₃
ii. V = IR
Calculation
From formula (i):
Rs = l + 2 + 3 = 6 Ω
From formula (ii),
1 = \(\frac {V}{R}\) = \(\frac {12}{6}\) = 2A
∴ V₁ = IR₁ = 2 × 1 = 2 V
∴ V₂ = IR₂ = 2 × 2 = 4 V
∴ V₃ = IR₃ = 2 × 3 = 6 V

Question 62.
A voltmeter is connected across a battery of emf 12 V and internal resistance of 10 Ω. If the voltmeter resistance is 230 Ω, what reading will be shown by the voltmeter? Answer:
Given: E = 12 volt, r = 10 Ω, R = 230 Ω
To find: Reading shown by voltmeter (V)
Formula: i. I = \(\frac {E}{R+r}\)
ii. V = E – Ir
Calculation
From formula (i),
I = \(\frac{12}{230+10}=\frac{12}{240}=\frac{1}{20} \mathrm{~A}\)
From formula (ii),
V= 12 – \(\frac {1}{20}\) × 10 = 12 – 0.5
= 11.5 volt

Question 63.
A battery of e.m.f. 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer:
Given: E = 10 V, r = 3 Ω, I = 0.5 A
To find: i. Resistance of resistor (R)
ii. Terminal voltage of battery (V)
Formula: I = \(\frac {E}{R+r}\)
Calculation: From formula, R = \(\frac {E}{I}\) – r
∴ R = \(\frac {10}{0.5}\)– 3 = 17 Ω
∴ V = IR = 0.5 × 17 = 8.5 volt

Question 64.
How many cells each of 1.5 V/500 mA rating would be required in series-parallel combination to provide 1500 mA at 3 V?
Answer:
21 = ………… = 1.5 V (given)
I₁ = I₂ = …………… = 500 mA (given)
1500 mA at 3 V is required.
To determine required number of cells:
For series V = V₁ + V₂ + ………….., and current remains same.
For parallel I = I₁ + I₂ + ………, and voltage remains same.
To achieve battery output of 3V, the cells should be connected in series.
If n are the number of cells connected in series, then
V = V₁ + V₂ + …………. + V_n
∴ V = nV₁
∴ 3 = n × 1.5
∴ n = 2 cells in series
The series combination of two cells in series will give a current 500 mA.
To achieve output of 1500 mA, the number of batteries (n) connected in parallel, each one having output 3V is,
I = I₁ + I₂ + ………. + I_n
∴ I = nI₁
∴ 1500 = n × 500
∴ n = 3 batteries each of two cells
∴ No of cells required are 2 × 3 = 6 .
∴ Number of cells = 6
The six cells must be connected as shown

Question 65.
Explain the concept of series combination of cells.
Answer:
i. In a series combination, cells are connected in single electrical path, such that the positive terminal of one cell is connected to the negative terminal of the next cell, and so on.

ii. The terminal voltage of batteiy/cell is equal to the sum of voltages of individual cells in series. Example: Given figure shows two 1.5 V cells connected in series. This combination provides total voltage,
V = 1.5 V + 1.5 V = 3 V.

iii. The equivalent emf of n number of cells in series combination is the algebraic sum of their individual emf.
\(\sum_{i} \mathrm{E}_{\mathrm{i}}\) = E₁ + E₂ + E₂+ …….. + E_n

iv. The equivalent internal resistance of n cells in a series combination is the sum of their individual internal resistance.
\(\sum_{i} \mathrm{r}_{\mathrm{i}}\) = r₁ + r₂ + r₃ + ……… + r_n

Question 66.
State advantages of cells in series.
Answer:

1. The cells connected in series produce a larger resultant voltage.
2. Cells which are damaged can be easily identified, hence can be easily replaced.

Question 67.
Explain combination of cells in parallel. Ans:
Answer:
i. Consider two cells which are connected in parallel. Here, positive terminals of all the cells are connected together and the negative terminals of all the cells are connected together.

ii. In parallel connection, the current is divided among the branches i.e. I₁ and I₂ as shown in figure.

iii. Consider points A and B having potentials V_A and V_B, respectively.

iv. For the first cell the potential difference across its terminals is, V = V_A – V_B = E₁ – I₁ r₁
∴ I₁ = \(\frac {E_1V}{r_1}\) ………. (1)

v. Point A and B are connected exactly similarly to the second cell.
Hence, considering the second cell,
V = V_A – V_B = E₂ – I₂r₂
∴ I₂ = \(\frac {E_2V}{r_2}\) ………. (2)

vi. Since, I = I₁ + I₂ ………….. (3)
Combining equations (1), (2) and (3),

viii. If we replace the cells by a single cell connected between points A and B with the emf E_eq and the internal resistance r_eq then,
V = E_eq– Ir_eq
From equations (4) and (5),

ix. For n number of cells connected in parallel with emf E₁, E₂, E₃, ………….., E_n and internal resistance r₁, r₂, r₃, …………, r_n
\(\frac{1}{\mathrm{r}_{\mathrm{rq}}}=\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\ldots \ldots \ldots+\frac{1}{\mathrm{r}_{\mathrm{n}}}\)
and \(\frac{\mathrm{E}_{\mathrm{eq}}}{\mathrm{r}_{\mathrm{rq}}}=\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\ldots \ldots \ldots+\frac{\mathrm{E}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}\)

Question 68.
State advantages and disadvantages of cells in parallel.
Answer:
Advantages:
For cells connected in parallel in a circuit, the circuit will not break open even if a cell gets damaged or open.

Disadvantages:
The voltage developed by the cells in parallel connection cannot be increased by increasing number of cells present in circuit.

Question 69.
State the basic categories of electrical cells.
Answer:
Electrical cells can be divided into several categories like primary cell, secondary cell, fuel cell, etc.

Question 70.
Write short note on primary cell.
Answer:

1. A primary cell cannot be charged again. It can be used only once.
2. Dry cells, alkaline cells are different examples of primary cells.
3. Primary cells are low cost and can be used easily. But these are not suitable for heavy loads.

Question 71.
Write short note on secondary cell.
Answer:

1. The secondary cells are rechargeable and can be reused.
2. The chemical reaction in a secondary cell is reversible.
3. Lead acid cell and fuel cell are some examples of secondary cells.
4. Lead acid battery is used widely in vehicles and other applications which require high load currents.
5. Solar cells are secondary cells that convert solar energy into electrical energy.

Question 72.
Write short note on fuel cells vehicles.
Answer:

1. Fuel cells vehicles (FCVs) are electric vehicles that use fuel cells instead of lead acid batteries to power the vehicles.
2. Hydrogen is used as a fuel in fuel cells. The by product after its burning is water.
3. This is important in terms of reducing emission of greenhouse gases produced by traditional gasoline fuelled vehicles.
4. The hydrogen fuel cell vehicles are thus more environment friendly.

Question 73.
What can be concluded from the following observations on a resistor made up of certain material? Calculate the power drawn in each case.

Case	Current (A)	Voltage (V)
A	0.2	1.6
B	0.4	3.2
C	0.6	4.8
d	0.8	6.4

Answer:
i. As the ratio of voltage and current different readings are same, hence ohm’s is valid i.e., V = IR.

ii. Electric power is given by, P = IV
∴ (a) P₁ = 0.2 × 1.6 = 0.32 watt
(b) P₂ = 0.4 × 3.2 = 1.28 watt
(c) P₃ = 0.6 × 4.8 = 2.88 watt
(d) P₄ = 0.8 × 6.4 = 5.12 watt

Question 74.
Answer the following questions from the circuit given below. [S₁, S₂, S₃, S₄, S₅ ⇒ Switches]. Calculate the current (I) flowing in the following cases:
i. S₁, S₄ → open; S₂, S₃, S₅ → closed.
ii. S₂, S₅ → open; S₁, S₃, S₄ → closed.
iii. S₃ → open; S₁, S₂, S₄, S₅ → closed.

Answer:
i. Here, the circuit can be represented as,

ii. Here, the circuit can be represented as,

iii. Here, the circuit can be represented as,

∴ As switch S3 is open, no current will flow in the circuit.

Question 75.
An electric circuit with a carton resistor and an electric bulb (60 watt, 300 Ω) are connected in series with a 230 V source.

i. Calculate the current flowing through the circuit.
ii. If the electric bulb of 60 watt is replaced by an electric bulb (80 watt, 300 Ω), will it glow? Justify your answer.
Answer:
Resistance of carbon resistor (R₁)
= 16 × 10 Ω = 160 Ω ….(using colour code)
Resistance of bulb (R₂) = 300 Ω
∴ Current through the circuit = \(\frac{V}{R_{1}+R_{2}}\)
∴ I = \(\frac{230}{(160+300)}=\frac{230}{460}\) = 0.5 A

ii. Power drawn through electric bulb
= I²R₂ = (0.5)² × 300 = 75 watt
Hence, if the bulb is replaced by 80 watt bulb, it will not glow.

Question 76.
From the graph given below, which of the two temperatures is higher for a metallic wire? Justify your answer.

Answer:
As R = \(\frac {V}{I}\)

For constant V,
I₂ > I₁
∴ R₁ > R₂
Now, for metallic wire,
R ∝ T
∴ T₁ > T₂
T₁ is greater than T₂.

Question 77.
If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d. of the combination and hence show that this is nearly n times that of a single cell.
Answer:
i. Let n identical cells, each of emf E and internal resistance r, be connected in series. Let the current supplied by this combination to an external resistance R be I.

ii. The equivalent emf of the combination,
E_eq = E + E + …….. (n times) = nE

iii. The equivalent internal resistance of the combination,
r_eq= r + r + … (n times)
= nr

iv. The terminal p.d. of the combination is
V = E_eq – Ir_eq = nE – Inr = n (E – Ir)
∴ V = n × terminal p.d. of a single cell
Thus, the terminal p.d. of the series combination is n times that of a single cell.

Question 78.
If n identical cells, each of emf E and internal resistance r, are connected in parallel, derive an expression for the current supplied by this combination to external resistance R. Prove that the combination supplies current almost n times the current supplied by a single cell, when the external resistance R is much smaller than the internal resistance of the parallel combination of the cells.
Answer:
i. Consider n identical cells, each of emf E and internal resistance r, connected in parallel.

ii. Let the current supplied by the combination to the external resistance R be I.
In this case, the equivalent emf of the combination is E.

iii. The equivalent internal resistance r’ of the combination is,
\(\frac{1}{\mathrm{r}^{\prime}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}\) + …………. (n terms)
∴ \(\frac{1}{\mathrm{r}^{\prime}}=\frac{\mathrm{n}}{\mathrm{r}} \Rightarrow \mathrm{r}^{\prime} \frac{\mathrm{n}}{\mathrm{r}}\)

iv. But V = IR is the terminal p.d. across each cell.

v. Hence, the current supplied by each cell,
I = \(\frac {E-V}{r}\)

vi. This gives the current supplied by the combination to the external resistance as
I = \(\frac {E-V}{r}\) + \(\frac {E-V}{r}\) + …….. (n terms) = n(\(\frac {E-V}{r}\))
Thus, current I = n × current supplied by a single cell
This proves that, the current supplied by the combination is n times the current supplied by a single cell.

Multiple Choice Questions

Question 1.
The drift velocity of the free electrons in a conductor is independent of
(A) length of the conductor.
(B) cross-sectional area of conductor.
(C) current.
(D) electric charge.
Answer:
(A) length of the conductor.

Question 2.
The direction of drift velocity in a conductor is
(A) opposite to that of applied electric field.
(B) opposite to the flow of positive charge.
(C) in the direction of the flow of electrons,
(D) all of these.
Answer:
(D) all of these.

Question 3.
The drift velocity of free electrons in a conductor is vd, when the current is flowing in it. If both the radius and current are doubled, the drift velocity will be
(A) \(\frac {v_d}{8}\)
(B) \(\frac {v_d}{4}\)
(C) \(\frac {v_d}{2}\)
(D) v_d
Answer:
(C) \(\frac {v_d}{2}\)

Question 4.
The drift velocity vd of electrons varies with electric field strength E as
(A) v_d ∝ E
(B) v_d ∝ \(\frac {1}{E}\)
(C) v_d ∝ E^1/2
(D) v_d × E^{\(\frac {1}{1/2}\)}
Answer:
(A) v_d ∝ E

Question 5.
When a current I is set up in a wire of radius r, the drift speed is v_d. If the same current is set up through a wire of radius 2r, then the drift speed will be
(A) v_d/4
(B) v_d/2
(C) 2v_d
(D) 4v_d
Answer:
(A) v_d/4

Question 6.
When potential difference is applied across an electrolyte, then Ohm’s law is obeyed at
(A) zero potential
(B) very low potential
(C) negative potential
(D) high potential.
Answer:
(D) high potential.

Question 7.
A current of 1.6 A is passed through an electric lamp for half a minute. If the charge on the electron is 1.6 × 10^-19 C, the number of electrons passing through it is
(A) 1 × 10¹⁹
(B) 1.5 × 10²⁰
(C) 3 × 10¹⁹
(D) 3 × 10²⁰
Answer:
(D) 3 × 10²⁰

Question 8.
The SI unit of the emf of a cell is
(A) V/m
(B) V/C
(C) J/C
(D) C/J
Answer:
(C) J/C

Question 9.
The unit of specific resistance is
(A) Ω m^-1
(B) Ω^-1 m^-1
(C) Ω m
(D) Ω m^-2
Answer:
(C) Ω m

Question 10.
If the length of a conductor is halved, then its conductivity will be
(A) doubled
(B) halved
(C) quadrupled
(D) unchanged
Answer:
(D) unchanged

Question 11.
The resistance of a metal conductor increases with temperature due to
(A) change in current carriers.
(B) change in the dimensions of the conductor.
(C) increase in the number of collisions among the current carriers.
(D) increase in the rate of collisions between the current carriers and the vibrating atoms of the conductor.
Answer:
(D) increase in the rate of collisions between the current carriers and the vibrating atoms of the conductor.

Question 12.
The resistivity of Nichrome is 10^-6 Ω-m. The wire of this material has radius of 0.1 mm with resistance 100 Ω, then the length will be
(A) 3.142 m
(B) 0.3142 m
(C) 3.142 cm
(D) 31.42 m
Answer:
(A) 3.142 m

Question 13.
Given a current carrying wire of non-uniform cross-section. Which of the following is constant throughout the length of the wire?
(A) Current, electric field and drift speed
(B) Drift speed only
(C) Current and drift speed
(D) Current only
Answer:
(D) Current only

Question 14.
A cell of emf E and internal resistance r is connected across an external resistance R (R >> r). The p.d. across R is A 1
(A) \(\frac {E}{R+r}\)
(B) E(I – \(\frac {r}{R}\))
(C) E(I + \(\frac {r}{R}\))
(D) E (R + r)
Answer:
(B) E(I – \(\frac {r}{R}\))

Question 15.
The e.m.f. of a cell of negligible internal resistance is 2 V. It is connected to the series combination of 2 Ω, 3 Ω and 5 Ω resistances. The potential difference across 3 Ω resistance will be
(A) 0.6 V
(B) 10 V
(C) 3 V
(D) 6 V
Answer:
(A) 0.6 V

Question 16.
A P.D. of 20 V is applied across a conductance of 8 mho. The current in the conductor is
(A) 2.5 A
(B) 28 A
(C) 160 A
(D) 45 A
Answer:
(C) 160 A

Question 17.
If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be
(A) 0.1%
(B) 0.2%
(C) 1 %
(D) 2 %
Answer:
(C) 1 %

Question 18.
If a certain piece of copper is to be shaped into a conductor of minimum resistance, its length (L) and cross-sectional area A shall be respectively
(A) L/3 and 4 A
(B) L/2 and 2 A
(C) 2L and A2
(D) L and A
Answer:
(A) L/3 and 4 A

Question 19.
A given resistor has the following colour scheme of the various strips on it: Brown, black, green and silver. Its value in ohm is
(A) 1.0 × 104 ± 10%
(B) 1.0 × 105 ± 10%
(C) 1.0 × 106 + 10%
(D) 1.0 × 107 ± 10%
Answer:
(C) 1.0 × 106 + 10%

Question 20.
A given carbon resistor has the following colour code of the various strips: Orange, red, yellow and gold. The value of resistance in ohm is
(A) 32 × 104 ± 5%
(B) 32 × 104 ± 10%
(C) 23 × 105 ± 5%
(D) 23 × 105 ± 10%
Answer:
(A) 32 × 104 ± 5%

Question 21.
A typical thermistor can easily measure a change in temperature of the order of
(A) 10^-3 °C
(B) 10^-2 °C
(C) 10² °C
(D) 10³ °C
Answer:
(A) 10^-3 °C

Question 22.
Thermistors are usually prepared from
(A) non-metals
(B) metals
(C) oxides of non-metals
(D) oxides of metals
Answer:
(D) oxides of metals

Question 23.
On increasing the temperature of a conductor, its resistance increases because
(A) relaxation time decreases.
(B) mass of the electron increases.
(C) electron density decreases.
(D) all of the above.
Answer:
(A) relaxation time decreases.

Question 24.
Which of the following is used for the formation of thermistor?
(A) copper oxide
(B) nickel oxide
(C) iron oxide
(D) all of the above
Answer:
(D) all of the above

Question 25.
Emf of a cell is 2.2 volt. When resistance R = 5 Ω is connected in series, potential drop across the cell becomes 1.8 volt. Value of internal resistance of the cell is
(A) 10/9 Ω
(B) 7/12 Ω
(C) 9/10 Ω
(D) 12/7 Ω
Answer:
(A) 10/9 Ω

Question 26.
A strip of copper, another of germanium are cooled from room temperature to 80 K. The resistance of
(A) copper strip decreases germanium decreases. and that of
(B) copper strip decreases germanium increases. and that of
(C) Both the strip increases.
(D) copper strip increases germanium decreases. and that of
Answer:
(B) copper strip decreases germanium increases. and that of

Question 27.
The terminal voltage of a cell of emf E on short circuiting will be
(A) E
(B) \(\frac {E}{2}\)
(C) 2E
(D) zero
Answer:
(D) zero

Question 28.
If a battery of emf 2 V with internal resistance one ohm is connected to an external circuit of resistance R across it, then the terminal p.d. becomes 1.5 V. The value of R is
(A) 1 Ω
(B) 1.5 Ω
(C) 2 Ω
(D) 3 Ω
Answer:
(D) 3 Ω

Question 29.
A hall is used 5 hours a day for 25 days in a month. It has 6 lamps of 100 W each and 4 fans of 150 W. The total energy consumed for the month is
(A) 1500 kWh
(B) 150 kWh
(C) 15 kWh
(D) 1.5 kWh
Answer:
(B) 150 kWh

Question 30.
The internal resistance of a cell of emf 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.5 V
(C) 1.95 V
(D) 2 V
Answer:
(C) 1.95 V

Question 31.
The emf of a cell is 12 V. When it sends a current of 1 A through an external resistance, the p.d. across the terminals reduces to 10 V. The internal resistance of the cell is
(A) 0.1 Ω
(B) 0.5 Ω
(C) 1 Ω
(D) 2 Ω
Answer:
(D) 2 Ω

Question 32.
Three resistors, 8 Ω, 4 Ω and 10 Ω connected in parallel as shown in figure, the equivalent resistance is

(A) \(\frac {19}{40}\) Ω
(B) \(\frac {40}{19}\) Ω
(C) \(\frac {80}{19}\) Ω
(D) \(\frac {34}{23}\) Ω
Answer:
(B) \(\frac {40}{19}\) Ω

Question 33.
A potential difference of 20 V is applied across the ends of a coil. The amount of heat generated in it is 800 cal/s. The value of resistance of the coil will be
(A) 12 Ω
(B) 1.2 Ω
(C) 0.12 Ω
(D) 0.012 Ω
Answer:
(C) 0.12 Ω

Question 34.
In a series combination of cells, the effective internal resistance will
(A) remain the same.
(B) decrease.
(C) increase.
(D) be half that of the 1st cell.
Answer:
(C) increase.

Question 35.
The terminal voltage across a cell is more than its e.m.f., if another cell of
(A) higher e.m.f. is connected parallel to it.
(B) less e.m.f. is connected parallel to it.
(C) less e.m.f. is connected in series with it.
(D) higher e.m.f. is connected in series with it.
Answer:
(A) higher e.m.f. is connected parallel to it.

Question 36.
A 100 W, 200 V bulb is connected to a 160 volt supply. The actual’ power consumption would be
(A) 64 W
(B) 125 W
(C) 100 W
(D) 80 W
Answer:
(A) 64 W

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 13 Electromagnetic Waves and Communication System

Question 1.
Describe Gauss’ law of electrostatics in brief.
Answer:
i. Gauss’ law of electrostatics states that electric flux through any closed surface S is equal to the total electric charge Q_in enclosed by the surface divided by so.
\(\int \vec{E} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{Q}_{\text {in }}}{\varepsilon_{0}}\)
where, \(\vec{E}\) is the electric field and e0 is the permittivity of vacuum. The integral is over a closed surface S.

ii. Gauss’ law describes the relation between an electric charge and electric field it produces.

Question 2.
Describe Gauss’ law of magnetism in brief.
Answer:
i. Gauss’ law for magnetism states that magnetic monopoles which are thought to be magnetic charges equivalent to the electric charges, do not exist. Magnetic poles always occur in pairs.

ii. This means, magnetic flux through a closed surface is always zero, i.e., the magnetic field lines are continuous closed curves, having neither beginning nor end.
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dS}}\) = 0
where, B is the magnetic field. The integral is over a closed surface S.

Question 3.
Describe Faraday’s law along with Lenz’s law.
Answer:
i. Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

ii. Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

iii. According to Faraday’s law with Lenz’s law,
\(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} l}=-\frac{\mathrm{d} \phi_{\mathrm{m}}}{\mathrm{dt}}\)
where, ø_m is the magnetic flux and the integral is over a closed loop.

Question 4.
What does Ampere’s law describe?
Answer:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Question 5.
Describe Ampere-Maxwell law in brief.
Answer:
According to Ampere-Maxwell law, magnetic field is generated by moving charges and also by varying electric fields.
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\varepsilon_{0} \mu_{0} \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\)
where, p0 and e0 are the permeability and permittivity of vacuum respectively and the integral is over a closed loop, I is the current flowing through the loop, E is the electric flux linked with the circuit.

Question 6.
What are Maxwell’s equations for charges and currents in vacuum?
Answer:
\(\int \vec{E} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{Q}_{\text {in }}}{\varepsilon_{0}}\)
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dS}}\) = 0
\(\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} l}=-\frac{\mathrm{d} \phi_{\mathrm{m}}}{\mathrm{dt}}\)
\(\int \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} l}=\mu_{0} \mathrm{I}+\varepsilon_{0} \mu_{0} \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\)

Question 7.
Explain the origin of displacement current?
Answer:

1. Maxwell pointed a major flaw in the Ampere’s law for time dependant fields.
2. He noticed that the magnetic field can be generated not only by electric current but also by changing electric field.
3. Therefore, he added one more term to the equation describing Ampere’s law. This term is called the displacement current.

Question 8.
In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write respectively against (i), (ii), and (iii).

Name of the Physicist	Work
i. H. Hertz	a. Existence of EM waves
ii. J. Maxwell	b. Properties of EM waves
iii. G. Marconi	c. Wireless communication
	d. Displacement current

Answer:
(i – a, b), (ii – d), (iii – c)

Question 9.
Varying electric and magnetic fields regenerate each other. Explain.
Answer:

1. According to Maxwell’s theory, accelerated charges radiate EM waves.
2. Consider a charge oscillating with some frequency. This produces an oscillating electric field in space, which produces an oscillating magnetic field which in turn is a source of oscillating electric field.
3. Thus, varying electric and magnetic fields regenerate each other.

Question 10.
Draw a neat diagram representing electromagnetic wave propagating along Z-axis.
Answer:

Question 11.
How can energy be transported in the form of EM waves?
Answer:

1. Maxwell proposed that an oscillating electric charge radiates energy in the form of EM wave.
2. EM waves are periodic changes in electric and magnetic fields, which propagate through space.
3. Thus, energy can be transported in the form of EM waves.

Question 12.
State the main characteristics of EM waves.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

iii. The \(\vec{E}\) and \(\vec{B}\) fields vary sinusoidally and are in phase.

iv. EM waves are produced by accelerated electric charges.

v. EM waves can travel through free space as well as through solids, liquids and gases.

vi. In free space, EM waves travel with velocity c, equal to that of light in free space.
c = \(\frac {1}{\sqrt{µ_0ε_0}}\) = 3 × 10⁸ m/s,
where µ₀ is permeability and ε₀ is permittivity of free space.

vii. In a given material medium, the velocity (v_m) of EM waves is given by v_m = \(\frac {1}{\sqrt{µε}}\)
where µ is the permeability and ε is the permittivity of the given medium.

viii. The EM waves obey the principle of superposition.

ix. The ratio of the amplitudes of electric and magnetic fields is constant at any point and is equal to the velocity of the EM wave.
\( \left|\overrightarrow{\mathrm{E}}_{0}\right|=\mathrm{c}\left|\overrightarrow{\mathrm{B}}_{0}\right| \text { or } \frac{\left|\overrightarrow{\mathrm{E}}_{0}\right|}{\left|\overrightarrow{\mathrm{B}_{0}}\right|}=\mathrm{c}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
where, |\(\vec{E_0}\)| and |\(\vec{B_0}\)| are the amplitudes of \(\vec{E}\) and \(\vec{B}\) respectively.

x. As the electric field vector (\(\vec{E_0}\)) is more prominent than the magnetic field vector (\(\vec{B_0}\)), it is responsible for optical effects due to EM waves. For this reason, electric vector is called light vector.

xi. The intensity of a wave is proportional to the square of its amplitude and is given by the equations
\(\mathrm{I}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}, \mathrm{I}_{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B}_{0}^{2}}{\mu_{0}}\)

xii. The energy of EM waves is distributed equally between the electric and magnetic fields. I_E = I_B.

Question 13.
Give reason: Electric vector is called light vector.
Answer:
As the electric field vector (\(\vec{E_0}\)) is more prominent than the magnetic field vector (\(\vec{B_0}\)), it is responsible for optical effects due to EM waves. For this reason, electric vector is called light vector.

Question 14.
Explain the equations describing an EM wave.
Answer:
i. In an EM wave, the magnetic field and electric field both vary sinusoidally with x.

ii. For a wave travelling along X-axis having \(\vec{E}\) along Y-axis and \(\vec{B}\) along the Z-axis,
E_y = E₀ sin (kx – ωt)
B_z = B₀ sin (kx – ωt)
where, E₀ is the amplitude of the electric field (E_y) and B₀ is the amplitude of the magnetic field (B_z).

iii. The propagation constant is given by k = \(\frac {2π}{λ}\) and λ is the wavelength of the wave. The angular frequency of oscillations is given by ω = 2πv, v being the frequency of the wave.
Hence, E_y = E₀ sin (\(\frac {2πx}{λ}\) – 2πvt)
B_z = B₀ sin (\(\frac {2πx}{λ}\) – 2πvt)

iv. Both the electric and magnetic fields attain their maximum or minimum values at the same time and at the same point in space, i.e., \(\vec{E}\) and \(\vec{B}\) oscillate in phase with the same frequency.

Question 15.
A radio wave of frequency of 1.0 × 10⁷ Hz propagates with speed 3 × 10⁸ m/s. Calculate its wavelength.
Answer:
Given: v= 1.0 × 10⁷ Hz, c = 3 × 10⁸ m/s
To find: Wavelength (λ)
Formula: vλ
Calculation: From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{1.0×10^7}\) = 30 m

Question 16.
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
Given: V₁ = 7.5 MHz = 7.5 × 10⁶ Hz,
V₁ = 12 MHz = 12 × 10⁶ Hz.
To find: Wavelength band
Formula: λ = \(\frac {c}{v}\)
Calculation: From formula,
V₁ = \(\frac {3×10^8}{7.5×10^6}\) = 40 m
V₁ = \(\frac {3×10^8}{12×10^6}\) = 25 m
∴ Wavelength band = 40 m to 25 m

Question 17.
Calculate the ratio of the intensities of the two waves, if amplitude of first beam of light is 1.5 times the amplitude of second beam of light.
Answer:
a₁ = 1.5 a₂
To find: \(\frac {I_1}{I-2}\)
Formula: I ∝ a²
Calculation: From formula,
\(\frac{I_{1}}{I_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}=\left(\frac{1.5 a_{2}}{a_{2}}\right)^{2}\) = (1.5)² = 2.25

Question 18.
A beam of red light has an amplitude 2.5 times the amplitude of second beam of the same colour. Calculate the ratio of the intensities of the two waves.
Answer:
a₁ = 2.5 a₂
To find: \(\frac {I_1}{I-2}\)
Formula: I ∝ a²
Calculation: From formula,
\(\frac{I_{1}}{I_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}=\left(\frac{2.5 a_{2}}{a_{2}}\right)^{2}\) = (2.5)² = 6.25

Question 19.
Calculate the velocity of EM waves in vacuum.
Answer:
Given: ε₀ = 8.85 × 10^-12 C²/Nm²
µ₀ = 4π × 10^-7 Tm/A
To find: Velocity of EM waves (c)
Formula: c = \(\frac {1}{\sqrt{µ_0ε_0}}\)
Calculation: From formula,

………… (Taking square roots using log table)
= 0.2998 × 10⁹ ≈ 3 × 10⁸ m/s

Question 20.
In free space, an EM wave of frequency 28 MHz travels along the X-direction. The amplitude of the electric field is E = 9.6 V/m and its direction is along the Y-axis. What is amplitude and direction of magnetic field B?
Answer:
Given: v = 28 MHz, E = 9.6 V/m,
c = 3 × 10⁸ m/s
To find:
i. Amplitude of magnetic field (B)
ii. Direction of B
Formula:
|B| = \(\frac {|E|}{c}\)
Calculation: From formula,
|B| = \(\frac {9.6}{3×10^8}\) = 3.2 × 10^-8 T
Since that E is along Y-direction and the wave propagates along X-axis. The magnetic induction, B should be in a direction perpendicular to both X and Y axes, i.e., along the Z-direction.

Question 21.
An EM wave of frequency 50 MHz travels in vacuum along the positive X-axis and \(\vec{E}\) at a particular point, x and at a particular instant of time t is 9.6 j V/m. Find the magnitude and direction of \(\vec{B}\) at this point x and at instant of time t.
Answer:
Given: \(\vec{E}\) = 9.6 j V/m
i. e., Electric field E is directed along +Y axis Magnitude of \(\vec{B}\).
|B| = \(\frac {|E|}{c}\) = \(\frac {9.6}{3×10^8}\) = 3.2 × 10^-8 T
As the wave propagates along +X axis and E is along +Y axis, direction of B will be along +Z-axis i.e. B = 3.2 × 10^-8 \(\hat{k}\)T.

Question 22.
A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
Since the electromagnetic waves are transverse in nature, the electric and magnetic field vectors are mutually perpendicular to each other as well as perpendicular to the direction of propagation of wave.
As the wave is travelling along Z-direction,

\(\vec{E}\) and \(\vec{B}\) are in XY plane.
For v = 30 MHz = 30 × 10⁶ Hz
Wavelength, λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{30×10^6}\) = 10 m

Question 23.
For an EM wave propagating along X direction, the magnetic field oscillates along the Z-direction at a frequency of 3 × 10¹⁰ Hz and has amplitude of 10^-9 T.
i. What is the wavelength of the wave?
ii. Write the expression representing the corresponding electric field.
Answer:
Given: v = 3 × 10¹⁰ Hz, B = 10^-9 T
i. For wavelength of the wave:
λ = \(\frac{\mathrm{c}}{\mathrm{v}}=\frac{3 \times 10^{8}}{3 \times 10^{10}}\) = 10^-2 m

ii. Since B acts along Z-axis, E acts along Y-axis. Expression representing the oscillating electric field is
E_y = E₀ sin (kx – ωt)
E_y = E₀ sin [(\(\frac {2π}{λ}\))x – (2πv)t]
E_y = E₀ sin 2π [\(\frac {x}{λ}\) – vt]
E_y = E₀ sin 2π [\(\frac {x}{10^{-2}}\) – 3 × 10¹⁰ t]
E_y = E₀ sin 2π [100x – 3 × 10¹⁰ t] V/m

Question 24.
The magnetic field of an EM wave travelling along X-axis is
\(\vec{B}\) = \(\hat{k}\) [4 × 10^-4 sin (ωt – kx)]. Here B is in tesla, t is in second and x is in m. Calculate the peak value of electric force acting on a particle of charge 5 µC travelling with a velocity of 5 × 10⁵ m/s along the Y-axis.
Answer:
Expression for EM wave travelling along
X-axis, \(\vec{B}\) = \(\hat{k}\) [4 × 10^-4 sin (ωt – kx)]
Here, B₀ = 4 × 10^-4
Given: q = 5 µC = 5 × 10^-6 C
v = 5 × 10⁵ m/s along Y-axis
∴ E₀ = cB₀ = 3 × 10⁸ × 4 × 10^-4
= 12 × 10⁴ N/C
Maximum electric force = qE₀
= 5 × 10^-6 × 12 × 10⁴
= 0.6 N

Question 25.
The amplitude of the magnetic field part of harmonic electromagnetic wave in vaccum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Given: B₀ = 510 nT = 510 × 10^-9 T
To find: Amplitude of electric field (E₀)
Formula: E₀ = B₀C
Calculation: From formula,
E₀ = 510 × 10^-9 × 3 × 10⁸
= 153V/m

Question 26.
Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz. (i) Determine, B₀, ω, k, and λ. (ii) Find expressions for \(\vec{E}\) and \(\vec{B}\).
Solution:
For E₀ = 120 N/C, v = 50 MHz = 50 × 10⁶ Hz
i. λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{50×10^6}\) = 6 m
B₀ = \(\frac {E_0}{v}\) = \(\frac {120}{3×10^8}\) = 4 × 10^-7 T = 400 nT
k = \(\frac {2π}{λ}\) = \(\frac {2π}{6}\) = 1.0472 rad/m
ω = 2πv = 2π × 50 × 10⁶
= 3.14 × 10⁸ rad/s.

ii. Assuming motion of em wave along X-axis, expression for electric field vector may lie along Y-axis,
∴ \(\vec{E}\) = E₀ sin (kx – ωt)
= 120 sin (1.0472 × – 3.14 × 10⁸ t) \(\hat{j}\) N/C
Also, magnetic field vector will lie along Z-axis, expression for magnetic field vector,
∴ \(\vec{E}\) = B₀ sin (kx – ωt)
= 4 × 10^-7 sin (1.0472 × – 3.14 × 10⁸ t) \(\hat{k}\) T.

Question 27.
What is electromagnetic spectrum?
Answer:
The orderly distribution (sequential arrangement) of EM waves according to their wavelengths (or frequencies) in the form of distinct groups having different properties is called the EM spectrum.

Question 28.
State various units used for frequency of electromagnetic waves.
Answer:

1. SI unit of frequency of electromagnetic waves is hertz (Hz).
2. Higher frequencies are represented by kHz, MHz, GHz etc.
   [Note: 1 kHz = 10³ Hz, 1 MHz =10⁶ Hz. 1 GHz = 10⁹ Hz]

Question 29.
State different units used for wavelength of electromagnetic waves.
Answer:

1. The SI unit of wavelength of electromagnetic waves is metre (m).
2. Small wavelengths are represented by micrometre (µm), angstrom (Å), nanometre (nm) etc.
   [Note:l A = 10^-10 m = 10^-8 cm, 1 µm = 10^-6 m, 1 nm = 10^-9 m.]

Question 30.
How are radio waves produced? State their properties and uses.
Answer:
Production:

1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Properties:

1. They have very long wavelengths ranging from a few centimetres to a few hundreds of kilometres.
2. The frequency range of AM band is 530 kHz to 1710 kHz. Frequency of the waves used for TV-transmission range from 54 MHz to 890 MHz, while those for FM radio band range from 88 MHz to 108 MHz.

Uses:

1. Radio waves are used for wireless communication purpose.
2. They are used for radio broadcasting and transmission of TV signals.
3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 31.
How are microwaves produced? State their properties and uses.
Answer:
Production:

1. Microwaves are produced by oscillator electric circuits containing a capacitor and an inductor.
2. They can be produced by special vacuum tubes.

Properties:

1. They heat certain substances on which they are incident.
2. They can be detected by crystal detectors.

Uses:

1. Used for the transmission of TV signals.
2. Used for long distance telephone communication.
3. Microwave ovens are used for cooking.
4. Used in radar systems for the location of distant objects like ships, aeroplanes etc,
5. They are used in the study of atomic and molecular structure.

Question 32.
How are infrared waves produced? State their properties and uses.
Answer:
Production:

1. All hot bodies are sources of infrared rays. About 60% of the solar radiations are infrared in nature.
2. Thermocouples, thermopile and bolometers are used to detect infrared rays.

Properties:

1. When infrared rays are incident on any object, the object gets heated.
2. These rays are strongly absorbed by glass.
3. They can penetrate through thick columns of fog, mist and cloud cover.

Uses:

1. Used in remote sensing.
2. Used in diagnosis of superficial tumours and varicose veins.
3. Used to cure infantile paralysis and to treat sprains, dislocations and fractures.
4. They are used in solar water heaters and solar cookers.
5. Special infrared photographs of the body called thermograms, can reveal diseased organs because these parts radiate less heat than the healthy organs.
6. Infrared binoculars and thermal imaging cameras are used in military applications for night vision.
7. Used to keep green house warm.
8. Used in remote controls of TV, VCR, etc.

Question 33.
Write short note on visible light.
Answer:

1. It is the most familiar form of EM waves.
2. These waves are detected by human eye. Therefore this wavelength range is called the visible light.
3. The visible light is emitted due to atomic excitations.
4. Visible light emitted or reflected from objects around us provides us information about those objects and hence about the surroundings.
5. Different wavelengths give rise to different colours as shown in the table given below.
   

   Colour	Wavelength
   Violet	380-450 nm
   Blue	450-495 nm
   Green	495-570 nm
   Yellow	570-590 nm
   Orange	590-620 nm
   Red	620-750 nm

Question 34.
How are ultraviolet rays produced? State their properties and uses.
Answer:
Production:

1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Properties:

1. They produce fluorescence in certain materials, such as ‘phosphors’.
2. They cause photoelectric effect.
3. They cannot pass through glass but pass through quartz, fluorite, rock salt etc.
4. They possess the property of synthesizing vitamin D, when skin is exposed to them.

Uses:

1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
2. Used in burglar alarms and security systems.
3. Used to distinguish real and fake gems.
4. Used in analysis of chemical compounds.
5. Used to detect forgery.

Question 35.
How are X-rays produced? State their properties and uses.
Answer:
Production:

1. German physicist W. C. Rontgen discovered X-rays while studying cathode rays. Hence, X-rays are also called Rontgen rays.
2. Cathode ray is a stream of electrons emitted by the cathode in a vacuum tube.
3. X-rays are produced when cathode rays are suddenly stopped by an obstacle.

Properties:

1. They are high energy EM waves.
2. They are not deflected by electric and magnetic fields.
3. X-rays ionize the gases through which they pass.
4. They have high penetrating power.
5. Their over dose can kill living plant and animal tissues and hence are harmful.

Uses:

1. Useful in the study of the structure of crystals.
2. X-ray photographs are useful to detect bone fracture. X-rays have many other medical uses such as CT scan.
3. X-rays are used to detect flaws or cracks in metals.
4. These are used for detection of explosives, opium etc.

Question 36.
X-rays are used in medicine and industry. Explain.
Answer:
X-rays have many practical applications in medicine and industry. Because X-ray photons are of such high energy, they can penetrate several centimetres of solid matter and can be used to visualize the interiors of materials that are opaque to ordinary light.

Question 37.
How are Gamma rays produced? State their properties and uses.
Answer:
Production:
Gamma rays are emitted from the nuclei of some radioactive elements such as uranium, radium etc.

Properties:

1. They are highest energy (energy range keV – GeV) EM waves.
2. They are highly penetrating.
3. They have a small ionising power.
4. They kill living cells.

Uses:

1. Used as insecticide and disinfectant for wheat and flour.
2. Used for food preservation.
3. Used in radiotherapy for the treatment of cancer and tumour.
4. They are used to produce nuclear reactions.

Question 38.
Identify the name and part of electromagnetic spectrum and arrange these wavelengths in ascending order of magnitude:
Electromagnetic waves with wavelength
i. λ₁ are used by a FM radio station for broad casting.
ii. λ₂ are used to detect bone fracture.
iii. λ₃ are absorbed by the ozone layer of atmosphere.
iv. λ₄ are used to treat muscular strain.
Answer:
i. λ₁ belongs to radiowaves.
ii. λ₂ belongs to X-rays.
iii. λ₃ belongs to ultraviolet rays.
iv. λ₄ belongs to infrared radiations.
Ascending order of magnitude of wavelengths:
λ₃ < λ₃ < λ₄ < λ₁

Question 39.
Explain how different types of waves emitted by stars and galaxies are observed?
Answer:
i. Stars and galaxies emit different types of waves. Radio waves and visible light can pass through the Earth’s atmosphere and reach the ground without getting absorbed significantly. Thus, the radio telescopes and optical telescopes can be placed on the ground.

ii. All other type of waves get absorbed by the atmospheric gases and dust particles. Hence, the y-ray, X-ray, ultraviolet, infrared, and microwave telescopes are kept aboard artificial satellites and are operated remotely from the Earth.

iii. Even though the visible radiation reaches the surface of the Earth, its intensity decreases to some extent due to absorption and scattering by atmospheric gases and dust particles. Optical telescopes are therefore located at higher altitudes.

Question 40.
In communication using radiowaves, how are EM waves propagated?
Answer:
In communication using radio waves, an antenna in the transmitter radiates the EM waves, which travel through space and reach the receiving antenna at the other end.

Question 41.
Draw a schematic structure of earth’s atmosphere describing different atmospheric layers.
Answer:

Question 42.
Draw a diagram showing different types of EM waves.
Answer:

Question 43.
Explain ground wave propagation.
Answer:

1. When a radio wave from a transmitting antenna propagates near surface of the Earth so as to reach the receiving antenna, the wave propagation is called ground wave or surface wave propagation.
2. In this mode, radio waves travel close to the surface of the Earth and move along its curved surface from transmitter to receiver.
3. The radio waves induce currents in the ground and lose their energy by absorption. Therefore, the signal cannot be transmitted over large distances.
4. Radio waves having frequency less than 2 MHz (in the medium frequency band) are transmitted by ground wave propagation.
5. This is suitable for local broadcasting only. For TV or FM signals (very high frequency), ground wave propagation cannot be used.

Question 44.
Explain space wave propagation.
Answer:
i. When the radio waves from the transmitting antenna reach the receiving antenna either directly along a straight line (line of sight) or after reflection from the ground or satellite or after reflection from troposphere, the wave propagation is called space wave propagation.

ii. The radio waves reflected from troposphere are called tropospheric waves.

iii. Radio waves with frequency greater than 30 MHz can pass through the ionosphere (60 km – 1000 km) after suffering a small deviation. Hence, these waves cannot be transmitted by space wave propagation except by using a satellite.

iv. Also, for TV signals which have high frequency, transmission over long distance is not possible by means of space wave propagation.

Question 45.
Explain the concept of range of the signal.
Answer:
i. The maximum distance over which a signal can reach is called its range.

ii. For larger TV coverage, the height of the transmitting antenna should be as large as possible. This is the reason why the transmitting and receiving antennas are mounted on top of high rise buildings.

iii. Range is the straight line distance from the point of transmission (the top of the antenna) to the point on Earth where the wave will hit while travelling along a straight line.

iv. Let the height of the transmitting antenna (AA’) situated at A be h. B represents the point on the surface of the Earth at which the space wave hits the Earth.

v. The triangle OA’B is a right angled triangle. From ∆OA’ B,
(OA’)² = A’B² + OB²
(R + h)² = d² + R²
or R² + h² + 2Rh = d² + R² As
h << R, neglecting h²
d ≈ \(\sqrt{2Rh}\)

vi. The range can be increased by mounting the receiver at a height h’ say at a point C on the surface of the Earth. The range increases to d + d’ where d’ is 2Rh’. Thus
Total range = d + d’ = \(\sqrt{2Rh}\) + \(\sqrt{2Rh’}\)

Question 46.
Explain sky wave propagation.
Answer:

1. When radio waves from a transmitting antenna reach the receiving antenna after reflection in the ionosphere, the wave propagation is called sky wave propagation.
2. The sky waves include waves of frequency between 3 MHz and 30 MHz.
3. These waves can suffer multiple reflections between the ionosphere and the Earth. Therefore, they can be transmitted over large distances.

Question 47.
What is critical frequency?
Answer:
Critical frequency is the maximum value of the frequency of radio wave which can be reflected back to the Earth from the ionosphere when the waves are directed normally to ionosphere.

Question 48.
What is skip distance (zone)?
Answer:
Skip distance is the shortest distance from a transmitter measured along the surface of the Earth at which a sky wave of fixed frequency (if greater than critical frequency) will be returned to the Earth so that no sky waves can be received within the skip distance.

Question 49.
A radar has a power of 10 kW and is operating at a frequency of 20 GHz. It is located on the top of a hill of height 500 m. Calculate the maximum distance upto which it can detect object located on the surface of the Earth.
(Radius of Earth = 6.4 × 10⁶ m)
Answer:
Given: h = 500 m, R = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
d = \(\sqrt{2Rh}\) = \(\sqrt{2×64×10^6×500}\)
= 8 × 10⁴
= 80 km

Question 50.
If the height of a TV transmitting antenna is 128 m, how much square area can be covered by the transmitted signal if the receiving antenna is at the ground level? (Radius of the Earth = 6400 km)
Answer:
Given: h = 128 m, R = 6400 km – 6400 × 10³ m
To find: Area covered (A)
Formulae: i. d = \(\sqrt{2Rh}\) ii. A = πd²
Calculation:
From formula (i),

= 4.048 × 10⁴
= 40.48 km
From formula (ii).
Area covered = 3.142 × (40.48)²
= antilog [log 3.142 + 2log 40.48]
= antilog [0.4972 + 2(1.6073)]
= antilog [3.7118]
= 5.150 × 10³
= 5150 km²

Question 51.
The height of a transmitting antenna is 68 m and the receiving antenna is at the top of a tower of height 34 m. Calculate the maximum distance between them for satisfactory transmission in line of sight mode. (Radius of Earth = 6400 km)
Answer:
Given: h_t = 68 m, h_r = 34 m,
R = 6400 km = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation:
From formula,

= 2.086 × 10⁴
= 20.86 km
d = d_t + d_r = 29.51 + 20.86 = 50.37 km

Question 52.
Explain block diagram of communication system.
Answer:
i. There are three basic (essential) elements of every communication system:

1. Transmitter
2. Communication channel
3. Receiver

ii. In a communication system, the transmitter is located at one place and the receiver at another place.

iii. The communication channel is a passage through which signals transfer in between a transmitter and a receiver.

iv. This channel may be in the form of wires or cables, or may also be wireless, depending on the types of communication system.

Question 53.
What are the two different modes of communication?
Answer:
i. There are two basic modes of communication:
a. point to point communication
b. broadcast communication

ii. In point to point communication mode, communication takes place over a link between a single transmitter and a receiver e.g. telephony.

iii. In the broadcast mode, there are large number of receivers corresponding to the single transmitter e.g., Radio and Television transmission.

Question 54.
Explain the following terms:
i. Signal
ii. Analog signal
iii. Digital signal
iv. Transmitter
v. Transducer
vi. Receiver
vii. Attenuation
viii. Amplification
ix. Range
x. Repeater
Answer:
i. Signal: The information converted into electrical form that is suitable for transmission is called a signal. In a radio station, music and speech are converted into electrical form by a microphone for transmission into space. This electrical form of sound is the signal. A signal can be analog or digital.

ii. Analog signal: A continuously varying signal (voltage or current) is called an analog signal. Since a wave is a fundamental analog signal, sound and picture signals in TV are analog in nature.

iii. Digital signal: A signal (voltage or current) that can have only two discrete values is called a digital signal. For example, a square wave is a digital signal. It has two values viz, +5 V and 0 V.

iv. Transmitter: A transmitter converts the signal produced by a source of information into a form suitable for transmission through a channel and subsequent reception.

v. Transducer: A device that converts one form of energy into another form of energy is called a transducer. For example, a microphone converts sound energy into electrical energy. Therefore, a microphone is a transducer. Similarly, a loudspeaker is a transducer which converts electrical energy into sound energy.

vi. Receiver: The receiver receives the message signal at the channel output, reconstructs it in recognizable form of the original message for delivering it to the user of information.

vii. Attenuation: The loss of strength of the signal while propagating through the channel is known as attenuation. It occurs because the channel distorts, reflects and refracts the signals as it passes through it.

viii. Amplification: Amplification is the process of raising the strength of a signal, using an electronic circuit called amplifier.

ix. Range: The maximum (largest) distance between a source and a destination up to which the signal can be received with sufficient strength is termed as range.

x. Repeater: It is a combination of a transmitter and a receiver. The receiver receives the signal from the transmitter, amplifies it and transmits it to the next repeater. Repeaters are used to increase the range of a communication system.

Question 55.
Explain the role of modulation.
Answer:

1. Low frequency signals cannot be transmitted over large distances. Because of this, a high frequency wave, called a carrier wave, is used.
2. Some characteristic (e.g. amplitude, frequency or phase) of this wave is changed in accordance with the amplitude of the signal. This process is known as modulation.
3. Modulation also helps avoid mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters.
4. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would have got mixed up.

Question 56.
Explain the different types of modulation.
Answer:

1. Modulation can be done by modifying the amplitude (amplitude modulation), frequency (frequency modulation), and phase (phase modulation) of the carrier wave in proportion to the intensity of the signal wave keeping the other two properties same.
2. The carrier wave is a high frequency wave while the signal is a low frequency wave.
3. Waveform (a) in the figure shows a carrier wave and waveform (b) shows the signal.
4. Amplitude modulation, frequency modulation and phase modulation of carrier waves are shown in waveforms (c), (d) and (e) respectively.
   

Question 57.
State advantages and disadvantages of amplitude modulation.
Answer:
Advantages:

1. It is simple to implement.
2. It has large range.
3. It is cheaper.

Disadvantages:

1. It is not very efficient as far as power usage is concerned.
2. It is prone to noise.
3. The reproduced signal may not exactly match the original signal.

In spite of this, these are used for commercial broadcasting in the long, medium and short wave bands.

Question 58.
State uses and limitations of frequency modulation.
Answer:

1. Frequency modulation (FM) is more complex as compared to amplitude modulation and, therefore is more difficult to implement.
2. However, its main advantage is that it reproduces the original signal closely and is less susceptible to noise.
3. This modulation is used for high quality broadcast transmission.

Question 59.
State benefits of phase modulation.
Answer:

1. Phase modulation (PM) is easier than frequency modulation.
2. It is used in determining the velocity of a moving target which cannot be done using frequency modulation.

Question 60.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
i. 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
ii. 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen known as Lamb Shift).
iii. 5890 A – 5896 A [double lines of sodium]
Answer:
i. Radio waves (short wavelength or high frequency end)
ii. Radio waves (short wavelength or high frequency end)
iii. Visible region (yellow light)

Question 61.
Vidhya and Vijay were studying the effect of certain radiations on flower plants. Vidhya exposed her plants to UV rays and Vijay exposed his plants to infrared rays. After few days, Vidhya’s plants got damaged and Vijay’s plants had beautiful bloom. Why did this happen?
Answer:
Frequency of UV rays is greater than infrared rays, hence UV rays are much more energetic than infrared rays. Plants cannot tolerate the exposure of high energy rays. As a result, Vidhya’s plants got damaged and Vijay’s plants had a beautiful bloom.

Multiple Choice Questions

Question 1.
Which of the following type of radiations are radiated by an oscillating electric charge?
(A) Electric
(B) Magnetic
(C) Thermoelectric
(D) Electromagnetic
Answer:
(D) Electromagnetic

Question 2.
If \(\vec{E}\) and \(\vec{B}\) are the electric and magnetic field vectors of e.m. waves, then the direction of propagation of e.m. direction of wave is along the
(A) \(\vec{E}\)
(B) \(\vec{B}\)
(C) \(\vec{E}\) × \(\vec{B}\)
(D) \(\vec{E}\) • \(\vec{B}\)
Answer:
(C) \(\vec{E}\) × \(\vec{B}\)

Question 3.
The unit of expression µ₀o ε₀ is
(A) m / s
(B) m² / s²
(C) s² / m²
(D) s / m
Answer:
(C) s² / m²

Question 4.
According to Maxwell’s equation the velocity of light in any medium is expressed as
(A) \(\frac {1}{\sqrt{µ_0ε_0}}\)
(B) \(\frac {22}{\sqrt{µε}}\)
(C) \(\sqrt{\frac {µ}{ε}}\)
(D) \(\sqrt{\frac {µ_0}{ε}}\)
Answer:
(B) \(\frac {22}{\sqrt{µε}}\)

Question 5.
The electromagnetic waves do not transport.
(A) energy
(B) charge
(C) momentum
(D) pressure
Answer:
(B) charge

Question 6.
In an electromagnetic wave, the direction of the magnetic induction \(\vec{B}\) is
(A) parallel to the electric field \(\vec{E}\).
(B) perpendicular to the electric field \(\vec{E}\).
(C) antiparallel to the pointing vector \(\vec{S}\).
(D) random.
Answer:
(B) perpendicular to the electric field \(\vec{E}\).

Question 7.
Which of the following electromagnetic waves have the longest wavelength?
(A) heat waves
(B) light waves
(C) radio waves
(D) microwaves.
Answer:
(C) radio waves

Question 8.
Radio waves do not penetrate in the band of
(A) ionosphere
(B) mesosphere
(C) troposphere
(D) stratosphere
Answer:
(A) ionosphere

Question 9.
Which of the following electromagnetic wave has least wavelength?
(A) Gamma rays
(B) X- rays
(C) Radio waves
(D) microwaves
Answer:
(A) Gamma rays

Question 10.
If E is an electric field and \(\vec{B}\) is the magnetic induction, then the energy flow per unit area per unit time in an electromagnetic field is given by
(A) \(\frac {1}{µ_0}\) \(\vec{E}\) × \(\vec{B}\)
(B) \(\vec{E}\).\(\vec{B}\)
(C) E² + B²
(D) \(\frac {E}{B}\)
Answer:
(A) \(\frac {1}{µ_0}\) \(\vec{E}\) × \(\vec{B}\)

Question 11.
Out of the X-rays, microwaves, ultra-violet rays, the shortest frequency wave is ……………
(A) X-rays
(B) microwaves
(C) ultra-violet rays
(D) γ-rays
Answer:
(B) microwaves

Question 12.
The part of electromagnetic spectrum used in operating radar is ……………
(A) y-rays
(B) visible rays
(C) infra-red rays
(D) microwaves
Answer:
(D) microwaves

Question 13.
The correct sequence of descending order of wavelength values of the given radiation source is …………..
(A) radio waves, microwaves, infra-red, γ- rays
(B) γ-rays, infra-red, radio waves, microwaves
(C) Infra-red, radio waves, microwaves, γ- rays
(D) microwaves, γ-rays, infra-red, radio waves
Answer:
(A) radio waves, microwaves, infra-red, γ- rays

Question 14.
The nuclei of atoms of radioactive elements produce ……………
(A) X-rays
(B) γ-rays
(C) microwaves
(D) ultra-violet rays
Answer:
(B) γ-rays

Question 15.
The electronic transition in atom produces
(A) ultra violet light
(B) visible light
(C) infra-red rays
(D) microwaves
Answer:
(B) visible light

Question 16.
When radio waves from transmitting antenna reach the receiving antenna directly or after reflection in the ionosphere, the wave propagation is called ………………
(A) ground wave propagation
(B) space wave propagation
(C) sky wave propagation
(D) satellite propagation
Answer:
(C) sky wave propagation

Question 17.
Basic components of a transmitter are ……………..
(A) message signal generator and antenna
(B) modulator and antenna
(C) signal generator and modulator
(D) message signal generator, modulator and antenna
Answer:
(D) message signal generator, modulator and antenna

Question 18.
The process of changing some characteristics of a carrier wave in accordance with the incoming signal is called …………..
(A) amplification
(B) modulation
(C) rectification
(D) demodulation
Answer:
(B) modulation

Question 19.
The process of superimposing a low frequency signal on a high frequency wave is …………….
(A) detection
(B) mixing
(C) modulation
(D) attenuation
Answer:
(C) modulation

Question 20.
A device that converts one form of energy into another form is termed as ……………
(A) transducer
(B) transmitter
(C) amplifier
(D) receiver
Answer:
(A) transducer

Question 21.
A microphone which converts sound into electrical signal is an example of .
(A) a thermistor
(B) a rectifier
(C) a modulator
(D) an electrical transducer
Answer:
(D) an electrical transducer

Question 22.
The process of regaining of information from carries wave at the receiver is called
(A) modulation
(B) transmission
(C) propagation
(D) demodulation
Answer:
(D) demodulation

Question 23.
Range of communication can be increased by
(A) increasing the heights of transmitting and receiving antennas.
(B) decreasing the heights of transmitting and receiving antennas.
(C) increasing height of transmitting antenna and decreasing the height of receiving antenna.
(D) increasing height of receiving antenna only.
Answer:
(A) increasing the heights of transmitting and receiving antennas.

Question 24
Ionosphere mainly consists of
(A) positive ions and electrons
(B) water vapour and smoke
(C) ozone layer
(D) dust particles
Answer:
(A) positive ions and electrons

Question 25.
The reflected waves from the ionosphere are
(A) ground waves.
(B) sky waves.
(C) space waves.
(D) very high frequency waves.
Answer:
(B) sky waves.

Question 26.
Communication is the process of
(A) keeping in touch.
(B) exchanging information.
(C) broadcasting.
(D) entertainment.
Answer:
(B) exchanging information.

Question 27.
The message fed to the transmitter are generally
(A) radio signals
(B) audio signals
(C) both (A) and (B)
(D) optical signals
Answer:
(B) audio signals

Question 28.
Line of sight propagation is also called as ……………. propagation.
(A) sky wave
(B) ground wave
(C) sound wave
(D) space wave
Answer:
(D) space wave

Question 29.
The ozone layer in the atmosphere absorbs
(A) only the radio waves.
(B) only the visible light.
(C) only the y rays.
(D) X-rays and ultraviolet rays.
Answer:
(D) X-rays and ultraviolet rays.

Question 30.
Modem communication systems consist of
(A) electronic systems
(B) electrical system
(C) optical system
(D) all of these
Answer:
(D) all of these

Question 31.
What determines the absorption of radio waves by the atmosphere?
(A) Frequency .
(B) Polarisation
(C) Interference
(D) Distance of receiver
Answer:
(A) Frequency .

Question 32.
The portion of the atmosphere closest to the earth’s surface is ……………
(A) troposphere
(B) stratosphere
(C) mesosphere
(D) ionosphere
Answer:
(A) troposphere

Question 33.
An antenna behaves as resonant circuit only when its length is ………………
(A) λ/2
(B) λ/4
(C) λ
(D) n λ/2
Answer:
(D) n λ/2

Question 34.
Space wave travels through …………………
(A) ionosphere
(B) mesosphere
(C) troposphere
(D) stratosphere
Answer:
(C) troposphere

Question 35.
Transmission lines start radiating
(A) at low frequencies
(B) at high frequencies.
(C) at both high and low frequencies.
(D) none of the above.
Answer:
(B) at high frequencies.

Question 36.
If ‘h_t‘ and ‘h_r’ are height of transmitting and receiving antennae and ‘R’ is radius of the earth, the range of space wave is
(A) \(\sqrt {2R}\) (h_t + h_r)
(B) 2R \(\sqrt {(h_t + h_r)}\)
(C) \(\sqrt {2R(h_t + h_r)}\)
(D) \(\sqrt {2R}\) (√h_t + √h_r)
Answer:
(D) \(\sqrt {2R}\) (√h_t + √h_r)

Question 37.
In a communication system, noise is most likely to affect the signal ………..
(A) at the transmitter
(B) in the transmission medium
(C) in the information source
(D) at the destination
Answer:
(B) in the transmission medium

Question 38.
The power radiated by linear antenna of length 7’ is proportional to (A = wavelength)
(A) \(\frac {λ}{l}\)
(B) (\(\frac {λ}{l}\))²
(C) \(\frac {l}{λ}\)
(D) (\(\frac {l}{λ}\))²
Answer:
(D) (\(\frac {l}{λ}\))²

Question 39.
For efficient radiation and reception of signal with wavelength λ, the transmitting antennas would have length comparable to ……………….
(A) λ of frequency used
(B) λ/2 of frequency used
(C) λ/3 of frequency used
(D) λ/4 of frequency used
Answer:
(A) λ of frequency used

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 14 Semiconductors Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 14 Semiconductors

Question 1.
What are the factors on which electrical conductivity of any solid depends?
Answer:
Electrical conduction in a solid depends on its temperature, the number of charge carriers, how easily these carries can move inside a solid (mobility), its crystal structure, types and the nature of defects present in a solid.

Question 2.
Why are metals good conductor of electricity?
Answer:
Metals are good conductors of electricity due to the large number of free electrons (≈ 10²⁸ per m³) present in them.

Question 3.
Give the formula for electrical conductivity of a solid and give significance of the terms involved.
Answer:
Electrical conductivity (σ) of a solid is given by a = nqµ,
where, n = charge carrier density (number of charge carriers per unit volume)
q = charge on the carriers
µ = mobility of carriers

Question 4.
Explain in brief temperature dependence of electrical conductivity of metals and semiconductors with the help of graph.
Answer:
i. The electrical conductivity of a metal decreases with increase in its temperature.

ii. When the temperature of a semiconductor is increased, its electrical conductivity also increases

Question 5.
Mention the broad classification of semiconductors along with examples.
Answer:
A broad classification of semiconductors can be:

1. Elemental semiconductors: Silicon, germanium
2. Compound Semiconductors: Cadmium sulphide, zinc sulphide, etc.
3. Organic Semiconductors: Anthracene, doped pthalocyanines, polyaniline etc.

Question 6.
What are some electrical properties of semiconductors?
Answer:

1. Electrical properties of semiconductors are different from metals and insulators due to their unique conduction mechanism.
2. The electronic configuration of the elemental semiconductors plays a very important role in their electrical properties.
3. They are from the fourth group of elements in the periodic table.
4. They have a valence of four.
5. Their atoms are bonded by covalent bonds. At absolute zero temperature, all the covalent bonds are completely satisfied in a single crystal of pure semiconductor like silicon or germanium.

Question 7.
Explain in detail the distribution of electron energy levels in an isolated atom with the help of an example.
Answer:

1. An isolated atom has its nucleus at the centre which is surrounded by a number of revolving electrons. These electrons are arranged in different and discrete energy levels.
2. Consider the electronic configuration of sodium (atomic number 11) i.e, 1s², 2s², 2p⁶, 3s¹. The outermost level 3s can take one more electron but it is half filled in sodium,
3. The energy levels in each atom are filled according to Pauli’s exclusion principle which states that no two similar spin electrons can occupy the same energy level.
4. That means any energy level can accommodate only two electrons (one with spin up state and the other with spin down state)
5. Thus, there can be two states per energy level.
6. Figure given below shows the allowed energy levels of a sodium atom by horizontal lines. The curved lines represent the potential energy of an electron near the nucleus due to Coulomb interaction.
   

Question 8.
Explain formation of energy bands in solid sodium with neat labelled energy band diagrams.
Answer:
i. For an isolated sodium atom (atomic number 11) the electronic configuration is given as 1s², 2s², 2p⁶, 3s¹. The outermost level 3s is half filled in sodium.

ii. The energy levels are filled according to Pauli’s exclusion principle.

iii. Consider two sodium atoms close enough so that outer 3s electrons can be considered equally to be part of any atom.

iv. The 3s electrons from both the sodium atoms need to be accommodated in the same level.

v. This is made possible by splitting the 3 s level into two sub-levels so that the Pauli’s exclusion principle is not violated. Figure given below shows the splitting of the 3 s level into two sub levels.

vi. When solid sodium is formed, the atoms come close to each other such that distance between them remains of the order of 2 – 3 Å. Therefore, the electrons from different atoms interact with each other and also with the neighbouring atomic cores.

vii. The interaction between the outer most electrons is more due to overlap while the inner most electrons remain mostly unaffected. Each of these energy levels is split into a large number of sub levels, of the order of Avogadro’s number due to number of atoms in solid sodium is of the order of this number.

viii. The separation between the sublevels is so small that the energy levels appear almost continuous. This continuum of energy levels is called an energy band. The bands are called 1 s band, 2s band, 2p band and so on. Figure shows these bands in sodium metal.

Question 9.
Explain concept of valence band and conduction band in solid crystal.
Answer:
A. Valence band (V.B):

1. The topmost occupied energy level in an atom is the valence level. The energy band formed by valence energy levels of atoms in a solid is called the valence band.
2. In metallic conductors, the valence electrons are loosely attached to the nucleus. At ordinary room temperature, some valence electrons become free. They do not leave the metal surface but can move from atom to atom randomly.
3. Such free electrons are responsible for electric current through conductors.

B. Conduction band (C.B):

1. The immediately next energy level that electrons from valence band can occupy is called conduction level. The band formed by conduction levels is called conduction band.
2. It is the next permitted energy band beyond valence band.
3. In conduction band, electrons move freely and conduct electric current through the solids.
4. An insulator has empty conduction band.

Question 10.
Draw neat labelled diagram showing energy bands in sodium. Why broadening of higher bands is different than that of the lower energy bands?
Answer:

Broadening of valence and higher bands is more since interaction of these electrons is stronger than the inner most electrons.

Question 11.
State the conditions when electrons of a semiconductor can take part in conduction.
Answer:

1. All the energy levels in a band, including the topmost band, in a semiconductor are completely occupied at absolute zero.
2. At some finite temperature T, few electrons gain thermal energy of the order of kT, where k is the Boltzmann constant.
3. Electrons in the bands between the valence band cannot move to higher band since these are already occupied.
4. Only electrons from the valence band can be excited to the empty conduction band, if the thermal energy gained by these electrons is greater than the band gap.
5. Electrons can also gain energy when an external electric field is applied to a solid. Energy gained due to electric field is smaller, hence only electrons at the topmost energy level gain such energy and participate in electrical conduction.

Question 12.
Define 1 eV.
Answer:
1 eV is the energy gained by an electron while it overcomes a potential difference of one volt. 1 eV= 1.6 × 10^-19 J.

Question 13.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors?
Answer:

1. The 4 valence electrons of C, Si or Ge lie respectively in the second, third and fourth orbit.
2. Energy required to take out an electron from these atoms (i.e., ionisation energy E_g) will be least for Ge, followed by Si and highest for C.
3. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for

Question 14.
What is intrinsic semiconductor?
Answer:
A pure semiconductor is blown as intrinsic semiconductor.

Question 15.
Explain characteristics and structure of silicon using a neat labelled diagram.
Answer:

1. Silicon (Si) has atomic number 14 and its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p².
2. Its valence is 4.
3. Each atom of Si forms four covalent bonds with its neighbouring atoms. One Si atom is surrounded by four Si atoms at the comers of a regular tetrahedron as shown in the figure.
   

Question 16.
Describe in detail formation of holes in ii. intrinsic semiconductor.
Answer:
i. In intrinsic semiconductor at absolute zero temperature, all valence electrons are tightly bound to respective atoms and the covalent bonds are complete.

ii. Electrons are not available to conduct electricity through the crystal because they cannot gain enough energy to get into higher energy levels.

iii. At room temperature, however, a few covalent bonds are broken due to heat energy produced by random motion of atoms. Some of the valence electrons can be moved to the conduction band. This creates a vacancy in the valence band as shown in figure.

iv. These vacancies of electrons in the valence band are called holes. The holes are thus absence of electrons in the valence band and they carry an effective positive charge.

Question 17.
How does electric conduction take place inside a pure silicon?
Answer:

1. There are two different types of charge carriers in a pure semiconductor. One is the electron and the other is the hole or absence of electron.
2. Electrical conduction takes place by transportation of both carriers or any one of the two carriers in a semiconductor.
3. When a semiconductor is connected in a circuit, electrons, being negatively charged, move towards positive terminal of the battery.
4. Holes have an effective positive charge, and move towards negative terminal of the battery. Thus, the current through a semiconductor is carried by two types of charge carriers moving in opposite directions.
5. Figure given below represents the current through a pure silicon.
   

Question 18.
Why do holes not exist in conductor?
Answer:

1. In case of semiconductors, there is one missing electron from one of the covalent bonds.
2. The absence of electron leaves an empty space called as hole; each hole carries an effective positive charge.
3. In case of an conductor, number of free electrons are always available for conduction. There is no absence of electron in it. Hence holes do not exist in conductor.

Question 19.
What is the need for doping an intrinsic semiconductor?
Answer:
The electric conductivity of an intrinsic semiconductor is very low at room temperature; hence no electronic devices can be fabricated using them. Addition of a small amount of a suitable impurity to an intrinsic semiconductor increases its conductivity appreciably. Hence, intrinsic semiconductors are doped with impurities.

Question 20.
Explain what is doping.
Answer:

1. The process of adding impurities to an intrinsic semiconductor is called doping.
2. The impurity atoms are called dopants which may be either trivalent or pentavalent. The parent atoms are called hosts.
3. The dopant material is so selected that it does not disturb the crystal structure of the host.
4. The size and the electronic configuration of the dopant should be compatible with that of the host.
5. Doping is expressed in ppm (parts per million), i.e., one impurity atom per one million atoms of the host.
6. Doping significantly increases the concentration of charge carriers.

Question 21.
What is extrinsic semiconductors?
Answer:
The semiconductor with impurity is called a doped semiconductor or an extrinsic semiconductor.

Question 22.
Draw neat diagrams showing schematic electronic structure of:
i. A pentavalent atom [Antimony (Sb)]
ii. A trivalent atom [Boron (B)]
Answer:

[Note: Electronic structure of antimony is drawn as per its electronic configuration in accordance with Modern Periodic Table.]

Question 23.
With the help of neat diagram, explain the structure of n-type semiconductor in detail.
Answer:
i. When silicon or germanium crystal is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony we get n-type semiconductor.

ii. When a dopant atom of 5 valence electrons occupies the position of a Si atom in the crystal lattice, 4 electrons from the dopant form bonds with 4 neighbouring Si atoms and the fifth electron from the dopant remains very weakly bound to its parent atom

iii. To make this electron free even at room temperature, very small energy is required. It is 0.01 eV for Ge and 0.05 eV for Si.

iv. As this semiconductor has large number of electrons in conduction band and its conductivity is due to negatively charged carriers, it is called n-type semiconductor.

v. The n-type semiconductor also has a few electrons and holes produced due to the thermally broken bonds.

vi. The density of conduction electrons (n_e) in a doped semiconductor is the sum total of the electrons contributed by donors and the thermally generated electrons from the host.

vii. The density of holes (n_h) is only due to the thermally generated holes of the host Si atoms.

viii. Thus, the number of free electrons exceeds the number of holes (n_e >> n_h). Thus, in n-type semiconductor electrons are the majority carriers and holes are the minority carriers.

Question 24.
What are some features of n-type semiconductor?
Answer:

1. These are materials doped with pentavalent impurity (donors) atoms.
2. Electrical conduction in these materials is due to majority charge carriers i.e., electrons.
3. The donor atom loses electrons and becomes positively charged ions.
4. Number of free electrons is very large compared to the number of holes, n_e >> n_h. Electrons are majority charge carriers.
5. When energy is supplied externally, negatively charged free electrons (majority charges carries) and positively charged holes (minority charges carries) are available for conduction.

Question 25.
With the help of neat diagram, explain the structure of p-type semiconductor in detail.
Answer:
i. When silicon or germanium crystal is doped with a trivalent impurity such as boron, aluminium or indium, we get a p-type semiconductor.

ii. The dopant trivalent atom has one valence electron less than that of a silicon atom. Every trivalent dopant atom shares its three electrons with three neighbouring Si atoms to form covalent bonds. But the fourth bond between silicon atom and its neighbour is not complete.

iii. The incomplete bond can be completed by another electron in the neighbourhood from Si atom.

iv. The shared electron creates a vacancy in its place. This vacancy or the absence of electron is a hole.

v. Thus, a hole is available for conduction from each acceptor impurity atom.

vi. Holes are majority carriers and electrons are minority carriers in such materials. Acceptor atoms are negatively charged ions and majority carriers are holes. Therefore, extrinsic semiconductor doped with trivalent impurity is called a p-type semiconductor.

vii. For a p-type semiconductor, n_h >> n_e^.

Question 26.
What are some features of p-type semiconductors?
Answer:

1. These are materials doped with trivalent impurity atoms (acceptors).
2. Electrical conduction in these materials is due to majority charge carriers i.e., holes.
3. The acceptor atoms acquire electron and become negatively charged-ions.
4. Number of holes is very large compared to the number of free electrons. n_h >> n_e. Holes are majority charge carriers.
5. When energy is supplied externally, positively charged holes (majority charge carriers) and negatively charged free electrons (minority charge carriers) are available for conduction.

Question 27.
What are donor and acceptor impurities?
Answer:

1. Every pentavalent dopant atom which donates one electron for conduction is called a donor impurity.
2. Each trivalent atom which can accept an electron is called an acceptor impurity.

Question 28.
Explain the energy levels of both donor and acceptor impurities with a schematic band structure.
Answer:
i. The free electrons donated by the donor impurity atoms occupy energy levels which are in the band gap and are close to the conduction band.

ii. The vacancies of electrons or the extra holes are created in the valence band due to addition of acceptor impurities. The impurity levels are created just above the valence band in the band gap.

Question 29.
Distinguish between p-type and n-type semiconductor.
Answer:

p-type semiconductor	n-type semiconductor
1. The impurity of some trivalent element like B, Al, In, etc. is mixed with semiconductor.	The impurity of some pentavalent element like P, As, Sb, etc. is mixed
2. The impurity atom accepts one electron hence the impurities	The impurity atom donates – one electron, hence the impurities added are known as donor impurities.
3. The holes are majority charge carriers and electrons are minority charge carriers.	The electrons are j majority charge carriers and holes are minority charge carriers.
4. The acceptor energy level is close to the valence band and far away from conduction band.	Donor energy level is close to the conduction band and far away from valence band.

Question 30.
What is the charge on a p-type and n-type semiconductor?
Answer:
n-type as well as p-type semiconductors are electrically neutral.

Question 31.
Explain the transportation of holes inside a p-type semiconductor.
Answer:
i. Consider a p-type semiconductor connected to terminals of a battery as shown.

ii. When the circuit is switched on, electrons at 1 and 2 are attracted to the positive terminal of the battery and occupy nearby holes at x and y. This creates holes at the positions 1 and 2 previously occupied by electrons.

iii. Next, electrons at 3 and 4 move towards the positive terminal and create holes in their previous positions.

iv. But, the holes are captured at the negative terminal by the electrons supplied by the battery.

v. In this way, holes are transported from one place to other and density of holes is kept constant so long as the battery is working.

Question 32.
A pure Si crystal has 4 × 10²⁸ atoms m^-3. It is doped by 1 ppm concentration of antimony. Calculate the number of electrons and holes. Given n₁ = 1.2 × 10¹⁶/m³.
Answer:
As, the atom is doped with 1 ppm concentration of antimony (Sb).
1 ppm = 1 parts per one million atoms. = 1/10⁶
∴ no. of Si atoms = \(\frac {Total no. of Si atoms}{10^6}\)
= \(\frac {4×10^{28}}{10^6}\) = 4 × 10²² m^-3
i.e., total no. of extra free electrons (n_e)
= 4 × 10²² m^-3
n_i² = n_e n_h
∴ n_h = \(\frac {n_i^2}{n_e}\) = \(\frac {(1.2×10^{16})^2}{4×10^{22}}\)
= \(\frac {144×10^{30}{4×10^{22}}\)
= 36 × 10^-8
= 3.6 × 10⁹ m^-3.

Question 33.
A pure silicon crystal at temperature of 300 K has electron and hole concentration 1.5 × 10¹⁶ m^-3 each. (n_e = n_h). Doping bv indium increases n_h to 4.5 × 10²² m^-3. Calculate ne for the doped silicon crystal.
Answer:
Given: At 300 K, n_i = n_e = n_h = 1.5 × 10¹⁶ m^-3
After doping n_h = 4.5 × 10²² m^-3
To find: Number density of electrons (n_e)
Formula: n_i² = n_e nn_h
Calculation From formula:
n_e = \(\frac {n_i^2}{n_h}\) = \(\frac {(1.5×10^{16})^2}{4×10^{22}}\)
= \(\frac {255×10^{30}{45×10^{21}}\)
= 5 × 10^-9 m^-3.

Question 34.
A Ge specimen is doped with A/. The concentration of acceptor atoms is ~10²¹ atoms/m³. Given that the intrinsic concentration of electron-hole pairs is ~10 ¹⁹/m³, calculate the concentration of electrons in the specimen.
Answer:
Given: At room temperature,
n_i = n_e = n_h = 10¹⁹ m^-3
After doping n_h = 10²¹ m^-3
To find: Number density of electrons (nc)
Formulae: n_i² = n_en_h
Calculation: From formula,
n_e = \(\frac {n_i^2}{n_h}\) = \(\frac {(10^{19})^2}{10^{21}}\)
= \(\frac {255×10^{30}{45×10^{21}}\)
= 10¹⁷ m^-3.

Question 35.
A semiconductor has equal electron and hole concentration of 2 × 10⁸ m^-3. On doping with a certain impurity, the electron concentration increases to 4 × 10¹⁰ m^-3, then calculate the new hole concentration of the semiconductor.
Answer:
Given: n_i = 2 × 10⁸ m^-3, n = 4 × 10¹⁰ m^-3
After doping n_h = 10²¹ m^-3
To find: Number density of holes (n_h)
Formulae: n_i ²= n_en_h
Calculation: From formula.
n_h = \(\frac {n_i^2}{n_e}\) = \(\frac {(2×10^{8})^2}{4×10^{10}}\) = 10⁶ m^-3

Question 36.
What is a p-n junction?
Answer:
When n-type and p-type semiconductor materials are fused together, the junction formed is called as p-n junction.

Question 37.
Explain the process of diffusion in p-n junction.
Answer:
i. The transfer of electrons and holes across the p-n junction is called diffusion.

ii. When an n-type and a p-type semiconductor materials are fused together, initially, the number of electrons in the n-side of a junction is very large compared to the number of electrons on the p-side. The same is true for the number of holes on the p-side and on the n-side.

iii. Thus, a large difference in density of carriers exists on both sides of the p-n junction. This difference causes migration of electrons from the n-side to the p-side of the

iv. They fill up the holes in the p-type material and produce negative ions.

v. When the electrons from the n-side of a junction migrate to the p-side, they leave behind positively charged donor ions on the n- side. Effectively, holes from the p-side migrate into the n-region.

vi. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

vii. The extent up to which the electrons and the holes can diffuse across the junction depends on the density of the donor and the acceptor ions on the n-side and the p-side respectively, of the junction.

Question 38.
Define potential barrier.
Answer:
The diffusion of carriers across the junction and resultant accumulation of positive and negative charges across the junction builds a potential difference across the junction. This potential difference is called the potential barrier.

Question 39.
Draw neat labelled diagrams for potentials barrier and depletion layer in a p-n junction.
Answer:

Question 40.
Explain in brief electric field across a p-n junction with a neat labelled diagram.
Answer:

1. When p-type semiconductor is fused with n-type semiconductor, a depletion region is developed across the junction.
2. The n-side near the boundary of a p-n junction becomes positive with respect to the p-side because it has lost electrons and the p-side has lost holes.
3. Thus, the presence of impurity ions on both sides of the junction establishes an electric field across this region such that the n-side is at a positive voltage relative to the p-side.
   

Question 41.
What is the need of biasing a p-n junction?
Answer:

1. Due to potential barrier across depletion region, charge carriers require extra energy to overcome the barrier.
2. A suitable voltage needs to be applied to the junction externally, so that these charge carriers can overcome the potential barrier and move across the junction.

Question 41.
Explain the mechanism of forward biased p-n junction.
Answer:

1. In forward bias, a p-n junction is connected in an electric circuit such that the p-region is connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
2. The external voltage effectively opposes the built-in potential of the junction. The width of potential barrier is thus reduced.
3. Also, negative charge carriers (electrons) from the n-region are pushed towards the junction.
4. A similar effect is experienced by positive charge carriers (holes) in the p-region and they are pushed towards the junction.
5. Both the charge carriers thus find it easy to cross over the barrier and contribute towards the electric current.
   

Question 42.
Explain the mechanism of reverse biased p-n junction.
Answer:
i. In reverse biased, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased.

ii. Also, the negative charge carriers (electrons) from the n-region are pulled away from the junction.

iii. Similar effect is experienced by the positive charge carriers (holes) in the p-region and they are pulled away from the junction.

iv. Both the charge carriers thus find it very difficult to cross over the barrier and thus do not contribute towards the electric current.

Question 43.
State some important features of the depletion region.
Answer:

1. It is formed by diffusion of electrons from n-region to the p-region. This leaves positively charged ions in the n-region.
2. The p-region accumulates electrons (negative charges) and the n-region accumulates the holes (positive charges).
3. The accumulation of charges on either sides of the junction results in forming a potential barrier and prevents flow of charges.
4. There are no charges in this region.
5. The depletion region has higher potential on the n-side and lower potential on the p-side of the junction.

Question 44.
What is p-n junction diode? Draw its circuit symbol.
Answer:
A p-n junction, when provided with metallic connectors on each side is called a junction diode

Question 45.
Explain asymmetrical flow of current in p-n junction diode in detail.
Answer:

i. The barrier potential is reduced in forward biased mode and it is increased in reverse biased mode.

ii. Carriers find it easy to cross the junction in forward bias and contribute towards current because the barrier width is reduced and they are pushed towards the junction and gain extra energy to cross the junction.

iii. The current through the diode in forward bias is large and of the order of a few milliamperes (10^-3 A) for a typical diode.

iv. When connected in reverse bias, width of the potential barrier is increased and the carriers are pushed away from the junction so that very few carriers can cross the junction and contribute towards current.

v. This results in a very small current through a reverse biased diode. The current in reverse biased diode is of the order of a few microamperes (10^-6 A).

vi. When the polarity of bias voltage is reversed, the width of the depletion layer changes. This results in asymmetrical current flow through a diode as shown in figure.

Question 46.
What is knee voltage?
Answer:
In forward bias mode, the voltage for which the current in a p-n junction diode rises sharply is called knee voltage.

Question 47.
What is a forward current in case of zero biased p-n junction diode?
Answer:
When the diode terminals are shorted together, some holes (majority carriers) in the p-side have enough thermal energy to overcome the potential barrier. Such carriers cross the barrier potential and contribute to current. This current is known as the forward current.

Question 48.
Define reverse current in zero biased p-n junction diode.
Answer:
When the diode terminals are shorted together some holes generated in the n-side (minority carriers), move across the junction and contribute to current. This current is known as the reverse current.

Question 49.
Explain the I-V characteristics of a reverse biased junction diode.
Answer:
i. The positive terminal of the external voltage is connected to the cathode (n-side) and negative terminal to the anode (p-side) across the diode.

ii. In case of reverse bias the width of the depletion region increases and the p-n junction behaves like a high resistance.

iii. Practically no current flows through it with an increase in the reverse bias voltage. However, a very small leakage current does flow through the junction which is of the order of a few micro amperes, (µA).

iv. When the reverse bias voltage applied to a diode is increased to sufficiently large value, it causes the p-n junction to overheat. The overheating of the junction results in a sudden rise in the current through the junction. This is because covalent bonds break and a large number of carries are available for conduction. The diode thus no longer behaves like a diode. This effect is called the avalanche breakdown.

v. The reverse biased characteristic of a diode is shown in figure.

Question 50.
Explain zero biased junction diode.
Answer:
i. When a diode is connected in a zero bias condition, no external potential energy is applied to the p-n junction.

li. The potential barrier that exists in a junction prevents the diffusion of any more majority carriers across it. However, some minority carriers (few free electrons in the p-region and few holes in the n-region) drift across the junction.

iii. An equilibrium is established when the majority carriers are equal in number (n_e = n_h) and both moving in opposite directions. The net current flowing across the junction is zero. This is a state of‘dynamic equilibrium’.

iv. The minority carriers are continuously generated due to thermal energy.

v. When the temperature of the p-n junction is raised, this state of equilibrium is changed.

vi. This results in generating more minority carriers and an increase in the leakage current. An electric current, however, cannot flow through the diode because it is not connected in any electric circuit

Question 51.
What is dynamic equilibrium?
Answer:
An equilibrium is established when the majority carriers are equal in number (n_e = n_h) and both moving in opposite directions. The net current flowing across the junction is zero. This is a state of‘dynamic equilibrium’.

Question 52.
Draw a neat diagram and state I-V characteristics of an ideal diode.
Answer:
An ideal diode offers zero resistance in forward biased mode and infinite resistance in reverse biased mode.

Question 53.
What do you mean by static resistance of a diode?
Answer:
Static (DC) resistance:

1. When a p-n junction diode is forward biased, it offers a definite resistance in the circuit. This resistance is called the static or DC resistance (R_g) of a diode.
2. The DC resistance of a diode is the ratio of the DC voltage across the diode to the DC current flowing through it at a particular voltage.
3. It is given by, R_g = \(\frac {V}{I}\)

Question 54.
Explain dynamic resistance of a diode.
Answer:

1. The dynamic (AC) resistance of a diode, r_g, at a particular applied voltage, is defined as
   r_g = \(\frac {∆V}{∆I}\)
2. The dynamic resistance of a diode depends on the operating voltage.
3. It is the reciprocal of the slope of the characteristics at that point.

Question 55.
Draw a graph representing static and dynamic resistances of a diode.
Answer:

Question 56.
Refer to the figure as shown below and find the resistance between point A and B when an ideal diode is (i) forward biased and (ii) reverse biased.

Answer:
We know that for an ideal diode, the resistance is zero when forward biased and infinite when reverse biased.
i. Figure (a) shows the circuit when the diode is forward biased. An ideal diode behaves as a conductor and the circuit is similar to two resistances in parallel.

∴ R_AB = (30 × 30)/(30 +30) = 900/60 = 15 Ω

ii. Figure (b) shows the circuit when the diode is reverse biased.

It does not conduct and behaves as an open switch along path ACB. Therefore, R_AB = 30 Ω. the only resistance in the circuit along the path ADB.

Question 57.
State advantages of semiconductor devices.
Answer:

1. Electronic properties of semiconductors can be controlled to suit our requirement.
2. They are smaller in size and light weight.
3. They can operate at smaller voltages (of the order of few mV) and require less current (of the order of pA or mA), therefore, consume lesser power.
4. Almost no heating effects occur, therefore these devices are thermally stable.
5. Faster speed of operation due to smaller size.
6. Fabrication of ICs is possible.

Question 58.
State disadvantages of semiconductor devices.
Answer:

1. They are sensitive to electrostatic charges.
2. Not very useful for controlling high power.
3. They are sensitive to radiation.
4. They are sensitive to fluctuations in temperature.
5. They need controlled conditions for their manufacturing.
6. Very few materials are semiconductors.

Question 59.
Explain applications of semiconductors.
Answer:
i. Solar cell:

1. It converts light energy into electric energy.
2. t is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photo resistor: It changes its resistance when light is incident on it.

iii. Bi-polar junction transistor:

1. These are devices with two junctions and three terminals.
2. A transistor can be a p-n-p or n-p-n transistor.
3. Conduction takes place with holes and electrons.
4. Many other types of transistors are designed and fabricated to suit specific requirements.
5. They are used in almost all semiconductor devices.

iv. Photodiode: It conducts when illuminated with light.

v. LED (Light Emitting Diode):

1. It emits light when current passes through it.
2. House hold LED lamps use similar technology.
3. They consume less power, are smaller in size and have a longer life and are cost effective.

vi. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

vii. Integrated Circuits (ICs): A small device having hundreds of diodes and transistors performs the work of a large number of electronic circuits.

Question 60.
Explain any four application of p-n junction diode.
Answer:
1. Solar cell:

1. It converts light energy into electric energy.
2. It is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photodiode: It conducts when illuminated with light.

iii. LED (Light Emitting Diode):

1. It emits light when current passes through it.
2. House hold LED lamps use similar technology.
3. They consume less power, are smaller in size and have a longer life and are cost effective.

iv. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

Question 61.
What is thermistor?
Answer:
Thermistor is a temperature sensitive resistor. Its resistance changes with change in its temperature.

Question 62.
What are different ty pes of thermistor and what are their applications?
Answer:
There are two types of thermistor:
i. NTC (Negative Thermal Coefficient) thermistor: Resistance of a NTC thermistor decreases with increase in its temperature. Its temperature coefficient is negative. They are commonly used as temperature sensors and also in temperature control circuits.

ii. PTC (Positive Thermal Coefficient) thermistor: Resistance of a PTC thermistor increases with increase in its temperature. They are commonly used in series with a circuit. They are generally used as a reusable fuse to limit current passing through a circuit to protect against over current conditions, as resettable fuses.

Question 63.
How are thermistors fabricated?
Answer:
Thermistors are made from thermally sensitive metal oxide semiconductors. Thermistors are very sensitive to changes in temperature.

Question 64.
Enlist any two features of thermistor.
Answer:

1. A small change in surrounding temperature causes a large change in their resistance.
2. They can measure temperature variations of a small area due to their small size.

Question 65.
Write a note on:
i. Electric devices
ii. Electronic devices
Answer:
i. Electric devices:

1. These devices convert electrical energy into some other form.
2. Examples: Fan, refrigerator, geyser etc. Fan converts electrical energy into mechanical energy. A geyser converts it into heat energy.
3. They use good conductors (mostly metals) for conduction of electricity.
4. Common working range of currents for electric circuits is milliampere (mA) to ampere.
5. Their energy consumption is also moderate to high. A typical geyser consumes about 2.0 to 2.50 kW of power.
6. They are moderate to large in size and are costly.

ii. Electronic devices:

1. Electronic circuits work with control or sequential changes in current through a cell.
2. A calculator, a cell phone, a smart watch or the remote control of a TV set are some of the electronic devices.
3. Semiconductors are used to fabricate such devices.
4. Common working range of currents for electronic circuits it is nano-ampere to µA.
5. They consume very low energy. They are very compact, and cost effective.

Question 66.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction?
Answer:

1. No. Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å).
2. Hence, continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 67.
What is Avalanche breakdown and zener breakdown?
Answer:
i. Avalanche breakdown: In high reverse bias, minority carriers acquire sufficient kinetic energy and collide with a valence electron. Due to collisions the covalent bond breaks. The valence electron enters conduction band. A breakdown occurring in such a manner is avalanche breakdown. It occurs with lightly doped p-n junctions.

ii. Zener breakdown: It occurs in specially designed and highly doped p-n junctions, viz., zener diodes. In this case, covalent bonds break directly due to application of high electric field. Avalanche breakdown voltage is higher than zener voltage.

Question 68.
Indicators on platform, digital clocks, calculators make use of seven LEDs to indicate a number. How do you think these LEDs might be arranged?
Answer:
i. The indicators on platforms, digital clocks, calculators are made using seven LEDs arranged in such a way that when provided proper signal they light up displaying desired alphabet or number.

ii. This arrangement of LEDs is called Seven Segment Display.

Multiple Choice Questions

Question 1.
The number of electrons in the valence shell of semiconductor is ……………
(A) less than 4
(B) equal to 4
(C) more than 4
(D) zero
Answer:
(B) equal to 4

Question 2.
If the temperature of semiconductor is increased, the number of electrons in the valence band will ……………….
(A) decrease
(B) remains same
(C) increase
(D) either increase or decrease
Answer:
(A) decrease

Question 3.
When N-type semiconductor is heated, the ……………..
(A) number of electrons and holes remains same.
(B) number of electrons increases while that of holes decreases.
(C) number of electrons decreases while that of holes increases.
(D) number of electrons and holes increases equally.
Answer:
(D) number of electrons and holes increases equally.

Question 4.
In conduction band of solid, there is no electron at room temperature. The solid is ……………
(A) semiconductors
(B) insulator
(C) conductor
(D) metal
Answer:
(B) insulator

Question 5.
In the crystal of pure Ge or Si, each covalent bond consists of …………..
(A) 1 electron
(B) 2 electrons
(C) 3 electrons
(D) 4 electrons
Answer:
(B) 2 electrons

Question 6.
A pure semiconductor is ……………..
(A) an extrinsic semiconductor
(B) an intrinsic semiconductor
(C) p-type semiconductor
(D) n-type semiconductor
Answer:
(B) an intrinsic semiconductor

Question 7.
For an extrinsic semiconductor, the valency of the donor impurity is …………..
(A) 2
(B) 1
(C) 4
(D) 5
Answer:
(D) 5

Question 8.
In a semiconductor, acceptor impurity is
(A) antimony
(B) indium
(C) phosphorous
(D) arsenic
Answer:
(B) indium

Question 9.
What are majority carriers in a semiconductor?
(A) Holes in n-type and electrons in p-type.
(B) Holes in n-type and p-type both.
(C) Electrons in n-type and p-type both.
(D) Holes in p-type and electrons in n-type.
Answer:
(D) Holes in p-type and electrons in n-type.

Question 10.
When a hole is produced in P-type semiconductor, there is ……………….
(A) extra electron in valence band.
(B) extra electron in conduction band.
(C) missing electron in valence band.
(D) missing electron in conduction band.
Answer:
(C) missing electron in valence band.

Question 11.
The number of bonds formed in p-type and n-type semiconductors are respectively
(A) 4,5
(B) 3,4
(C) 4,3
(D) 5,4
Answer:
(B) 3,4

Question 12.
The movement of a hole is brought about by the valency being filled by a ………………..
(A) free electrons
(B) valence electrons
(C) positive ions
(D) negative ions
Answer:
(B) valence electrons

Question 13.
The drift current in a p-n junction is
(A) from the p region to n region.
(B) from the n region to p region.
(C) from n to p region if the junction is forward biased and from p to n region if the junction is reverse biased.
(D) from p to n region if the junction is forward biased and from n to p region if the junction is reverse biased.
Answer:
(B) from the n region to p region.

Question 14.
If a p-n junction diode is not connected to any circuit, then
(A) the potential is same everywhere.
(B) potential is not same and n-type side has lower potential than p-type side.
(C) there is an electric field at junction direction from p-type side to n-type side.
(D) there is an electric field at the junction directed from n-type side to p-type side.
Answer:
(D) there is an electric field at the junction directed from n-type side to p-type side.

Question 15.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(A) free electrons in the n-region attract them.
(B) they move across the junction by the potential difference.
(C) hole concentration in p-region is more as compared to n-region.
(D) all the above.
Answer:
(C) hole concentration in p-region is more as compared to n-region.

Question 16.
The width of depletion region ……………
(A) becomes small in forward bias of diode
(B) becomes large in forward bias of diode
(C) is not affected upon by the bias
(D) becomes small in reverse bias of diode
Answer:
(A) becomes small in forward bias of diode

Question 17.
For p-n junction in reverse bias, which of the following is true?
(A) There is no current through P-N junction due to majority carriers from both regions.
(B) Width of potential barriers is small and it offers low resistance.
(C) Current is due to majority carriers.
(D) Both (B) and (C)
Answer:
(A) There is no current through P-N junction due to majority carriers from both regions.

Question 18.
In the circuit shown below Di and D2 are two silicon diodes. The current in the circuit is …………….

(A) 2 A
(B) 2 mA
(C) 0.8 mA
(D) very small (approx 0)
Answer:
(D) very small (approx 0)

Question 19.
For an ideal junction diode,
(A) forward bias resistance is infinity.
(B) forward bias resistance is zero.
(C) reverse bias resistance is infinity.
(D) both (B) and (C).
Answer:
(D) both (B) and (C).

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound

Question 1.
State the different types of waves.
Answer:

1. Waves which require a material medium for their propagation are called mechanical waves. Example: Sound waves, string waves, seismic waves, etc.
2. Waves which do not require material medium for their propagation are called electromagnetic waves. Example: Light waves, radio waves, y-rays, etc.
3. The wave associated with any object when it is in motion is called as matter wave.
4. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
5. Waves are also of stationary type.

Question 2.
Define the following terms. Give their SI units.
i. Period
ii. Frequency
iii. Velocity
Answer:
i. Period (T):
The time taken by the particle of a medium to complete one vibration is called period of the wave.
SI unit: second (s)

ii. Frequency (n):
The number of vibrations performed by a particle per second is called frequency of a wave.
SI unit: hertz (Hz)

iii. Velocity (v):
The distance covered by a wave per unit time is called the velocity of the wave.
SI unit: m/s

Question 3.
State the properties that should be possessed by a medium for a mechanical wave to propagate through it.
Answer:

1. The medium must be continuous and elastic so that it can regain its original state as soon as the deforming forces are removed.
2. The medium should possess inertia. It must be capable of storing energy and transferring it in the form of waves.
3. The frictional resistance of the medium should be negligible to avoid damped oscillations.

Question 4.
What are two types of progressive waves? State two characteristics of progressive waves.
Answer:
Progressive waves are classified into two types:
a. Transverse progressive waves
b. Longitudinal progressive waves.

Characteristics of progressive waves:
1. All the vibrating particles of medium have same amplitude, period and frequency.
2.. State of oscillation i.e., phase changes from particle to particle.

Question 5.
A violin string emits sound of frequency 510 Hz. How far will the sound waves reach when string completes 250 vibrations? The velocity of sound is 340 m/s.
Answer:
Given: n = 510 Hz, v = 340 m/s,
number of vibrations = 250
To find: Distance
Formula: v = nλ
Calculation:
From formula,
λ = \(\frac {v}{n}\) = \(\frac {340}{510}\) = \(\frac {2}{3}\) m
Distance covered in 1 vibration = \(\frac {2}{3}\) m
∴ Distance covered in 250 vibration
= \(\frac {2}{3}\) × 250 = 166.7 m 3
Answer: The distance covered by sound waves is 166.7 m

Question 6.
The speed of sound in air is 330 m/s and that in glass is 4500 m/s. What is the ratio of the wavelength of sound of a given frequency in the two media?
Answer:
Given: v_air = 330 m/s, v_glass = 4500 m/s
To find: \(\frac {λ_{air}}{λ_{glass}}\)
Formula: v = nλ
Calculation: From formula,
v_air = n λ_air
v_glass = n λ_glass
∴ \(\frac {λ_{air}}{λ_{glass}}\) = \(\frac {v_{air}}{v_{glass}}\) = \(\frac {330}{4500}\) = 7.33 × 10^-2

Question 7.
The velocity of sound in gas is 498 m/s and in air is 332 m/s. What is the ratio of wavelength of sound waves in gas to air?
Answer:
v_g = 498 m/s, v_a = 332 m/s
To find: Ratio of wavelengths (\(\frac {λ_g}{λ_a}\))
Formula: v = nλ
Calculation:
Frequency of sound wave remains same.
From formula,
For gas λ_g = \(\frac {v_g}{n}\) and for air λ_ag = \(\frac {v_a}{n}\)
∴ \(\frac {v_g}{v_a}\) = \(\frac {v_g}{v_a}\) = \(\frac {498}{332}\) = \(\frac {3}{2}\)
∴ \(\frac {v_g}{v_a}\) = 3 : 2

Question 8.
A human ear responds to sound waves of frequencies in the range of 20 Hz to 20 kHz. What are the corresponding wavelengths, if the speed of sound in air is 330 m/s? Answer:
Given: v₁ = v_g = 330 m/s, n₁ = 20 Hz,
n₂ = 20 kHz = 20 × 10³ Hz
To find: Wavelength (λ₁ and λ₂)
Formula: v = nλ
Calculation:
From formula,
λ₁ = \(\frac {v_1}{n_1}\) = \(\frac {330}{20}\) = 16.5 m
λ₂ = \(\frac {v_2}{n_2}\) = \(\frac {330}{20×10^3}\) = 16.5 × 10^-3 = 0.0165 m

Question 9.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (i) the reflected sound, (ii) the transmitted sound? Speed of sound in air is 340 m s^-1 and in water is 1486 m s^-1
Answer:
Given: n = 1000 kHz = 10⁶ Hz,
v_a = 340 m/s,
v_w = 1486 m/s
To find: Reflected wavelength (λ_R),
Transmitted wavelength (λ_T),
Formula: v = nλ
Calculation:
In different medium, frequency of sound wave remains same.
From formula,
Sound is reflected in air,
i. ∴ λ_R = \(\frac {v_a}{n}\) = \(\frac {330}{10^6}\) = 3.4 × 10^-4 m
Sound is transmitted in water,
ii. ∴ λ_T = \(\frac {v_w}{n}\) = \(\frac {1486}{10^6}\) = 1.486 × 10^-3 m

Question 10.
The wavelength of a sound note is 1 m in air and 2.5 m in a liquid. Find the speed of sound in the liquid, if the speed of the sound in air is 330 m/s.
Answer:
Given: λ_a = 1 m, λ_l = 2.5 m, v_a = 330 m/s,
To find: Speed of sound (v_l)
Formula: v = nλ
Calculation:
From formula,
Since the frequency n remains the same,
v_a = nλ_a and v_l = nλ_l
∴ \(\frac {v_l}{v_a}\) = \(\frac {λ_l}{λ_a}\)
∴ v_l = v_a \(\frac {λ_l}{λ_a}\) = 330 × \(\frac {2.5}{1}\) = 825 m/s

Question 11.
Define a polarised wave.
Answer:
A wave in which the vibrations of all the particles along the path of a wave are constrained to a single plane is called a polarised wave.

Question 12.
Write down the main characteristics of longitudinal waves.
Answer:
Characteristics of longitudinal waves:

1. All the particles of medium in the path of wave vibrate in a direction parallel to the direction of propagation of wave with same period and amplitude.
2. When longitudinal wave passes through a medium, the medium is divided into alternate compressions (high pressure zone) and rarefactions (low pressure zone).
3. A compression and adjacent rarefaction form one cycle of a longitudinal wave. The distance between any two consecutive points having same phase (successive compression or rarefactions) is called wavelength of the wave.
4. For propagation of longitudinal waves, the medium should possess the property of elasticity of volume (bulk modulus). Thus, longitudinal waves can travel through solids, liquids and gases. Longitudinal wave cannot travel through vacuum or empty space.
5. The compressions and rarefactions advance in the medium and are responsible for transfer of energy.
6. When longitudinal wave advances through medium, there is periodic variations in pressure and density along the path of wave and with time.
7. Since the direction of vibration of particles and direction of propagation of wave are same or parallel, longitudinal waves cannot be polarised.

Question 13.
State the mathematical expression for a transverse progressive wave travelling along the positive and negative x-axis.
Answer:
i. Consider a transverse progressive wave whose particle position is described by x and displacement from equilibrium position is described by y.
Such a sinusoidal wave can be written as follows:
∴ y (x, t) = a sin (kx – ωt + ø) ……… (1)
where a, k, ω and ø are constants,
y (x, t) = displacement as a function of position (x) and time (t)
a = amplitude of the wave,
ω = angular frequency of the wave
(kx₀ – ωt + ø) = argument of the sinusoidal wave and is the phase of the particle at x at time t.

ii. At a particular instant, t = t₀,
y (x, t₀) = a sin (kx – ωt₀ + ø)
= a sin (kx + constant)
Thus at t = t₀, shape of wave as a function of x is a sine wave.

iii. At a fixed location x = x₀
y(x₀, t) = a sin (kx₀ – ωt + ø)
= a sin (constant – ωt)
Hence the displacement y, at x = x₀ varies as a sine function.

iv. This means that the particles of the medium, through which the wave travels, execute simple harmonic motion around their equilibrium position.

v. For (kx – ωt + ø) to remain constant, x must increase in the positive direction as time t increases. Thus, the equation (1) represents a wave travelling along the positive x axis.

vi. Similarly, a wave travelling in the direction of the negative x axis is represented by,
y(x, t) = a sin (kx + ωt +ø) …….(2)

Question 14.
Explain the Laplace’s correction to the Newton’s formula for the velocity of sound in air.
Answer:
Laplace’s correction:
Laplace suggested that the compression or rarefaction takes place too rapidly. Heat produced during compression and lost during rarefaction does not get sufficient time for dissipation. Due to this, the whole heat content remains same. Thus, it is an adiabatic process.

According to Laplace, E is the adiabatic modulus of elasticity of air medium which is given by,
E = γP ….(1)
where P = pressure of the air medium γ = ratio of specific heat of air at constant pressure (c_p) and specific heat of air at constant volume (c_v). i.e., γ = c_p/c_v.

iii. Using equation, v = \(\sqrt{\frac {E}{ρ}}\), we have velocity of sound in air,
v = \(\sqrt{\frac {γP}{ρ}}\), …. [From equation (1)]
For air, γ = 1.41
At NTP, P = 0.76 × 13600 × 9.8 N/m²
ρ = 1.293 kg/m³.
∴ v = \(\sqrt{\frac {1.41×0.76×13600×9.8}{1.293}}\) = 332.35 m/s
This value is in close agreement with the experimental value.

Question 15.
What is the effect of temperature on the velocity of sound in air?
Answer:
Effect of temperature on velocity of sound:
i. Let v₀ and v be the velocity of sound in air at T₀ and T Kelvin respectively. Let ρ₀ and p be the densities of gas at temperature T0 and T respectively.

ii. Considering the number of moles n = 1 for the gas, we have,

iii. From above formula, we can conclude that velocity of sound in air increases with increase in temperature.

Question 16.
Show that for 1 °C rise in temperature, the velocity of sound in air increases by 0.61 m/s.
Answer:
Let v₀ = velocity of sound at 0 °C or 273 K
v = velocity of sound at t °C or (273 + 1) K
we have, v ∝ √T

Hence, velocity of sound increases by 0.61 m/s when temperature increases by 1 °C.

Question 17.
Suppose you are listening to an out-door live concert sitting at a distance of 150 m from the speakers. Your friend is listening to the live broadcast of the concert in another country and the radio signal has to travel 3000 km to reach him. Who will hear the music first and what will be the time difference between the two? Velocity of light = 3 × 10⁸ m/s and that of sound is 330 m/s.
Answer:
Distance between me and the speakers
(s₁) = 150 m, distance radio signal has to travel (S₂) = 3000 km, v_sound 330 m/s, v_light = 3 × 10⁸ m/s
Time taken by sound to reach me,
= \(\frac {s_1}{v_sound}\) = \(\frac {150}{330}\) = 0.4546 s
Time taken by the broadcasted sound (done by
EM waves = \(\frac {s_2}{v_light}\) = \(\frac {3000km}{30×10^5km/s}\) = \(\frac {3×1^30}{3×10^5}\) = 10^-2 s
∴My friend will hear the sound first.
The time difference will be = 0.4546 – 0.01
= 0.4446 s.

Question 18.
Consider a closed box of rigid walls so that the density’ of the air inside it is constant. On heating, the pressure of this enclosed air is increased from P₀ to P. It is now observed that sound travels 1.5 times faster than at pressure P₀. Calculate P/P₀.
Answer:
Given: v_P = 1.5 v_P0
To find: Ratio of pressure (\(\frac {p}{p_0}\))

Question 19.
The densities of nitrogen and oxygen at NTP are 1.25 kg/m³ and 1.43 kg/m³ respectively. If the speed of sound in oxygen at NTP is 320 m/s, calculate the speed in nitrogen, under the same conditions of temperature and pressure, (γ for both the gases is 1.4)
Answer:
Given: ρ_N = 1.25 kg/m³, ρ = 1.43 kg/m³,
v₀ = 320 m/s,
To find: Speed in nitrogen (v_N)
Formula: v = (\(\sqrt{\frac {γP}{ρ}}\) )
Calculation: From formula,

Question 20.
Find the temperature at which the velocity of sound in air will be 1.5 times its velocity at 0 °C
Answer:
Given: \(\frac {p}{p_0}\) = 1.5, T₀ = 0 °C = 273 K
To find: Temperature (T)
Formula: \(\frac {v}{v_0}\) = \(\sqrt{\frac {T}{T_0}}\)
Calculation:
From formula,
\(\frac {T}{T_0}\) = (\(\frac {v}{v_0}\))²
∴ T = T₀ (\(\frac {v}{v_0}\))²
∴ T = 273 (1.5)² = 614.25 K = 341.25 °C

Question 21.
The velocity of sound in air at 27 °C is 340 m/s. Calculate the velocity of sound in air at 127 °C.
Answer:
Given: T₁ = 27 °C = 27 + 273 = 300 K,
v₁ = 340 m/s,
T₂ = 127 °C = 127 + 273 = 400 K
To find: Velocity (v₂)
Formula: \(\frac {v_1}{v_2}\) = \(\sqrt{\frac {T_1}{T_2}}\)
Calculation: From formula,
v₂ = v₁ \(\sqrt{\frac {T_2}{T_1}}\) = 340, \(\sqrt{\frac {400}{300}}\)
= 340 × 1.1547
∴ v₂ = 392.6 m/s

Question 22.
The wavelength of a note is 27 m in air when the temperature is 27 °C. What is the wavelength when the temperature is increased to 37 °C?
Answer:
Given: λ₁ = 27 m,
T₁ = 27 °C = 273 + 27 = 300 K,
T₂ = 37 °C = 273 + 37 = 310 K
To find: Wavelength (λ₂)

Question 23.
We cannot hear an echo at every place. Give reason.
Answer:

1. Echo of sound depends upon the temperature of the surrounding and distance between source and reflecting surface.
2. To hear a distinct echo at 22 °C, the minimum distance required between the source of sound and reflecting surface should be 17.2 metre.
3. The velocity of sound depends on the temperature of air. Thus, the minimum distance will change with temperature. Hence, we cannot hear an echo at every place.

Question 24.
Write a short note on reverberation.
Answer:

1. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
2. Sound wave gets reflected multiple times if the distance between reflecting surface and source of sound is less than 15 m.
3. During reverberation, the time interval between the successive reflections of a sound is small.
4. As a result, the reflected sound waves overlap and produce a continuously increasing loud sound which is at times difficult to understand. Measures to decrease reverberation:
5. Reverberation can be decreased by making the walls and roofs rough and by using curtains in the hall to avoid reflection of sound.
6. Chairs and wall surfaces should be covered with sound absorbing materials.
7. Porous cardboard sheets, perforated acoustic tiles, gypsum boards, thick curtains etc. should be used on the ceilings and walls.

Question 25.
Define acoustics.
Answer:
The branch of physics which deals with the study of production, transmission and reception of sound is called acoustics.

Question 26.
State the conditions that must be satisfied for proper acoustics in an auditorium along-with their remedies.
Answer:
i. Acoustics of an auditorium should be such that the sound is heard sufficiently loudly at all the points in the auditorium. The surface behind the speaker should be parabolic with the speaker at its focus for uniform distribution of sound in the auditorium. Reflection of sound helps to maintain good loudness through the entire auditorium.

ii. Echoes and reverberations should be reduced. More absorptive reflecting surfaces and full auditoriums help in reducing echoes.

iii. Unnecessary focusing of sound, poor audibility zone or region of silence should be avoided. Curved surface of the wall or ceiling should be avoided for this purpose.

iv. Echelon effect which arises due to the mixing of sound produced in the hall by the echoes of sound produced in front of regular structure like stairs should be reduced. Stair type construction in the hall must be avoided for this purpose.

v. To avoid outside stray sound from entering, the auditorium should be sound-proof when closed.

vi. Inside fittings, seats, etc. should not produce any sound for proper acoustics. Air conditioners instead of fans and soft action door closers should be used.

Question 27.
State the applications of acoustics observed in nature.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Question 28.
State the medical applications of acoustics.
Answer:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Question 29.
State the underwater applications of acoustics.
Answer:

1. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
2. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
3. Motion and position of submerged objects like submarine can be measured with the help of this system.

Question 30.
State the applications of acoustics in environmental and geological studies.
Answer:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

ii. Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics.

iii. This is useful in studying the properties of the Earth. Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 31.
A man shouts loudly close to a high wall. He hears an echo. If the man is at 40 m from the wall, how long after the shout will the echo be heard? (speed of sound in air = 330 m/s)
Answer:
Given s = 40m, v = 330 m/s
To Find: time (t)
Formula: Time = distance \(\frac {distence}{speed}\)
Calculation:
The distance travelled by the sound wave
= 2 × distance from man to wall.
= 2 × 40 = 80 m.
From formula,
∴ Time taken to travel the distance
\(\frac {distence}{speed}\) = \(\frac {80}{30}\) = 0.24 s

Question 32.
Write a short note on pitch of sound note.
Answer:

1. Pitch refers to the sharpness or shrillness of sound.
2. Increase in frequency of sound results in increase in the pitch and the sound is said to be sharper.
3. Tone refers to a single frequency of a wave.
4. A note may contain single or multiple tones.
5. High frequency is generally referred as high pitch or high tone.
6. Generally, speech of the men is of low pitch (shrill) and that of the women is of high pitch (sharp). Tones of an acoustic guitar are sharper than that of a base guitar. Sound of table is sharper than that of a dagga.

Question 33.
Write a short note on quality (timbre) of sound note.
Answer:
i. Timbre of a sound refers to the quality of the sound which depends upon the mixture of tones and overtones in the sound. Same sound played on different musical instruments feels significantly different and the musical instrument from which the sound generated can be easily identified.

Question 34.
Write a short note on loudness of sound.
OR
Explain how loudness affects the characteristics of sound.
Answer:
Loudness:
i. Loudness depends upon the intensity of vibration.

ii. Intensity of a wave is proportional to square of the amplitude (I ∝ A²) and is measured in the (SI) unit ofW/m²

iii. The human response to intensity is not linear, i.e., a sound of double intensity is louder but not doubly loud.

iv. Under ideal conditions, for a perfectly healthy human ear, the least audible intensity is I₀ = 10^-12 W/m².

v. Loudness of a sound of intensity I (measured in unit bel) is given by,
L₂ = log₁₀ (\(\frac {I}{I_0}\)) ………….. (1)

vi. Decibel is the commonly used unit for loudness.

vii. As, 1 decibel or 1 dB = 0.1 bel.
∴ 1 bel = 10 dB. Thus, loudness in dB is 10 times loudness in bel.
∴ L_dB = 10L_bel = 10 log₁₀ (\(\frac {I}{I_0}\))
For sound of least audible intensity I₀
L_dB = 10 log₁₀ (\(\frac {I_0}{I_0}\)) = 10 log₁₀ (1) = 0 ………… (2)
This corresponds to threshold of hearing.

viii. For sound of 10 dB,
10 = 10 log₁₀ (\(\frac {I}{I_0}\))
∴ (\(\frac {I}{I_0}\)) = 10 ¹ or I = 10 I₀
For sound of 20 dB,
20 = 10 log₁₀ (\(\frac {I}{I_0}\))
= (\(\frac {I}{I_0}\)) = 10² or I = 100 I₀ and so on.

ix. This implies, loudness of 20 dB sound is felt double that of 10 dB, but its intensity is 10 times that of the 10 dB sound. Similarly, 40 dB sound is left twice as loud as 20 dB sound but its intensity is 100 times as that of 20 dB sound and 10000 times that of 10 dB sound. This is the power of logarithmic or exponential scale. If we move away from a (practically) point source, the intensity of its sound varies inversely with square of the distance, i.e., I ∝ \(\frac {1}{r^2}\).

Question 35.
When heard independently, two sound waves produce sensations of 60 dB and 55 dB respectively. How much will the sensation be if those are sounded together, perfectly in phase?
Answer:
L₁ = 60 dB = 10 log₁₀ \(\frac {I_1}{I_0}\)
∴ \(\frac {I_1}{I_0}\) = 10⁶
∴ I₁ = 10⁶I₀
Similarly, I₂ = 10^5.5 I₀
As the waves combine perfectly in phase, the vector addition of their amplitudes will be given by A² = (A₁ + A₂)² = A\(_1^2\) + A\(_2^2\) + 2A₁, A₂ As intensity is proportional to square of the amplitude.
∴ I = I₁ + I₂ + 2\(\sqrt {I_1I_2}\) = 10⁵ I₀ (10¹ +10^0.5 + 2\(\sqrt {10^{1.5}}\))
= 10⁵I₀(10 + 3.1623 +2 × 10^0.75)
= 24.41 × 10⁵I₀ = 2.441 × 10⁶I₀
∴ L = 10 log₁₀ (\(\frac {I}{I_0}\)) = 10 log₁₀ (2.441 × 10⁶)
= 10[log₁₀ (2.441) + log₁₀(10⁶)]
= 10(0.3876 + 6)
L = 63.876 dB ~ 64 dB

Question 36.
The noise level in a class-room in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increases to 100?
Answer:
Loudness of sound is given as,

∴ L_B – L_A = 0.301 × 10 = 3.01
∴ L_B = L_A + 3.01 = 53.01 dB

Question 37.
Calculate the decibel increase if there is a two-fold increase in the intensity of a wave. (Given: log₁₀ 2 = 0.3010)
Answer:
L = 10 log₁₀ \(\frac {I}{I_0}\) decibel
L’ = 10 log₁₀ \(\frac {2I}{I_0}\) decibel
L’ – L = 10 (log₁₀ \(\frac {2I}{I_0}\) – log₁₀ (\(\frac {I}{I_0}\))
= 10 log₁₀ 2
= 10 × 0.3010
∴ L’ – L = 3.01 dB

Question 38.
Derive the expression for apparent frequency when listener is stationary and source is moving away from the listener.
Answer:

i. Consider a source of sound S moving away from a stationary listener L with velocity v_s. Let the speed of sound with respect to the medium be v (always positive). The listener uses a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point Si which is at a distance d from the listener, emit a crest. This crest reaches the listener at time t₁, given as, t₁ = d/v. …………(1)

iii. Let T₀ be the time period at which the waves are emitted.
At t = T₀, distance travelled by the source away from the stationary listener to reach point S₂ = v_sT₀.
∴ Distance of point S₂ from the listener = d + v_sT₀.
At S₂, The source emits second crest. This crest reaches the listener at t₂, given as,
t₂ = T₀ + (\(\frac {d+v_sT_0}{v}\)) …………. (2)

iv. Similarly, the time taken by the (p+1)^th crest (where, p is an integer, p = 1, 2, 3,…), emitted by the source at time pT₀, to reach the listener is given as,
t_p+1 = pT₀ + (\(\frac {d+pv_sT_0}{v}\)) …………. (3)
∴ the listener’s detector counts p crests in the time interval,
t_p+1 – t₁ = pT₀ + (\(\frac {d+pv_sT_0}{v}\)) – \(\frac {d}{v}\)
The period of wave as recorded by the listener is,

Where, n = frequency recorded by the listener (apparent frequency)
n₀ = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving away from the listener.

Question 39.
Derive an expression for apparent frequency when listener is stationary and source is moving towards the listener.
Answer:

i. Consider a source of sound S moving towards a stationary listener L with velocity v_s. Let the speed of sound with respect to the medium be v (always positive). The listener use a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point S₁ which is at a distance d from the listener, emit a Crest. This crest reaches the listener at time t₁, given as,
∴ t₁ = d/v. ……….(1)

iii. Let T₀ be the time period at which the waves are emitted.
At t = T₀, distance travelled by the source away from the stationary listener to reach point S₂ = v_sT₀.
Distance of S₂ from the listener = d – v_sT₀.
At S₂, The source emits second crest. This crest reaches the listener at
t₂ = T₀ + (\(\frac {d-v_sT_0}{v}\)) ………….. (2)

iv. Similarly, the time taken by the (p+1)^th crest (where, p = 1,2,3,…), emitted by the source at time pT₀, to reach the listener is given as,
t_p+1 = pT₀ + (\(\frac {d-pv_sT_0}{v}\)) ……………. (3)
∴ the listener’s detector counts p crests in the time interval,
t_p+1 – t₁ = pT₀ + (\(\frac {d-pv_sT_0}{v}\)) – \(\frac {d}{v}\)
∴ the period of wave as recorded by the listener is,

Where, n = frequency recorded by the listener (apparent frequency)
n₀ = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving towards the listener.

Question 40.
Derive the expression for apparent frequency when the source is stationary and the listener is moving towards the source.
Answer:

i. Consider a listener approaching a stationary source S with velocity v_L as shown in figure. Let the speed of sound with respect to the medium be v (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L₁ is at an initial distance d from the source.
At time t = t₁ the listener receives the first wave at the position L₂.
Distance travelled by the listener towards the stationary source during time t₁ = v_Lt₁.
Distance travelled by the sound wave during time t₁ = d – v_Lt₁
∴ time taken by the sound wave to travel this distance, t₁ = \(\frac {d-v_Lt_1}{v}\)
∴ t₁ = \(\frac {d}{v+v_L}\) ………….. (1)

iii. Let at time t = T₀ (time period of the waves emitted by the source), the source emits a second wave.
At t = t₂, the listener receives the second wave. Distance travelled by the listener towards the stationary source during time t₂ = v_Lt₂.
Distance travelled by the sound wave during time t₂ = d – v_Lt₂
∴ time taken by the sound wave to travel this distance = \(\frac {d-v_Lt_2}{v}\)
However, this time should be counted after T₀, as the second wave was emitted at t = T₀.
∴ t₂ = T₀ + \(\frac {d-v_Lt_2}{v}\)
∴ t₂ = \(\frac {vT_0+d}{v+v_L}\) …………. (2)

iv. Similarly, for the third wave, we get,
t₃ = 2T₀ + \(\frac {d-v_Lt_3}{v}\)
∴ t₃ = \(\frac {2vT_0+d}{v+v_L}\) …………. (3)

v. Extending this argument to the (p + 1)^th wave, we get,
t_p+1 = pT₀ + \(\frac {d-v_Lt_{p+1}}{v}\)
∴ t_p+1 = \(\frac {pvT_0+d}{v+v_L}\) …………. (4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.

This is the expression for apparent frequency when the source is stationary and the listener is moving towards the source.

Question 41.
Derive the expression for apparent frequency when the source is stationary and the listener is moving away from the source.
Answer:

i. Consider a listener moving away a stationary source S with velocity V_L. Let the speed of sound with respect to the medium be y (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L₁ is at an initial distance d from the source.
At time t = t₁ the listener receives the first wave at the position L₂.
Distance travelled by the listener away from the stationary source during time t₁ = v_Lt₁.
Distance travelled by the sound wave during time t₁ = d + v_Lt₁
∴ time taken by the sound wave to travel this distance,
t₁ = \(\frac {d+v_Lt_1}{v}\)
∴ t₁ = \(\frac {d}{v-v_L}\) ………….. (1)

iii. Let at time t = T₀ (time period of the waves emitted by the source), the source emits a second wave.
At t = t₂, the listener receives the second wave. Distance travelled by the listener away from the stationary source during time t₂ = v_Lt₂.
∴ Distance travelled by the sound wave during time t₂ = d + v_Lt₂.
∴ time taken by the sound wave to travel this distance = \(\frac {d+v_Lt_2}{v}\)
However, this time should be counted after T₀, as the second wave was emitted at t = T₀.
∴ t₂ = T₀ \(\frac {d+v_Lt_2}{v}\) ………….. (2)
∴ t₂ = \(\frac {vT_0+d}{v-v_L}\)

iv. Similarly for the third wave, we get
t₃ = 2T₀ \(\frac {d+v_Lt_3}{v}\)
∴ t₃ = \(\frac {2vT_0+d}{v-v_L}\) …………..(3)

v. Extending this argument to the (p + 1)^th wave, we get,
t_p+1 = pT₀ + \(\frac {d+v_Lt_{p+1}}{v}\)
∴ t_p+1 = \(\frac {pvT_0+d}{v-v_L}\) …………..(4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.

This is the expression for apparent frequency when the source is stationary and the listener is moving away from the source.

Question 42.
State the common properties between Doppler effect of sound and light.
Answer:
i. The recorded frequency is different than the emitted frequency in case of relative motion between listener (or observer) and source (of sound or light waves).

ii. In case of relative approach, recorded frequency > emitted frequency.

iii. In case of relative recede, recorded frequency < emitted frequency.

iv. For values of listener velocity (v_L) or source velocity (v_s) much smaller then wave speed (speed of sound or light).
n = n₀ (1±\(\frac {v_r}{v}\))
Where, vr = relative velocity
n = actual frequency of the source
n₀ = apparent frequency of the source
v = velocity of sound in air.
(upper sign is used during relative approach and lower sign is during relative recede.)

v. If velocities of source and observer (listener) are along different lines, their respective components along the line joining them should be chosen for longitudinal Doppler effect and the same mathematical treatment is applicable.

Question 43.
State the major difference between Doppler effects of sound and light.
Answer:

1. Speed of light being absolute, only relative velocity between the observer and the source matter irrespective of who is in motion. However, for obtaining exact Doppler shift for sound waves, it is absolutely important to know who is in motion.
2. In case of light, classical and relativistic Doppler effects are different while sound only has classical doppler effect.
3. Presence of wind affects the velocity of sound which affects the Doppler shift in sound.

Question 44.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s^-1 (b) recedes from the platform with a speed of 10 m s^-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given: v_s = 10 m/s, v = 340 m/s, n₀ = 400 Hz
Apparent frequency (n), velocity of sound (v_s) in each case
Formulae:
i. n = n₀ (\(\frac {v}{v-v_s}\))
ii. n = n₀ \(\frac {v}{v+v_s}\)
Calculation:
a. As the train approaches the platform, using formula (i),
n = 400 (\(\frac {340}{340-10}\)) = 421.12 Hz

b. As the train recedes from the platform, using formula (ii),
n = 400 (\(\frac {340}{340+10}\)) = 388.57 Hz

ii. The relative motion of source and observer results in the apparent change in the frequency but has no effect on the speed of sound. Hence, the speed of sound remains unchanged in both the cases.

Question 45.
A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the observer with the speed of 72 km/hour. (Speed of sound in air = 340 m/s)
Answer:
Given: v_s = 72 km/ hr = 20 m/s, n₀ = 640 Hz,
v = 340 m/s
To find: Difference in apparent frequencies
(n_A – n’_A)
Formulae:
i. When the train moves towards the stationary observer then,
n_A = n₀ (\(\frac {v}{v-v_s}\))
ii. When the train moves away the stationary observer then,
n’_A = n₀ (\(\frac {v}{v+v_s}\))
Calculation: From formula (i),
n_A = 640 (\(\frac {340}{340-20}\))
∴ n_A = 680 Hz
From formula (ii),
n’_A = 640 (\(\frac {340}{340+20}\))
∴ n’_A = 604.4 Hz
Difference between n_A and n’_A
= n_A – n’_A = 75.6 Hz

Question 46.
The speed limit for a vehicle on road is 120 km/hr. A policeman detects a drop of 10% in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? (Given: Velocity of sound=340 m/s).
Answer:
Given: Speed limit, v_L = 120 km/hr
n’_A = n_A – \(\frac {10}{100}\) n_A = 0.9 n_A
Velocity of sound, v = 340 m/s
To Find: Velocity of source (v_s)
i. n_A = (\(\frac {v}{v-v_s}\))n
ii. n’_A = (\(\frac {v}{v+v_s}\))n
Calculation:
Dividing formula (i) by (ii),

Question 47.
A stationary source produces a note of frequency 350 Hz. An observer in a car moving towards the source measures the frequency of sound as 370 Hz. Find the speed of the car. What will be the frequency of sound as measured by the observer in the car if the car moves away from the source at the same speed? (Assume speed of sound = 340 m/s)
Answer:
Given: n₀ = 350 Hz, v = 340 m/s,
n_A = 370 Hz
To find: Speed (v_L), Frequency (n_A)
Formulae:
i. When the car moves towards the stationary source then,
n_A = n₀ (\(\frac {v+v_s}{v}\))

ii. When the car moves away from the stationary source then,
n_A = n₀ (\(\frac {v-v_L}{v}\))
Calculation: From formula (i),
370 = 35o (\(\frac {340+v_L}{340}\))
∴ 359.43 =340 +v_L
∴ v_L = 19.43 m/s
From formula (ii),
∴ n_A = 35o (\(\frac {340-20}{340}\)) = \(\frac {35}{34}\) × 320
∴ n_A = 329.41 Hz

Question 48.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Blowing of wind changes the velocity of sound. As the wind is blowing in the direction of sound, effective speed of sound v_e = v + v_w = 340 + 10 = 350 m/s
As the source and listener both are at rest, frequency is unchanged, i.e., n = 400 Hz.
∴ wavelength, λ = \(\frac {v_e}{n}\) = \(\frac {350}{400}\) = 0.875 m
For still air, v_w = 0 and v_e = v = 340 m/s
Also, as observer runs towards the stationary train v_L = 10 m/s and v_s = 0
The frequency now heard by the observer,
n = n₀ (\(\frac {v+v_L}{v}\)) = 400 (\(\frac {340+10}{340}\))
= 411.76 Hz
As the source is at rest, wavelength does not change i.e., λ’ = λ = 0.875 m
Comparing the answers, it can be stated that, the situations in two cases are different.

Question 49.
A SONAR system fixed in a submarine operates at a frequency 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Frequency of SONAR (source)
n = 40 kHz = 40 × 10³ Hz
Speed of sound waves, v = 1450 m s^-1
Speed of the listener, v_L = 360 km h^-1
= 360 × \(\frac {5}{18}\) ms^-1
= 100 m s^-1
Since, the source is at rest and the observer moves towards the source (SONAR).
We have,
n = n₀ (\(\frac {v+v_L}{v}\)) = 40 × 10³ × (\(\frac {1450+100}{14540}\))
∴ n = 4.276 × 10⁴ Hz
This frequency n’ is reflected by the enemy ship and is observed by the SONAR (which now acts as observer). Therefore, in this case v_s = 100 m s^-1
Apparent frequency,
n = n₀ (\(\frac {v}{v-v_s}\))
= 4.276 × 10⁴ × (\(\frac {1450}{1450-100}\)) = 4.59 × 10⁴ Hz
∴ n = 45.9 kHz

Question 50.
A rocket is moving at a speed of 220 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1200 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (i) the frequency of the sound as detected by the target and (ii) the frequency of the echo as detected by the rocket (velocity of sound = 330 m/s)
Answer:
Given: v_s = 220 m/s, v_L = 0 m/s, n = 1200 Hz
To find: Apparent frequency (n)
i. n = n₀ (\(\frac {v}{v-v_s}\))
ii. n = n₀ \(\frac {v+v_L}{v}\)
Calculation: For first case, observer is stationary and source i.e., rocket is moving towards the target.
Hence, using formula (i),
frequency of sound as detected by the target,
n = 1200 (\(\frac {330}{330-220}\)) = 3600 Hz
For second case, target acts as a source with frequency 3600 Hz as it is the source of echo. While rocket detector acts as an observer. This means, v_s = 0 and V_L = 220 m/s
Using formula (ii),
frequency of echo as detected by the rocket,
n = 3600 (\(\frac {330+220}{330}\)) = 600 Hz

Question 51.
A bat is flying about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, frequency of sound emitted by bat,
n = 40 kHz
Velocity of bat, v_s = 0.03 v
where v is velocity of sound.
The bat is moving towards the flat wall. This is the case of source in motion and the observer at rest.
Therefore, the frequency of sound reflected at the wall is,
n = n_s (\(\frac {v}{v-v_s}\)) = n × (\(\frac {v}{v-0.03v}\))
= n × \(\frac {1}{0.97}\) = \(\frac {n}{0.97}\)
The frequency n’ is reflected by the wall and is again received by the bat moving towards the wall. This is the case of an observer moving towards the source with velocity v_L = 0.03 v.
The frequency observed by bat,

Question 52.
A bat, flying at velocity V_B = 12.5 m/s, is followed by a car running at velocity V_c = 50 m/s. Actual directions of the velocities of the car and the bat are as shown in the figure below, both being in the same horizontal plane (the plane of the figure). To detect the car, the bat radiates ultrasonic waves of frequency 36 kHz. Speed of sound at surrounding temperature is 350 m/s.

There is an ultrasonic frequency detector fitted in the car. Calculate the frequency recorded by this detector. The ultrasonic waves radiated by the bat are reflected by the car. The bat detects these waves and from the detected frequency, it knows about the speed of the car. Calculate the frequency of the reflected waves as detected by the bat. (sin 37° = cos 53° ~ 0.6, sin 53° = cos 37° ~ 0.8)
Answer:
The components of velocities of the bat and the car, along the line joining them, are
v_c cos 53° ~ 50 × 0.6 = 30 m s^-1 and
v_B cos 37° ~ 12.5 × 0.8 = 10 m s^-1

Doppler shifted frequency n = n₀ (\(\frac {v±v_1}{v±v_s}\))
upper signs to be used during approach, lower signs during recede.
Case I: Frequency radiated by the bat
n₀ = 36 × 10³ Hz,
The source (bat) is receding, while the listener (car) is approaching
v_S = v_B cos 37° = 10 m/s and
V_L = v_C cos 53° = 30 m/s
∴ Frequency detected by the detector in the car,
n = n₀ (\(\frac {v+v_L}{v+v_s}\))
∴ n = 36 × 10³ (\(\frac {350+30}{350+10}\)) = 36 × 10³ × \(\frac {38}{36}\)
∴ n = 38 × 10³ Hz = 38 kHz

Case II: The car is the source.
Emitted frequency by the car, is given as,
n₀ = 38 × 10³ Hz,
Car, the source, is approaching the listener (bat).
Now bat-the listener is receding while car the source is approaching,
∴ v_s = v_c cos 53° = 30 m/s
∴ v_L = v_B cos 37° = 10 m/s
∴ n = n₀ (\(\frac {v-v_L}{v-v_s}\))
∴ n = 38 × 10³ (\(\frac {350-10}{350-30}\)) = 38 × 10³ × \(\frac {34}{33}\)
= 39.15 × 10³ Hz
∴ n = 39.15 kHz

Question 53.
Source of sound is placed at one end of a copper bar of length 1 km. Two sounds are heard at the other end at an interval of 2.75 seconds, (speed of sound in air = 330 m/s)
i. Why do we hear two sounds?
ii. Find the velocity of sound in copper.
Answer:
i. Two sounds are heard because sound travels through air as well as through copper.

ii. In air, t₁ = \(\frac {distence}{time}\) = \(\frac {1000}{330}\) = 3.03 s
As the time interval is 2.75 seconds and sound travels faster in copper.
∴ In copper, t₂ = 3.03 – 2.75 = 0.28 s
∴ velocity of sound in copper = \(\frac {1000}{0.28}\) = 3571 m/s

Question 54.
If all the persons mentioned in the table below are listening to a match commentary on the same channel at their respective locations positioning at same distance from television, then will they hear the same line of the commentary at same instant of time? Justify your answer.

Name of a person	Location	Humidity
Aijun	Bangalore	65 %
Virendra	Hyderabad	56%
Vikas	Delhi	54%
Nilesh	Mumbai	75%

Answer:
As the order of humidity for the above locations is Mumbai > Bangalore > Hyderabad > Delhi.
As velocity of sound increases with increase in humidity, the order of velocity of sound at their respective locations is Mumbai > Bangalore > Hyderabad > Delhi.
Hence, the order of persons who would listen the line of commentary first to last is Nilesh, Arjun, Virendra, Vikas.

Question 55.
Speed of sound is greater during day than at night. True or False? Justify your answer.
Answer:
True. At night, the amount of CO₂ in atmosphere increases the density of atmosphere. Since, Speed of sound is inversely proportional to the square root of density. Hence, speed of sound is greater during day than in night.

Question 56.
Case I: During summer (33 °C), Prakash was waiting for a train at the platform, train arrived tt seconds after he heard train’s whistle.
Case II: During winter (19 °C), train arrived t₂ seconds after Prakash heard the sound of train’s whistle.
i. Will t₂ be equal to t₁? Justify your answer.
ii. Calculate the velocity of sound in both the cases.
(velocity of sound in air at 0 °C = 330 m/s)
Answer:
i. Velocity of sound is directly proportional to square root of absolute temperature.
Hence, whistle’s sound will be first heard by Prakash in summer than in winter.
Therefore, the time interval between sound and train reaching Prakash in summer will be more than in winter.
i.e.,t₁ > t₂

ii. When t = 33 °C
∴ v₁ = v₀ + 0.61t
= 330 + 0.61 × 33
∴ v₁ = 350.13 m/s
When t = 19 °C
v₂ = v₀ + 0.61t
= 330 + 0.61 × 19
v₂ = 341.59 m/s

Question 57.
You are at a large outdoor concert, seated 300 m from speaker system. The concert is also being broadcast live. Consider a listener 5000 km away who receives the broadcast. Who will hear the music first, you or listener and by what time difference? (Speed of light = 3 × 10⁸ m/s and speed of sound in air = 343 m/s)
Answer:
s₁ = 300 m,
v₁ = 343 m/s,
∴ t₁ = \(\frac {s_1}{v_1}\) = \(\frac {300}{343}\) = 0.8746 s
Now,
s₂ = 5000 km = 5 × 10⁶ m,
v₂ = c = 3 × 10⁸ m/s
∴ t₂ = \(\frac {s_2}{c}\) = \(\frac {5×10^6}{3×10^8}\) = 0.0167 s
∴ t₂ < t₁
∴ Listener will hear the music first.
Time difference = t₁ – t₂
= 0.8746 – 0.0167
= 0.8579 s
The listener will hear the music first, about 0.8579 s before the person present at the concert.

Question 58.
When a source of sound moves towards a stationary observer, then the pitch increases. Give reason.
Answer:
When a source of sound moves towards a stationary observer, then the increase in pitch is due to actual or apparent change in wavelength. When the source of sound moves towards an observer at rest, waves get compressed and the effective velocity of sound waves relative to source becomes less than the actual velocity. Hence the wavelength of sound waves an decreases which results into increase in pitch.

Question 59.
In the examples given below, state if the wave motion is transverse, longitudinal or a combination of both?

1. Light waves travelling from Sun to Earth.
2. ultrasonic waves in air produced by a vibrating quartz crystal.
3. Waves produced by a motor boat sailing in water.

Answer:

1. Light waves (from Sun to Earth) are electromagnetic waves which are transverse in nature.
2. Ultrasonic waves in air are basically sound waves of frequency greater than the audible frequencies. Therefore, these waves are longitudinal.
3. The water surface is cut laterally and pushed backwards by the propeller of motor boat. Therefore, the waves produced are a mixture of longitudinal and transverse waves.

Question 60.
Internet my friend
Answer:
https://hyperphysics.phys-astr.gsu.edu/ hbase/hframe.html
[Students are expected to visit the above mentioned website and collect more information about sound.]

Multiple Choice Questions

Question 1.
Water waves are …………….
(A) longitudinal
(B) transverse
(C) both longitudinal and transverse
(D) neither longitudinal nor transverse
Answer:
(C) both longitudinal and transverse

Question 2.
Sound travels fastest in ……………..
(A) water
(B) air
(C) steel
(D) kerosene oil
Answer:
(C) steel

Question 3.
At room temperature, velocity of sound in air at 10 atmospheric pressure and at 1 atmospheric pressure will be in the ratio ……………..
(A) 10 : 1
(B) 1 : 10
(C) 1 : 1
(D) cannot say
Answer:
(C) 1 : 1

Question 4.
In a gas, velocity of sound varies directly as ………………
(A) square root of isothermal elasticity.
(B) square of isothermal elasticity.
(C) square root of adiabatic elasticity.
(D) adiabatic elasticity.
Answer:
(C) square root of adiabatic elasticity.

Question 5.
At a given temperature, velocity of sound in oxygen and in hydrogen has the ratio …………………
(A) 4 : 1
(B) 1 : 4
(C) 1 : 1
(D) 2 : 1
Answer:
(B) 1 : 4

Question 6.
With decrease in water vapour content in air, velocity of sound …………………..
(A) increases
(B) decreases
(C) remains constant
(D) cannot say
Answer:
(B) decreases

Question 7.
The temperature at which speed of sound in air becomes double its value at 0 °C is ……………….
(A) 546 °C
(B) 819 °C
(C) 273 °C
(D) 1092 °C
Answer:
(B) 819 °C

Question 8.
The velocity of sound in air at NTP is 330 m/s. What will be its value when temperature is doubled and pressure is halved?
(A) 330 m/s
(B) 165 m/s
(C) 330 √2 m/s
(D) \(\frac {330}{√2}\) m/s
Answer:
(D) \(\frac {330}{√2}\) m/s

Question 9.
A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second have wave velocity
(A) 2.5 m/s
(B) 5.0 m/s
(C) 8.0 m/s
(D) 10.0 m/s
Answer:
(D) 10.0 m/s

Question 10.
A radio station broadcasts at 760 kHz. What is the wavelength of the station?
(A) 395 m
(B) 790 m
(C) 760 m
(D) 197.5 m
Answer:
(A) 395 m

Question 11.
If the bulk modulus of water is 2100 MPa, what is the speed of sound in water?
(A) 1450 m/s
(B) 2100 m/s
(C) 0.21 m/s
(D) 21 m/s
Answer:
(A) 1450 m/s

Question 12.
If speed of sound in air at 0°C is 331 m/s. What will be its value at 35° C?
(A) 331 m/s
(B) 366 m/s
(C) 351.6 m/s
(D) 332 m/s.
Answer:
(C) 351.6 m/s

Question 13.
For a progressive wave, in the usual notation
(A) v = λT
(B) n = \(\frac {v}{λ}\)
(C) T = λv
(D) λ = \(\frac {1}{n}\)
Answer:
(B) n = \(\frac {v}{λ}\)

Question 14.
At normal temperature, for an echo to be heard the reflecting surface should be at a minimum distance of ………………. m.
(A) 34.4
(B) 17.2
(C) 10
(D) 20
Answer:
(B) 17.2

Question 15.
In a transverse wave, are regions of negative displacement.
(A) rarefactions
(B) compressions
(C) crests
(D) troughs
Answer:
(D) troughs

Question 16.
If pressure of air gets doubled at constant temperature then velocity of sound in air ……………….
(A) gets doubled
(B) remains unchanged
(C) √2 times initial velocity
(D) becomes half
Answer:
(B) remains unchanged

Question 17.
Wave motion has ……………………
(A) single periodicity.
(B) double periodicity.
(C) only periodicity in space.
(D) only periodicity in time.
Answer:
(B) double periodicity.

Question 18.
The speed of the mechanical wave depends upon ………………
(A) elastic properties of the medium only.
(B) density of the medium only.
(C) elastic properties and density of the medium
(D) initial speed.
Answer:
(C) elastic properties and density of the medium

Question 19.
Longitudinal waves CANNOT be …………………
(A) reflected
(B) refracted
(C) scattered
(D) polarised
Answer:
(D) polarised

Question 20.
Wavelength of the transverse wave is 30 cm. If the particle at some instant has displacement 2 cm, find the displacement of the particle 15 cm away at the same instant.
(A) 2 cm
(B) 17 cm
(C) -2 cm
(D) -17 cm
Answer:
(C) -2 cm

Question 21.
The wavelength of sound in air is 1.5 m and that in liquid is 2 m. If the velocity of sound in air is 330 m/s, the velocity of sound in liquid is
(A) 330 m/s
(B) 440 m/’s
(C) 495 m/s
(D) 660 m/s
Answer:
(B) 440 m/’s

Question 22.
The velocity of sound in a gas is 340 m/s at the pressure P, what will be the velocity of the gas when only pressure is doubled and temperature same?
(A) 170 m/s
(B) 243 m/s
(C) 340 m/s
(D) 680 m/s
Answer:
(C) 340 m/s

Question 23.
Choose the correct statement.
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
(B) For 1 °C rise in temperature, velocity of sound decreases by 0.61 m/s.
(C) For 1 °C rise in temperature, velocity of sound decreases by \(\frac {1}{273}\) m/s.
(D) For 1 °C rise in temperature, velocity of sound increases by \(\frac {1}{273}\) m/s.
Answer:
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.

Question 24.
A sound note emitted from a certain source has a velocity of 300 m/s in air and 1050 m/s in water. If the wavelength of sound note in air is 2 m, the wavelength in water is …………
(A) 2 m
(B) 6 m
(C) 7 m
(D) 12 m
Answer:
(C) 7 m

Question 25.
A thunder clap was heard 6 seconds after a lightening flash was seen. If the speed of sound in air is 340 m/s at the time of observation, the distance of the listener from the thunder clap is ………………
(A) 56.6 m
(B) 346 m
(C) 1020 m
(D) 2040 m
Answer:
(D) 2040 m

Question 26.
The speed of sound in air at NTP is 330 m/s. The period of sound wave of wavelength 66 cm is …………………
(A) 0.2 s
(B) 0.1
(C) 0.1 × 10^-2 s
(D) 0.2 × 10^-2 s
Answer:
(D) 0.2 × 10^-2 s

Question 27.
If the velocity of sound in hydrogen is 1248 m/s, the velocity of sound in oxygen is [Given: M_O = 32 and M_H = 2]
(A) 1248 m/s
(B) 624 m/s
(C) 312 m/s
(D) 300 m/s
Answer:
(C) 312 m/s

Question 28.
If the source is moving away from the observer, then the apparent frequency …………..
(A) will increase
(B) will remain the same
(C) will be zero
(D) will decrease
Answer:
(D) will decrease

Question 29.
The working of SONAR is based on …………………
(A) resonance
(B) speed of a star
(C) Doppler effect
(D) speed of rotation of sun
Answer:
(C) Doppler effect

Question 30.
The formula for speed of a transverse wave on a stretched spring is ……………… (m = linear mass density, T = tension in Spring)
(A) v = \(\sqrt{\frac {m}{T}}\)
(B) v = \(\sqrt{\frac {T}{m}}\)
(C) v = (\(\frac {m}{T}\))^{\(\frac {3}{2}\)}
(D) v = (\(\frac {T}{m}\))^{\(\frac {3}{2}\)}
Answer:
(B) v = \(\sqrt{\frac {T}{m}}\)

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics

Question 1.
Express the speed of EM waves in terms of permittivity and permeability of the medium.
Answer:
In a material medium, the speed of EM waves is given by, c = \(\sqrt{\frac {1}{εµ}}\)
Where, ε = Permittivity and µ = permeability.
These constants depend on the electric and magnetic properties of the medium.

Question 2.
How can one classify commonly observed phenomena of light on the basis of nature of light?
Answer:
Commonly observed phenomena concerning light can be broadly split into three categories:

1. Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.
2. Wave optics or physical optics: Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
3. Particle nature of light: Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.

Question 3.
State the fundamental laws on which ray optics is based.
Answer:
Ray optics is based on the following fundamental laws:
i. Light travels in a straight line in a homogeneous and isotropic medium.

ii. Two or more rays can intersect at a point without affecting their paths beyond that point.

iii. Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).

iv. Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.

b. Angle of incidence (90 and angle of refraction (62) are related by Snell’s law, given by,
n₁ sin θ₁ = n₂ sin θ₂
where, n₁, n₂ = refractive indices of medium 1 and medium 2 respectively.

Question 4.
Explain Cartesian sign conventions using a graph.
Answer:
According to Cartesian sign conventions:
i. All distances are measured from the optical centre or pole.

ii. Figures should be drawn in such a way that the incident rays travel from left to right. Thus, a real object should be shown to the left and virtual object or image to the right of pole (or optical centre).

iii. X-axis can be conveniently chosen as the principal axis with origin at the pole.

iv. Distances to the left of the pole are negative and those to the right of the pole are positive.

v. Distances above the principal axis (X-axis) are positive while those below it are negative.

Question 5.
Define and represent in a neat diagram the following terms:
i. Diverging beam
ii. Converging beam
Answer:
i. A diverging beam of light corresponds to rays of light coming from real point object.

ii. A converging beam corresponds to rays of light directed to a virtual point object or image.

Question 6.
Thickness of the glass of a spectacle is 2 mm and refractive index of its glass is 1.5. Calculate time taken by light to cross this thickness. Express your answer with most convenient prefix attached to the unit ‘second’.
Answer:
Speed of light in vacuum, c = 3 × 10⁸ m/s
Given that:
Refractive index, (n_glass) = 1.5
Thickness of the glass = 2 mm
= 2 × 10sup>-3 m
∵ Re fractive index (n_glass) = \(\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in glass (v) }}\)
∵ v = \(\frac{\mathrm{c}}{\mathrm{n}_{\text {glass }}}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 10⁸ m/s
As v = \(\frac {s}{t}\)
time taken (t) to cross the thickness (s),
t = \(\frac{\mathrm{s}}{\mathrm{v}}=\frac{2 \times 10^{-3}}{2 \times 10^{8}}\) = 1 × 10^-11 s
Most convenient unit to express this small time is nano second. (1 ns = 10^-9 s)
∴ t = 0.01 × 10^-9 s = 0.01 ns

Question 7.
Explain the properties of the image formed after reflection of light from a plane surface.
Answer:

1. The image of an object kept in front of a plane reflecting surface is virtual and laterally inverted.
2. Image is of the same size as that of the object.
3. It is situated at the same distance as that of object but on the other side of the reflecting surface.

Question 8.
Explain the formula to find the number of images formed when an object is placed in between two plane mirrors inclined at an angle θ.
Answer:

1. If an object is kept between two plane mirrors inclined at an angle θ, multiple images (N) are formed due to multiple reflections from both the mirrors.
2. The number of images can be calculated using formula n = \(\frac {360}{θ}\)
3. Exact number of images seen (N) depends upon the angle between the mirrors and position of the object.
4. When n is an even integer, for all positions of the object the number of images formed are N = n – 1.
5. When n is an odd integer:
   a. For an object placed at the angle bisector of the mirrors: N = n- 1
   b. For an object placed off the angle bisector of the mirrors: N = n
6. If n is not an integer, N = m, where m is integral part of n.

Question 9.
Define radius of curvature of a spherical mirror.
Answer:
Radius of the sphere of which a mirror is a part is called as radius of curvature of the mirror.

Question 10.
What is the focal length of a spherical mirror? Give its relation with the radius of curvature.
Answer:
i. For a concave mirror focal length is the distance at which parallel incident rays converge. For a convex mirror, it is the distance from where parallel rays appear to be diverging after reflection.

ii. In case of spherical mirrors, half of radius of curvature is focal length of the mirror,
f = \(\frac {R}{2}\)

Question 11.
Show with the help of a ray diagram that focal length of convex mirror is positive while that of concave mirror is negative.
Answer:
i. According to sign conventions, the incident rays are drawn from left of the mirror to the right as shown in the ray diagrams below.

ii. As the rays incident on convex mirror appear to converge at a point on the positive side of the origin, the focal length of the convex mirror is positive.

iii. However, in case of concave mirror, the rays converge at a point on negative side of the origin, the focal length of the concave mirror is negative.

Question 12.
Give relation between focal length, object distance and image distance for a small spherical mirror.
Answer:
For a point object or for a small finite object, the focal length of a small spherical mirror is related to object distance and image distance as,
\(\frac {1}{f}\) = \(\frac {1}{v}\) + \(\frac {1}{u}\)

Question 13.
What is lateral magnification? How does it vary in different types of spherical mirrors?
Answer:
i. Ratio of linear size of image to that of the object, measured perpendicular to the principal axis, is defined as the lateral magnification.
∴ m = \(\frac {h_2}{h_1}\) = \(\frac {v}{u}\) (for spherical lenses)
m = –\(\frac {v}{u}\) (for spherical mirrors)

ii. For any position of the object, a convex mirror always forms virtual, erect and diminished image. Thus, lateral magnification for convex lens is always m < 1.

iii. In the case of a concave mirror, it depends upon the position of the object.

Question 14.
Complete the following table for a concave mirror?

Position of object	Position of image	Lateral magnification
U = ∞	v = f	m = 0
u > 2f	…………….	m < 1
u = f	V = ∞	………………
……………	|v| > |u|	m > 1
2f > u > f	………………	m > 1

Answer:

Position of object	Position of image	Lateral magnification
U = ∞	v = f	m = 0
u > 2f	2f > v > f	m < 1
u = f	V = ∞	m = ∞
u < f	|v| > |u|	m > 1
2f > u > f	v > 2f	m > 1

Question 15.
Explain with proper diagram why parabolic mirrors are preferred over spherical one.
Answer:
i. Unlike spherical shape, every point on a parabola is equidistant from a straight line and a point.

ii. Consider given parabola having RS as directrix and F as the focus. Points A, B, C on it are equidistant from line RS and point F.

iii. Hence A’A = AF, B’B = BF, C’C = CF, and so on.

iv. If rays of equal optical path converge at a point, that point is the location of real image corresponding to that beam of rays.

v. From figure, the paths A”AA’, B”BB’. C”CC’, etc., are equal paths when mirror is neglected.

vi. If the parabola ABC is a mirror then by definition of parabola the respective optical paths,
A”AF = B”BF = C”CF

vii. Thus, F is the single point focus for entire beam of rays parallel to the axis and there is no spherical aberration.
Hence, parabolic mirrors are preferred over spherical one as there is no spherical aberration.

Question 16.
A small object is kept symmetrically between two plane mirrors inclined at 38°. This angle is now gradually increased to 41°, the object being symmetrical all the time. Determine the number of images visible during the process.
Answer:
The object is kept symmetrically between two plane mirrors. This implies the object is placed at angle bisector.
Thus, for θ = 38°,
n = \(\frac {360}{38}\) = 9.47
As it is not integral, N = 9 (the integral part of n)
∴ For going from 38° to 41°, the mirrors go through angles 39° and 40°. Number of images formed will remain 9 for all angles between 38° and 40°.
For angles > 40°, the n goes on decreasing and when θ = 41°,
n = \(\frac {360}{41}\) = 8.78 i.e., N = 8

Question 17.
A thin pencil of length 20 cm is kept along the principal axis of a concave mirror of curvature 30 cm. Nearest end of the pencil is 20 cm from the pole of the mirror. What will be the size of image of the pencil?
Answer:
For the pencil kept along the principal axis and the end of the pencil closest to pole is at 20 cm,
say, u₁ = -20 cm
Flence, the other end of the stick is at distance, u₂ = (u₁ + 20) = -40 cm from pole of the mirror.
As R = -30 cm, F = \(\frac {R}{2}\) = -15 cm

∴ v₂ = -24 cm
Here, negative signs indicate that images are formed on the left of the mirror.

The length of the image formed is given by,
v = v₂ – v₁ = -24 – (-60) = 36 cm.

Question 18.
An object is placed at 15 cm from a convex mirror having radius of curvature 20 cm. Find the position and kind of image formed by it.
Answer:
Given: u = – 15 cm,
f = \(\frac {R}{2}\) = + \(\frac {20}{2}\) = + 10cm
To find: Nature and position of image (v)
Formula: \(\frac {1}{v}\) + \(\frac {1}{u}\) = \(\frac {1}{f}\)
Calculation:
From formula,
∴ \(\frac {1}{v}\) = \(\frac {1}{f}\) – \(\frac {1}{u}\)
= \(\frac {1}{+10}\) – \(\frac {1}{-15}\) = \(\frac {1}{10}\) + \(\frac {1}{15}\)
= \(\frac {2+3}{30}\) = \(\frac {5}{30}\) = \(\frac {1}{6}\)
∴ v = 6 cm

Question 19.
Prove that refractive index of a glass slab is given by the formula,
n = \(\frac {Real depth}{Apparent depth}\)
Answer:
i. Consider a plane parallel slab of a transparent medium of refractive index n.

ii. A point object O at real depth R appears to be at I at apparent depth A, when seen from outside (air).

iii. Consider incident ray OA and OB as shown in figure.

iv. Assuming i and r to be very small, we can write,
tan r = \(\frac {x}{A}\) ≈ sin r and tan i = \(\frac {x}{R}\) ≈ sin i

v. According to Snell’s law, for a ray travelling from denser medium to rarer medium,
n = \(\frac{\sin \mathrm{r}}{\sin \mathrm{i}} \approx \frac{\left(\frac{\mathrm{x}}{\mathrm{A}}\right)}{\left(\frac{\mathrm{x}}{\mathrm{D}}\right)}=\frac{\mathrm{R}}{\mathrm{A}}=\frac{\text { Real depth }}{\text { Apparent depth }}\)

Question 20.
The depth of a pond is 10 m. What is the apparent depth for a person looking normally to the water surface? n_water = 4/3.
Answer:
Given: Real depth of pond, d_real = 10 m,
n_w = \(\frac {4}{3}\)
To find: Apparent depth
Formula: n = \(\frac {Realdepth}{Apparent depth}\)
Calculation: From formula,
∴ Apparent depth = \(\frac {Realdepth}{n}\) = \(\frac {10}{(\frac{4}{3})}\)
= \(\frac {10×3}{4}\) = 7.5 m

Question 21.
A crane flying 6 m above a still, clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the water surface. How much deep should the crane immerse its beak to pick that fish?
For the fish, how much above the water surface does the crane appear? Refractive index of water = 4/3.
Answer:
For crane, apparent depth of fish = 6 cm,
Given that refractive index (n_w) = \(\frac {4}{3}\)
n_w = \(\frac {Realdepth}{Apparent depth}\)
∴ Apparent depth = \(\frac {4}{3}\) × 6 = 8 cm
Similarly, for fish, real height of crane = 6 m and
\(\frac{\mathrm{n}_{\mathrm{air}}}{\mathrm{n}_{\mathrm{w}}}=\frac{1}{\mathrm{n}_{\mathrm{w}}}=\frac{\text { Real height }}{\text { Apparent height }}\)
\(\frac {3}{4}\) = \(\frac {6}{Apparent height}\)
i.e., Apparent height = \(\frac {4×6}{3}\) = 8 m

Question 22.
Write a short note on Periscope.
Answer:
i. Instrument used to see the objects on the surface of a water body from inside of water is called periscope.
ii. It consists of two right angled prisms. The incident rays of light are reflected twice through these prisms.
iii. Total internal reflections occur inside these prisms and a clear view of the surface of water is obtained.

Question 23.
A ray of light passes from glass (n_g = 1.52) to water (n_w = 1.33). What is the critical angle of incidence?
Answer:
Given: n_g = 1.52, n_w = 1.33
To find: Critical angle (i_c)
formula: sin i_c = \(\frac {n_2}{n_1}\) = \(\frac {n_w}{n_g}\)
Calculation:
From formula,
i_c = sin^-1 (\(\frac {1.33}{1.52}\)) = sin^-1 (0.875) = 61°2′

Question 24.
There is a tiny LED bulb at the center of the bottom of a cylindrical vessel of diameter 6 cm. Height of the vessel is 4 cm. The beaker is filled completely with an optically dense liquid. The bulb is visible from any inclined position but just visible if seen along the edge of the beaker. Determine refractive index of the liquid.
Answer:

As the bulb is just visible from the edge, the angle of incidence formed by ray OP must be equal to critical angle.
∴ refractive index (n) = \(\frac {1}{sin i_c}\)
From Figure,
tan i_c = \(\frac {PQ}{OQ}\) = \(\frac {4}{3}\)
∴ sin i_c = \(\frac {OQ}{OP}\) = \(\frac {3}{5}\)
∴ n_liquid = \(\frac {5}{3}\)

Question 25.
What are convex and concave lenses? For which condition, convex lens will have negative focal length?
Answer:

1. A lens is said to be convex if it is thicker in the middle and narrowing towards the periphery. According to Cartesian sign convention, its focal length is positive.
2. Convex lens is visualized to be internal cross section of two spheres (or one sphere or a plane surface).
3. A lens is concave if it is thicker at periphery and narrows down towards centre and has negative focal length.
4. Concave lens is visualized to be external cross section of two spheres.
5. For lenses of material optically rarer than the medium in which those are kept, convex lenses will have negative focal length and they will diverge the incident rays.

Question 26.
Which lenses can be considered as thin lenses?
Answer:
Lenses for which the maximum thickness is at least 50 times smaller than all the other distances are considered as thin lenses.

Question 27.
Give the expression for the focal length of combination of lenses when
i. Lenses are kept in contact with each other
ii. Two lenses kept at a distance d apart from each other.
Answer:
i. For thin lenses kept in contact:
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}+\frac{1}{\mathrm{f}_{3}}\) + ………

ii. For two lenses kept distance d apart:
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}\)

Question 28.
An object is placed infront of a convex surface separating two media of refractive index 1.1 and 1.5. The radius of curvature is 40 cm. Where is the image formed when an object is placed at 220 cm from the refracting surface?
Solution:
Given: n₁ = 1.1, n2 = 1.5, R = + 40 cm,
u = -220 cm
To find: Position of image (v)
Formula: \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R}\)
Calculation: From formula,

Question 29.
A glass paper-weight (n = 1.5) of radius 3 cm has a tiny air bubble trapped inside it. Closest distance of the bubble from the surface is 2 cm. Where will it appear when seen from the other end (from where it is farthest)?
Answer:
From figure, distance OR = 2 cm
∴ Distance OP = 4 cm

According to sign conventions,
OP = u = -4 cm and CP = R = -3 cm
For refraction at curved surface,

Question 30.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:
Given: n = 1.55, f = 20 cm,
R₁ = R and R₂ = – R
(By sign convention)
To Find: Radius of curvature (R)
Formula: \(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula,
\(\frac {1}{20}\) = (1.55 – 1) \(\left[\frac{1}{R}-\left(-\frac{1}{R}\right)\right]=0.55 \times \frac{2}{R}\)
∴ R = \(\frac {1.10}{1}\) × 20 = 22 cm

Question 31.
A dense glass double convex lens (n = 2) designed to reduce spherical aberration has |R₁| : |R₂| = 1:5. If a point object is kept 15 cm in front of this lens, it produces its real image at 7.5 cm. Determine R₁ and R₂.
Answer:
Given: |R₁| : |R₂| = 1 : 5, u = -15 cm,
v = +7.5 cm, n = 2
To Find: Radii of curvature of double convex lens (R₁) and (R₂)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. \(\frac {1}{f}\) = (n – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula (i),
\(\frac{1}{f}=\frac{1}{7.5}-\frac{1}{(-15)}=\frac{1}{5}\)
∴ f = +5 cm
Substituting this value in formula (ii), we get,
\(\frac{1}{5}=(2-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
∴ \(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}=\frac{1}{5}\)
By sign conventions,
\(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\left(-\mathrm{R}_{2}\right)}=\frac{1}{5}\) ………….. (1)
Also \(\frac{\left|R_{1}\right|}{\left|R_{2}\right|}=\frac{1}{5}\)
∴ |R₂| = 5 |R₁| …………… (2)
Substituting in equation (1),
∴ \(\frac{1}{R_{1}}-\frac{1}{\left(-5 R_{1}\right)}=\frac{1}{5}\)
∴ \(\frac {6}{5R_1}\) = \(\frac {1}{5}\)
∴ R₁ = 6 cm
Using in equation (2),
R₂ = 5 × 6 = 30 cm

Question 32.
Why are prism preferred for dispersion over two parallel surfaces? Explain its construction in brief.
Answer:
i. In case of two parallel surfaces, for dispersion to be easily detectable, they must be separated over a large distance.
ii. In order to have appreciable and observable dispersion, two parallel surfaces are not useful. In such case we use prisms, in which two refracting surfaces inclined at an angle.
iii. Commonly used prisms have three rectangular surfaces forming a triangle.
iv. Two of which take part in refraction at a time. The one, not involved in refraction is called base of the prism.
v. Any section of prism perpendicular to the base is called principal section of the prism. Commonly all the rays considered during refraction lie in this plane.

Question 33.
Draw neat labelled diagrams showing refraction of a monochromatic light and white light through a prism.
Answer:

Question 34.
For a prism prove that i + e = A + δ where the symbols have their usual meanings.
Answer:
i. Consider a principal section ABC of a prism of absolute refractive index n kept in air as shown in figure.

ii. Let A be the refracting angle of prism and surface BC be the base.

iii. A monochromatic ray PQ obliquely strikes first reflecting surface AB such that, angle of incidence ∠PQM at Q is i.

iv. After refraction at Q, the ray deviates towards the normal and strikes second refracting surface AC at R which is the point of emergence.

v. Angles of refraction at Q (∠NQR) and at R (∠QRN) are r₁ and r₂ respectively.

vi. After R. the ray deviates away from normal and finally emerges along RS making e as the angle of emergence.

vii. Emergent ray RS meets an extended incident ray QT at X if traced backward. In this case, ∠TXS gives the angle of deviation.

viii. From figure,
∠AQN = ∠ARN = 90°
∴ From quadrilateral AQNR,
A + ∠QNR = 180° ………. (1)
From ∆ QNR,
r₁ + r₂ + ∠QNR = 180° ………. (2)
∴ A = r₁ + r₂ ……… (3)

ix. Angle δ forms an exterior angle for ∆ XQR.
∴ ∠XQR + ∠XRQ = δ
∴ (i – r₁) + (e – r₂) = δ
∴ (i + e) – (r₁ + r₂) = δ
From equation (3),
δ = i + e – A
∴ i + .e = A + δ

Question 35.
Explain δ versus i curve for refraction of light through a prism.
Answer:
i. Variation of angle of incidence i with angle of deviation δ is as shown in figure.

ii. It shows that, with increasing values of i, the angle of deviation δ decreases initially to a certain minimum (δ_m) value and then increases.

iii. The curve shown in the figure is not a symmetric parabola, but the slope in the part towards right is less.

iv. Except at δ = δ_m there are two values of i for any given δ. From principle of reversibility of light, we can conclude that if one of these values is i, the other must be e and vice versa. Thus at δ = δ_m, we have i = e.

Question 36.
Show that, at condition of minimum deviation, n = \(\frac{\sin \left(\frac{\mathbf{A}+\boldsymbol{\delta}_{\mathrm{m}}}{\mathbf{2}}\right)}{\sin \left(\frac{\mathbf{A}}{\mathbf{2}}\right)}\)

i. For every angle of deviation except angle of minimum deviation, there are two values of angle of incidence.

ii. However, at angle of minimum deviation there is only one corressponding angle of incidence.

iii. From principle of reversibility in path PQRS, the values of i and e are interchangeable for every δ. Thus, at minimum deviation, i = e.

iv. This implies the angles of refraction r₁ and r₂ are also equal. Also, A = r₁ + r₂
∴ A = 2 r i.e., r = \(\frac {A}{2}\) ……. (1)

v. In case of minimum deviation, QR is parallel to base BC and the figure is symmetric.

vi. Using i + e = A + δ,
i + i = A + δ_m
∴ i = \(\frac {A+δ_m}{2}\) …………….(2)

vii. According to Snell’s law,
n = \(\frac {sin i}{sin r}\)
Thus, using equations (1) and (2),
n = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
This is the prism formula.

Question 37.
For grazing emergence of a ray in a prism, find out minimum possible values for angle of incidence (i) and angle of refraction (r₁) for commonly used glass prism.
Answer:

i. At the point of emergence in prism, the ray travels from a denser medium into rarer.
Thus, if r₂ = sin^-1 (\(\frac {1}{n}\)) is the critical angle, the angle of emergence e = 90°. This is called grazing emergence.

ii. Angle of prism A is constant for a given prism and A = r₁ + r₂. Hence the corresponding r₁ and i will have their minimum possible values.

iii. For commonly used glass prisms,
n = 1.5 (r₂)_max = sin^-1 (\(\frac {1}{n}\)) = sin^-1 (\(\frac {1}{1.5}\)) = 41.49°

iv. If, prism is symmetric (equilateral),
A = 60°
∴ r₁ = 60° – 41°49′ = 18°11′
∴ n = 1.5 = \(\frac{\sin \left(\mathrm{i}_{\min }\right)}{\sin \left(18^{\circ} 11^{\prime}\right)}\)
sin (i_min) = 1.5 × sin (18°11′)
∴ ii_min = 27°55′ ≅ 28°.

Question 38.
Derive the formula for angle of deviation for thin prisms.
OR
Show that in a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).
Answer:
For thin prisms (refracting angle < sin 10°). sin θ ≈ θ
∴ Refractive index, n = \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}_{1}} \approx \frac{\mathrm{i}}{\mathrm{r}_{1}}\)
Also n = \(\frac{\sin e}{\sin \mathrm{r}_{2}} \approx \frac{\mathrm{e}}{\mathrm{r}_{2}}\)
∴ i ≈ n r₁ and e ≈ nr₂

ii. Substituting this in, i + e = A + δ, we get,
i + e = n (r₁ + r₂) = nA = A + δ
∴ S = A(n – 1)
A and n are constant for a given prism. Thus, for a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).

Question 39.
Give the expression for mean deviation for a beam of white light.
Answer:
For a beam of white light, yellow colour is practically chosen to be the mean colour for violet and red.
This gives mean deviation as,
δ_VR = \(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\) ≈ δ_Y = A(n_Y – 1)
Where, n_Y = refractive index for yellow colour.

Question 40.
A fine beam of white light is incident upon the longer side of a plane parallel glass slab of breadth 5 cm at angle of incidence 60°. Calculate lateral deviation of red and violet rays and lateral dispersion between them as they emerge from the opposite side. Refractive indices of the glass for red and violet are 1.51 and 1.53 respectively.
Answer:
As shown in figure,
VM = L_V = lateral deviation for violet colour,
RT = L_R = lateral deviation for red colour,
∴ Lateral dispersion between these colours, L_VR = L_V – L_R

∴ r_R = sin^-1 (0.5735) ≈ 35°
Similarly,
sin r_V = \(\frac{\sin 60^{\circ}}{1.53}=\frac{\sqrt{3}}{2 \times 1.53}=\frac{1.732}{3.06}\)
= antilog {log (1.732) – log (3.06)}
= antilog {0.2385 – 0.4857}
= antilog 11.7528}
= 0.5659
∴ sin r_V = 0.566
∴ r_V = sin^-1 (0.566) = 34°28′
∴ Angle of deviation for red colour
= i – r_R = 60° – 35° = 25°
and that for violet colur = i – r_V = 60° – 34°28′
= 25°32′
From figure, in ∆ANR,
AR = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{R}}}=\frac{5}{\cos \left(35^{\circ}\right)}=\frac{5}{0.8192}\) = 6.104 cm
Similarly ∆ANV,
AV = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{V}}}=\frac{5}{\cos \left(34^{\circ} 28^{\prime}\right)}=\frac{5}{0.8244}\) = 6.065 cm
∴ For red colour, L_R = RT = AR [sin(i – r_R)]
= AR [sin (25°)]
= 6.104 × 0.4226
= 2.58 cm
For violet colour, L_V = VM
= AV [sin (i – r_V)]
= AV × sin (25° 32′)
= 6.065 × 0.431
= 2.61 cm
∴ L_VR = L_V – L_R = 2.61 – 2.58 = 0.03 cm
= 0.3 mm

Question 41.
For a glass (n = 1.5) prism having refracting angle 60°, determine the range of angle of incidence for which emergent ray is possible from the opposite surface and the corresponding angles of emergence. Also calculate the angle of incidence for which i = e. How much is the corresponding angle of minimum deviation?
Answer:
For an equilateral prism of glass, the minimum angle of incidence for which the emergent ray just emerges is i_min = 27° 55′. Corresponding angle of emergence is, e_max = 90°.
From the principle of reversibility of light, i_max = 90° and e_min = 27°55′
Also, for equilateral glass prism at minimum deviation,

Also, from prism formula,
i + e = A + δ
At minimum deviation,
∴ i + i = 60 + 37°10′ = 97°10′
∴ i = 48°35′

Question 42.
For a dense flint glass prism of refracting angle 10°, obtain angular deviation for extreme colours and dispersive power of dense flint glass. (n_red = 1.712, n_violet = 1.792)
Answer:
Given: A = 10°, n_R = 1.712, n_V = 1.792
To find:
i. Angular deviation for extreme colours (δ_V and δ_R)
ii. Dispersive power of flint glass (ω)
Formulae:
i. δ = A(n – 1)
ii. ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)

Question 43.
The refractive indices of the material of the prism for red and yellow colour are 1.620 and 1.635 respectively. Calculate the angular dispersion and dispersive power, if refracting angle is 8°.
Solution:
Given: n_R = 1.620, n_Y = 1.635, A = 8°
To find:
i. Angular dispersion (δ_V – δ_R)
ii. Dispersive power (ω)
Formulae:
i. δ_v – δ_r = A(n_V – n_R)
ii. ω = \(\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)
Calculation: Since, n_Y = \(\frac {n_V+n_R}{2}\)
∴ n_V = 2n_Y – n_R
n_V = 2 × 1.635 – 1.620 = 3.27 – 1.620
∴ n_V = 1.65
From formula (i),
δ_V – δ_R = 8(1.65 – 1.620)
= 8 × 0.03 = 0.24°
∴ δ_V – δ_R = 0.24°
From formula (ii),
ω = \(\frac{1.65-1.620}{1.635-1}=\frac{0.03}{0.635}\) = 0.0472

Question 44.
What could be the possible reasons for the upward bending of the light ray during hot days?
Answer:
Possible reasons for the upward bending at the road could be:
i. Angle of incidence at the road is glancing. At glancing incidence, the reflection coefficient is very large which causes reflection.
ii. Air almost in contact with the road is not steady. The non-uniform motion of the air bends the ray upwards and once it has bent upwards, it continues to do so.

Question 45.
State some properties of rainbow.
Answer:

1. It is seen during rains and on the opposite side of the Sun.
2. It is seen only during mornings and evenings and not throughout the day.
3. In the commonly seen rainbow red arch is outside and violet is inside.
4. In the rarely occurring concentric secondary rainbow, violet arch is outside and red is inside.
5. It is in the form of arc of a circle.
6. Complete circle can be seen from a higher altitude, i.e. from an aeroplane.
7. Total internal reflection is not possible in this case.

Question 46.
Why is total internal reflection not possible during formation of a rainbow ?
Answer:
i. For total internal reflection, the angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

ii. The relative refractive index of air with the water drop is just less than 1 and hence the critical angle is almost equal to 90°.

iii. Angle of incidence i in air, at the water drop, can’t be greater than 90°.

iv. As a result, angle of refraction r in water will always be less than the critical angle.

v. The figures shown indicates that this angle r itself acts as an angle of incidence at any point for one or more internal reflections. But this does not indicate the total internal reflection.

Question 47.
Rainbow is seen only for a definite angle range with respect to the ground. Justify.
Answer:
i. For clear visibility of rainbow, a beam must have enough intensity.

ii. The curve for angle of deviation and angle of incidence is almost parallel to X-axis near minimum deviation i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same and those rays form a beam of enough intensity.

iii. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. Hence, the rainbow is seen only for a definite angle range with respect to the ground for which the intensity of the beam is enough for the visibility.

Question 48.
How is the range of angles for which rainbows can be observed calculated?
Answer:
i. Angle of deviation for the final emergent ray, can be shown to be equal to δ = π + 2i – 4r for primary rainbow and δ = 2π + 2i – 6r for the secondary rainbow.
ii. Using these relations along with Snell’s law, sin i = n sin r, derivatives of angle of deviation (δ) is obtained.
iii. Second derivative of δ comes out to be negative, which shows that it is the minima condition.
iv. Equating first derivative to zero corresponding values of i and r are obtained. Thus, from the figures shown, the corresponding angles θ_R and θ_V at the horizontal are obtained. These angles give the visible angular position for the rainbow.

Question 49.
When can one see complete circle of a rainbow? Explain in detail.
Answer:
i. Figure given below illustrates formation of primary and secondary rainbows with their common centre O. It is the point where the line joining the sun and the observer meets the Earth when extended.

ii. P is location of the observer. Different colours of rainbows are seen on arches of cones of respective angles.

iii. Smallest half angle refers to the cone of violet colour of primary rainbow, which is 41°.

iv. As the Sun rises, the relative position of common centre of the rainbows with respect to observer shifts down. Hence as the Sun comes up, smaller and smaller part of the rainbows will be seen. If the Sun is above 41°, violet arch of primary rainbow cannot be seen.

v. Beyond 53°, no rainbows will be visible. That is why rainbows are visible only during mornings and evenings and in the shape of a bow.

vi. However, if observer moves up (may be in an aeroplane), the line PO itself moves up making lower part of the arches visible. After a certain minimum elevation, entire circle for all the cones can be visible.

Question 50.
Define following terms:
i. Longitudinal chromatic aberration
ii. Circle of least confusion
iii. Transverse chromatic aberration
Answer:
i. Longitudinal chromatic aberration:
Due to different refractive indices and angle of deviations, violet and red colours of a white light converge at different focal points, f_V and f_R. The distance between f_V and f_R is measured as the longitudinal chromatic aberration.

ii. Circle of least confusion:
In presence of aberration the image is not a single point but always a circle. At particular location on the screen, this circle has minimum diameter. This is called circle of least confusion.

iii. Transverse chromatic aberration:
Radius of the circle of least confusion is called the transverse chromatic aberration.

Question 51.
After Cataract operation, a person is recommended with concavo-convex spectacles of curvatures 10 cm and 50 cm. Crown glass of refractive indices 1.51 for red and 1.53 for violet colours is used for this. Calculate the lateral chromatic aberration occurring due to these glasses.
Answer:
For a concavo-convex lens, with convex shape facing the object, both the radii of curvature are positive as shown in the figure.

= (1.53 – 1) × 0.08 = 0.0424
∴ f_v = 23.58 cm
∴ Longitudinal (lateral) chromatic aberration
= f_V – f_R = 24.51 – 23.58 = 0.93 cm

Question 52.
Why do we need optical instruments for?
Answer:
i. Due to the limitation for focusing the eye lens it is not possible to take an object closer than a certain distance. This distance is called least distance of distance vision D. For a normal, unaided human eye D = 25cm.
ii. If an object is brought closer than this, we cannot see it clearly.
iii. If an object is too small the corresponding visual angle from 25 cm is not enough to see it and if we bring it closer than that, its image on the retina is blurred.
iv. Also, the visual angle made by cosmic objects far away from us such as stars is too small to make out minor details and we cannot bring those closer.
v. In such cases we need optical instruments like microscope and telescopes to observe these things clearly.

Question 53.
A convex lens has focal length of 2.0 cm. Find its magnifying power if image is formed at DDV.
Answer:
Given: f = 2 cm, v = D = 25 cm
To find: Magnifying power (M.P.)
Formula: M.P = 1 + \(\frac {D}{f}\)
Calculation:
From formula,
M.P = 1 + \(\frac {25}{2}\)
M.P. = 1 + 12.5 = 13.5

Question 54.
A magnifying glass of focal length 10 cm is used to read letters of thickness 0.5 mm held 8 cm away from the lens. Calculate the image size. How big will the letters appear? Can you read the letters if held 5 cm away from the lens? If yes, of what size would the letters appear? If no, why not?
Answer:
Given that, f = +10 cm, u = -8 cm,
From thin lens formula,
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
∴ \(\frac{1}{10}=\frac{1}{\mathrm{v}}-\frac{1}{-8}\)
∴ v = 40 cm
Magnification of a lens is,
m = \(\frac {v}{u}\) = \(\frac {Object size h-i}{Object size h-0}\)
∴ \(\frac {40}{8}\) = \(\frac {h_1}{0.5}\)
∴ h₁ = 2.5 cm
This implies the height of the image is 5 times that of the object.
Magnifying power,
M = \(\frac {D}{u}\) = \(\frac {25}{8}\) = 3.125
∴ Image will appear to be 3.125 times bigger,
i.e., 3.125 × 0.5 = 1.5625 cm
For u = -5 cm, v will be -10 cm
For an average human being to see clearly, the image must be at or beyond 25 cm. Thus it will not possible to read the letters if held 5 cm away from the lens.

Question 55.
A compound microscope has a magnification of 15. If the object subtends an angle of 0.5° to eye, what will be the angle subtended by the image at the eye?
Answer:
Given: M.P = 15, α = 0.5°
To Find: Angle(β)
Formula: M.P = \(\frac {β}{α}\)
Calculation:
From formula,
β = M.P × α = 15 × 0.5 = 7.5°

Question 56.
A compound microscope has a magnifying power of 40. Assume that the final image is formed at DDV(25 cm). If the focal length of eyepiece 10 cm, calculate the magnification produced by objective.
Answer:
M.P = 40, D = 25 cm, f_e = 10 cm
To Find: Magnification (m₀)
Formula: M.P = m₀ × M_e
Calculation:
From the formula,

Question 57.
The pocket microscope used by a student consists of eye lens of focal length 6.25 cm and objective of focal length 2 cm. At microscope length 15 cm, the final image appears biggest. Estimate distance of the object from the objective and magnifying power of the microscope.
Answer:
Given: f_e = 6.25 cm, f₀ = 2 cm, L = 15 cm
As image appears biggest, V_e = -25 cm.
To find:
i. Distance of object from objective (u₀)
ii. Magnifying power (M)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. L = |v₀| + |u_e|
iii. M = \(\left(\frac{v_{o}}{u_{o}}\right)\left(\frac{D}{u_{e}}\right)\)
Calculation: For eyelens, using formula (i),

∴ u_e = 5 cm
From formula (ii),
|v₀| = L – |u_e|
= 15 – 5 = 10 cm
Using formula (i) for objective,

Question 58.
Focal length of the objective of an astronomical telescope is 1 m. Under normal adjustment, length of the telescope is 1.05 m. Calculate focal length of the eyepiece and magnifying power under normal adjustment.
Answer:
Given: f₀ = 1 m, L = 1.05 m
To find:
i. Focal length of eyepiece (f_e)
Magnifying power under normal adjustment (M)
Formula:
i. L = f₀ + f_e
ii. M = \(\frac {f_0}{f_e}\)
Calculation. From formula (i),
f_e = L – f₀ = 1.05 – 1 = 0.05 m
From formula (ii),
M = \(\frac {1}{0.05}\) = 20

Question 59.
Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?
Answer:
Given: M.P = 12, v = D, f₀ = 90 cm,
To find: Focal length of eye piece (f_e)
Formula: M.P = \(\frac {f_0}{f_e}\) (1 + \(\frac {f_e}{D}\))
Calculation:
From formula.
12 = \(\frac {90}{f_e}\) (1 + \(\frac {f_e}{25}\))
∴ f_e = 10.71 cm

Question 60.
Two convex lenses of an astronomical telescope have focal length 1.3 m and 0.05 m respectively. Find the magnifying power and the length of the telescope.
Answer:
Given: f₀ = 1.3 m, f_e = 0.05 m
To find:
i. Magnifying power of telescope (M.P.)
ii. Length of telescope (L)
Formulae:
i. M.P = \(\frac {f_0}{f_e}\)
ii. L = f₀ + f_e
Calculation: From formula (i),
M.P = \(\frac {1.3}{0.05}\) = 26
From formula (ii),
L = 1.3 + 0.05 = 1.35 m

Question 61.
What is the angle of deviation of reflected ray if ray of light is incident on a plane mirror at an incident angle θ?
Answer: When a ray of light is incident on a plane mirror at an angle θ, the reflected ray gets deviated by an angle of (π – 2θ).

Question 62.
Does nature of the image depend upon size of the mirror?
Answer:
No, nature of the image is independent of size of the mirror.

Question 63.
If a convex mirror is held in air and then dipped in oil, then what will be the change in its focal length?
Answer:
Focal length of spherical mirrors are independent of the medium.

Question 64.
When ray of light falls normally on a mirror, its angle of incidence is 90°. True or false? Justify your answer.
Answer:
False, when light falls normally on a mirror, its angle of incidence is zero degree.

Question 65.
In one of the performances, a magician keeps a gold ring beneath a thick glass slab (µ = \(\frac {3}{2}\)) Then he keeps a flask filled with water (µ = \(\frac {4}{3}\)), over the slab. Now when spectators one by one observe from the open end of the flask, the ring disappears at a certain angle of viewing.
i. What could be the reason behind the disappearance?
ii. At what angle of viewing the ring vanishes?
Answer:
i. The ring disappears due to total internal reflection of the light at water-air interface.

ii. _aµ_w = \(\frac {4}{3}\)
sin i_c = \(\frac {1}{µ}\)
∴ i_c = sin^-1 (\(\frac {1}{_aµ_w}\)) = sin^-1 (\(\frac {3}{4}\)) = 48.6°
Hence, for angle of viewing for which angle of incidence of ring from water to air is greater that 48.6°, the ring will vanish.

Question 66.
Why dispersion of light is not observed in glass slab but it is observed in prism?
Answer:
When a light passes from one medium to another, at one interface, it changes its speed. The glass slab and prism both have two glass-air interfaces. Hence, the light undergoes refraction twice in both the cases. When the two interfaces are parallel to each other, although the colours are separated at first interface, they all travel the same path after refracting from second interface. However, in prism, the two interfaces are not parallel. Therefore, the colours separated at first interface do not travel the same path after second refraction but emerge out at different wavelengths producing spectrum.

Question 67.
A prism manufacturer is planning to build a dispersive prism out of following materials with the refracting angles as given.
i. Glass (µ = 1.5), A = 60°
ii. Plastic (µ = 1.4), A = 90°
iii. Fluorite (µ = 1.45), A = 64°
If he desires to give the prism following relations of i and δ, then which of the above combinations can be used to construct the prism?

Answer:
From the given values of i and δ, δm = 37°
From prism formula, µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)

i. For Glass (µ = 1.5), A = 60°;
µ = \(\frac{\sin \left(\frac{60^{\circ}+37^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= 1.5
Hence, this combination can be used for fabricating the desired prism.

ii. For Plastic (µ = 1.4), A = 90°;
µ = \(\frac{\sin \left(\frac{90^{\circ}+37^{\circ}}{2}\right)}{\sin \left(45^{\circ}\right)}\)
= 1.26
As P_piastic = 1-4, this combination cannot be used.

iii. For Fluorite (µ = 1.45), A = 64°
µ = \(\frac{\sin \left(\frac{64^{\circ}+37^{\circ}}{2}\right)}{\sin \left(32^{\circ}\right)}\)
= 1.45
Hence, this combination can also be used for fabrication of prism.

Question 68.
Find the refractive index of material of following prism if the ray of light incident at angle 45° suffers minimum deviation through the prism.

Answer:

A = 60°
Also, ray of light suffers minimum deviation.
∴ 2i = A + δ_m
∴ δ_m = 2i – A = 90° – 60° = 30°
From prism formula,
µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
= \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= √2
Hence, refractive index of material of prism is √2.

Multiple Choice Questions

Question 1.
Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
(A) 4 × 10^-11 s
(B) 2 × 10^-11 ns
(C) 16 × 10^-11 s
(D) 8 × 10^-10 s
Answer:
(A) 4 × 10^-11 s

Question 2.
If mirrors are inclined to each other at an angle of 90°, the total number of images seen for a symmetric position of an object will be
(A) 3
(B) 4
(C) 5
(D) 3 or 4
Answer:
(A) 3

Question 3.
In case of a convex mirror, the image formed is
(A) always on opposite side, virtual, erect.
(B) always on the same side, virtual, erect.
(C) always on opposite side, real, inverted.
(D) dependent on object distance.
Answer:
(A) always on opposite side, virtual, erect.

Question 4.
A glass slab is placed in the path of a beam of convergent light. The point of convergence of light
(A) moves towards the glass slab.
(B) moves away from the glass slab.
(C) remains at the same point.
(D) undergoes a lateral shift.
Answer:
(A) moves towards the glass slab.

Question 5.
For a person seeing an object placed in optically rarer medium,
(A) apparent depth of the object is more than real depth
(B) apparent depth is smaller than the real depth.
(C) apparent depthe might be smaller or greater depending on the position of the person.
(D) nothing can be concluded about the depth of object from given data.
Answer:
(A) apparent depth of the object is more than real depth

Question 6.
Light travels from a medium of refractive index µ₁ to another of refractive index µ₂ (µ₁ > µ₂). For total internal reflection of light, which is NOT true?
(A) Light must travel from medium of refractive index µ₁ to µ₂.
(B) Angle of incidence must be greater than the critical angle.
(C) There is no refraction of light.
(D) Light must travel from the medium of refractive index µ₂ to µ₁.
Answer:
(D) Light must travel from the medium of refractive index µ₂ to µ₁.

Question 7.
Optical fibre is based on which of the following phenomenon?
(A) Reflection.
(B) Refraction.
(C) Total internal reflection.
(D) Dispersion.
Answer:
(C) Total internal reflection.

Question 8.
Commonly used glass have refractive index of 1.5. What is the critical angle for such glass?
(A) 49°
(B) 42°
(C) 45°
(D) 40°
Answer:
(B) 42°

Question 9.
If the refractive index of water is 4/3 and that of glass slab is 5/3. Then the critical angle of incidence for which a light ray tending to go from glass to water is totally reflected, is
(A) sin^-1 (\(\frac {3}{4}\))
(B) sin^-1 (\(\frac {3}{5}\))
(C) sin^-1 (\(\frac {2}{3}\))
(D) sin^-1 (\(\frac {4}{5}\))
Answer:
(D) sin^-1 (\(\frac {4}{5}\))

Question 10.
While deriving prism formula, which of the following condition is NOT satisfied?
(A) I = e
(B) r₁ = r₂
(C) r = \(\frac {A}{2}\)
(D) δ_m = i + e + r
Answer:
(D) δ_m = i + e + r

Question 11.
If the critical angle for the material of a prism is C and the angle of the prism is A, then there will be no emergent ray when
(A) A < 2 C
(B) A = 2 C
(C) A > 2 C
(D) A < \(\frac {C}{2}\)
Answer:
(C) A > 2 C

Question 12.
Chromatic aberrations is caused due to
(A) spherical shape of lens
(B) spherical shape of mirrors
(C) angle of deviation for violet light being more than that for red light.
(D) refractive index for violet light being less than that for red light
Answer:
(C) angle of deviation for violet light being more than that for red light.

Question 13.
In normal adjustment, magnifying powser of a astronomical telescope is given by
(A) \(\frac {D}{f_0}\) \(\frac {L}{f_e}\)
(B) \(\frac {L}{D}\) \(\frac {f_e}{f_0}\)
(C) \(\frac {f_0}{f_e}\)
(D) \(\frac {f_e}{f_0}\)
Answer:
(C) \(\frac {f_0}{f_e}\)