Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: Simple Harmonic Motion
Q2: A cylindrical plastic bottle of negligible mass is filled with 310 mL of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency. If the radius of the bottle is 2.5 cm then ω is close to (density of water = 10^3 kg/m^3 )
Options: (a) 2.50 rad/s
(b) 3.75 rad/s
(c) 5.00 rad/s
(d) 1.25 rad/s
Answer: None of the options given are correct
Solution: Restoring force due to pressing the bottle with amount x,
F = - (ρAx)g
a = - (ρAg/m)x
Therefore, ω^2 = ρAg/m = [ρ(πr^2)g] / m
ω = √(10^3 * π * (2.5 * 10^-2)^2 * 10 / 310 * 10^-3)
Topic: Simple Harmonic Motion
Q3: A particle undergoing simple harmonic motion has time-dependent displacement given by x(t) = Asin(πt / 90). The ratio of kinetic to the potential energy of this particle at t = 210 s will be
Options: (a) 1 / 3
(b) 2
(c) 1
(d) 3
Answer: Answer: (a) 1 / 3
Solution: The maximum kinetic energy of the particle, K.E = (1 / 2)mA^2ω^2cos^2(ωt)
The potential energy of the particle at any time t, U = (1 / 2)mA^2ω^2sin^2(ωt)
The ratio of kinetic energy to potential energy, r = K.E/U = [(1 / 2)mA^2ω^2cos^2(ωt)] / [(1 / 2)mA^2ω^2sin^2(ωt)]
r = cos^2(ωt) / sin^2(ωt)
r = cot^2(ωt)
The angular frequency is given as ω = π / 90
Time, t = 210 s
Therefore, r = cot^2(π / 90) * 210
r = cot^2(7π / 3)
r = cot^2(2π + π / 3)
We know,
cot(2nπ + θ) = cotθ
Therefore, r = cot^2(π / 3)
r = (1 / √3)^2
r = 1 / 3
Topic: Simple Harmonic Motion
Q4: A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then
Options: (a) K2 = K1
(b) K2 = K1 / 2
(c) K2 = 2K1
(d) K2 = K1 / 4
Answer: Answer: (c) K2 = 2K1
Solution: Maximum kinetic energy = (1 / 2)mω^2A^2
ω = √(g / L)
A = Lθ
K.E = (1 / 2)mgLθ^2
K1 = (1 / 2)mgLθ^2
If the length is doubled
K2 = (1 / 2)mg2Lθ^2
From (1) and (2)
(K1 / K2) = (1 / 2)mgLθ^2 / (1 / 2)mg2Lθ^2
= 1 / 2
K2 = 2K1
Topic: Simple Harmonic Motion
Q5: A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^-2 m. The relative change in the angular frequency of the pendulum is best given by
Options: (a) 10^-5 rad/s
(b) 10^-1 rad/s
(c) 1 rad/s
(d) 10^-3 rad/s
Answer: Answer: (d) 10^-3 rad/s
Solution: Angular frequency of the pendulum
ω = √(g / l)
(Δω / ω) = (1 / 2)(Δgeff / geff)
(Δω / ω) = (1 / 2)(2Aω^2 / g)
Δω / ω = (Aω^2 / g)
= (1^2 * 10^-2 / 10) = 10^-3
Topic: Simple Harmonic Motion
Q6: A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 / 1000 of the original amplitude is close to
Options: (a) 100 s
(b) 10 s
(c) 20 s
(d) 50 s
Answer: Answer: (c) 20 s
Solution: Frequency of damped oscillation, f = 5 Hz
For A = A / 2
t1 = 2 s
Also, A = A0 e^-(b / 2m)t or 1 / 2 = e^-(b / 2m)2
(b / m) = ln2
For A = A / 1000, t2 = ?
1 / 1000 = e^-(b / 2m)t2
Or 10^-3 = e^-(b / 2m)t2
(b / 2m)t2 = 3ln10 or t2 = 6ln10 / ln2
t2 = 20 s
Topic: Simple Harmonic Motion
Q7: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
Options: (a) 335 m/s
(b) 322 m/s
(c) 328 m/s
(d) 341 m/s
Answer: Answer: (c) 328 m/s
Solution: Due to the jagged end
(λ1 / 4) = 11 cm, so v / (512 * 4) = 11 cm------(1)
(λ2 / 4) = 27 cm, so v / (256 * 4) = 27 cm------(2)
(2) - (1)
v / (512 * 4) = 0.16
v = 328 m/s
Topic: Simple Harmonic Motion
Q9: In an engine, the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston is close to
Options: (a) 0.7 Hz
(b) 1.9 Hz
(c) 1.2 Hz
(d) 0.1 Hz
Answer: Answer: (b) 1.9 Hz
Solution: The amplitude of S.H.M., A = 7 cm = 0.07 m
As the washer does not stay in contact with the piston, at some particular frequency i.e. normal force on the washer = 0
Maximum acceleration of washer = Aω^2 = g
ω = √(g / A) = √(10 / 0.07) = √(1000 / 7)
Frequency of the piston, v = ω / 2π
= 1.9 Hz
Topic: Simple Harmonic Motion
Q10: A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to
Options: (a) 680 Hz
(b) 510 Hz
(c) 340 Hz
(d) 170 Hz
Answer: Answer: (d) 170 Hz
Solution: vair = 340 m/s, v = 5 m/s
f1 - f2 = 5 beats per second
Apparently frequency heard by the observer on reflection from the wall,
f1 = [v / (v - vs)]f = [340 / (340 - 5)]f = (340 / 335)f
f2 = [v / (v + vs)]f = [340 / (340 + 5)]f = (340 / 345)f
Since f1 - f2 = 5
[(340 / 335) - (340 / 345)]f = 5 ⇒ f = (5 / 340)([335 * 345] / 10) = 169.96 Hz = 170 Hz
Topic: Simple Harmonic Motion
Q11: A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^12 s^-1. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver 108 and Avogadro number 6.02 * 10^23 gm mole^-1 )
Options: (a) 6.4 N m^-1
(b) 7.1 N m^-1
(c) 2.2 N m^-1
(d) 5.5 N m^-1
Answer: Answer: (b) 7.1 N m^-1
Solution: Frequency of a particle executing SHM,
f = 1 / (2π) √(k / m)
k = 4π^2 * f^2 * m
Here, f = 10^12 s^-1, m = [108 / (6.02 * 10^23)] * 10^-3 Kg
k = 4 * (3.14)^2 * (10^12)^2 * [108 * 10^-3 / 6.02 * 10^23] = 7.1 Nm^-1
Topic: Simple Harmonic Motion
Q12: The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s^-1. At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is π / 4
Options: (a) 500 √2 m/s^2
(b) 500 m/s^2
(c) 750 √2 m/s^2
(d) 750 m/s^2
Answer: Answer: (a) 500 √2 m/s^2
Solution: For simple harmonic motion,
Maximum acceleration/Maximum velocity = 10 ⇒ ω^2a / ωa = 10 or ω = 10
At t = 0; displacement, x = 5
x = asin(ωt + Φ)
5 = asin(0 + π / 4) or 5 = asin(π / 4) or a = 5√2 m
Maximum acceleration = ω^2a = 10^2 * 5√2 = 500√2 ms^-2
Topic: Simple Harmonic Motion
Q13: A child swinging on a swing in sitting position, stands up, then the time period of the swing will
Options: (a) increase
(b) decrease
(c) remains the same
(d) increases if the child is long and decreases if the child is short
Answer: Answer: (b) decrease
Solution: The time period will decrease.
When the child stands up, the centre of gravity is shifted upwards and so the length of swing decreases. T = 2π√(l / g)
Topic: Sound Waves
Q1: A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to
Options: 666 Hz
753 Hz
500 Hz
333 Hz
Answer: 666 Hz
Solution: Frequency of the sound produced by the open flute
f=2(v / 21)=(2 x 330) /(2 x 0.5)=660 Hz
Velocity of observer, v0=10 x(5 / 18)=(25 / 9) m/s
As the source is moving towards the observer, according to the Doppler effect.
Frequency detected by observer
f'={((v+v0) / v)} f={((25 / 9)+330) / 330} 660
=2((25 / 9)+330)
f'=665.55 ≈ 666 Hz
Topic: Sound Waves
Q2: A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source when it is moving away from the observer after crossing him? (Take the velocity of sound in air as 350 m/s)
Options: 750 Hz
857 Hz
1143 Hz
807 Hz
Answer: 750 Hz
Solution: When source is moving towards a stationary observer,
fapp= fsource {(V-0) / (V-50)}
1000=fsource (350 / 300)
When source is moving away from observer
f'=fsource {350 / (350+50)}
f'={(1000 x 300) / 350} x(350 / 400)
f'=750 Hz
Topic: Sound Waves
Q3: Two sources of sound S1 and S2 produce sound waves of the same frequency 660 Hz. A listener is moving from source S1 towards S2 with constant speed u m/s and he hears 10 beats. The velocity of sound is 330 m/s. Then u is equals
Options: 5.5 m/s
15.0 m/s
2.5 m/s
10.0 m/s
Answer: 2.5 m/s
Solution: f1=f[(v-v0) / v]
f2=f[(v+v0) / v]
Frequency f2-f1=fx(2v0 / v)
10=660 x(2u / 330)
u=2.5 m/s
Topic: Sound Waves
Q4: A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in the air. A tuning fork of frequency 512 Hz produces the first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces the first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to
Options: 322 ms^-1
341 ms^-1
335 ms^-1
328 ms^-1
Answer: 328 ms^-1
Solution: (λ / 4)=0.11+e
{V /(512) 4}=0.11+e
{V /(256) 4}=0.27+e
After solving (1) and (2) we get
V=328 ms^-1
Topic: Sound Waves
Q5: A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? (Given reference intensity of sound is 10^-12 W/m^2)
Options: 40 cm
20 cm
10 cm
30 cm
Answer: 40 cm
Solution: Using β=10 log10(I / I0)
Or 120=10 log 10(I / 10^-12)
I / 10^-12=10^12
I=1
Also I=1=P / 4πr^2=2 / 4πr^2
r^2=2 / 4πI=2 / 4π
r^2=2 / 4(3.14)
On solving the above equations, we get
r=40 cm
Topic: Sound Waves
Q6: Two cars A and B are moving away from each other in opposite directions. Both the cars are moving at a speed of 20 m/s with respect to the ground. If an observer in car A detects a frequency of 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B ? (speed of sound in air = 340 m/s)
Options: 2250 Hz
200 Hz
2300 Hz
2150 Hz
Answer: 2250 Hz
Solution: f'=f{(v-v0) / (v+vs)}
2000=f{(340-20) / (340+20)}
f=2250 Hz
Topic: Sound Waves
Q7: A person standing on an open ground hears the sound of a jet aeroplane, coming from the north at an angle 60 degrees with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, the speed of the plane is
Options: (√3 / 2) v
2v / √3
V
v / 2
Answer: v / 2
Solution: Distance, PQ=vp xt (Distance = speed x time)
Distance, QR=V . t
Cos 60=PQ / QR
1 / 2=(vp x t) / (v.t)
vp=v / 2
Topic: Sound Waves
Q8: Three sound waves of equal amplitudes have frequencies(f-1, f, f+1). They superpose to give beats. The number of beats produced per second is
Options: 2
1
4
3
Answer: 2
Solution: Beat produced between(f-1) and f is 1 .
Beat produced between f and f+1 is 1 .
Beat produced between(f-1) and (f+1) is 2
Therefore No of beats produced per second will be 2
Topic: Sound Waves
Q9: A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in the air is 340 m/s.
Options: 6
4
12
8
Answer: 6
Solution: Fundamental frequency of the closed organ pipe is
f=v / 4 L
f=340 /(4 x 0.85)=100 Hz
The natural frequencies of the closed organ pipe is
fn=(2n-1) f=f, 3f, 5f, 7f, 9f, 11f, 13f ...
So the possible frequencies below 1250 Hz are
fn=100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz
Number of frequencies =6
Topic: Sound Waves
Q10: An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light =3 x 10^8 m/s)
Options: 17.3 GHz
15.3 GHz
10.1 GHz
12.1 GHz
Answer: 17.3 GHz
Solution: Doppler effect in light (speed of observer is not very small compared to speed of light)
f'=f[(1+v / c) / (1-v / c)]^{1 / 2}
Here, frequency (v / c)=1 / 2
So, f'=f [(3 / 2) / (1 / 2)]^{1 / 2}
f'=10 x √3=17.3 GHz
Topic: Wave Optics
Q1: In Young's double-slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle 1/40 rad by using the light of wavelength λ1. When the light of wavelength λ2 is used a bright fringe is seen at the same angle in the same setup. Given that 1 and 2 are in the visible range (380 nm to 740 nm), their values are
Options: (a) 400 nm, 500 nm
(b) 625 nm, 500 nm
(c) 380 nm, 525 nm
(d) 380 nm, 500 nm
Answer: (b) 625 nm, 500 nm
Solution: Path difference = dsinθ = d × θ = (0.1 mm)(1 / 40) = 2.5 × 10^{-3} mm = 2500 nm
For bright fringes, path difference = nλ
So, 2500 = nλ1 = mλ2
n = 4, m = 5
or λ1 = 2500 / 4 = 625 nm
λ2 = 2500 / 5 = 500 nm
Topic: Wave Optics
Q2: In Young's double-slit experiment, the path difference, at a certain point on the screen, between two interfering waves is (1/8)th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
Options: (a) 0.80
(b) 0.94
(c) 0.85
(d) 0.74
Answer: (c) 0.85
Solution: The phase difference between two waves is given as
(Δx) x (2π/λ) = (λ/8) x (2π/λ) = π/4
So, the intensity at this point is
I = I0 cos²(Φ/2)
I = I0 cos²(π/8)
I = I0[(1 + cos(π/4)) / 2] = I0[(1 + (1/√2)) / 2] = 0.85 I0
Topic: Wave Optics
Q3: In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 m and a width of 4.05 m. The number of bright fringes between the first and the second diffraction minima is
Options: (a) 10
(b) 04
(c) 05
(d) 09
Answer: (c) 05
Solution: λg = 5303 Å, d = 19.44 m, a = 4.05 m
For diffraction location of first minima and second minima
y1 = Dλ/a, y2 = 2Dλ/a
y2 - y1 = (2Dλ/a) - (Dλ/a) = Dλ/a
β = Dλ/d
Number of bright fringes = (y2 - y1) / β
= (Dλ/a) x (d/Dλ)
= d/a
= 19.44 / 4.05 = 5
Topic: Wave Optics
Q4: In an interference experiment the ratio of amplitudes of coherent waves is (a1 / a2) = (1 / 3). The ratio of maximum and minimum intensities of fringes will be
Options: (a) 4
(b) 9
(c) 2
(d) 18
Answer: (a) 4
Solution: (a1 / a2) = (1 / 3)
Imax = a1 + a2
Imin = a1 - a2
(Imax / Imin) = [(a1 + a2)² / (a1 - a2)²] = [(1 + 1/3)² / (1 - 1/3)²] = (4/2)² = 4
Topic: Wave Optics
Q5: Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.
Options: (a) 610 x 10^{-9} radian
(b) 152.5 x 10^{-9} radian
(c) 457.5 x 10^{-9} radian
(d) 305 x 10^{-9} radian
Answer: (d) 305 x 10^{-9} radian
Solution: The limit of resolution,
θΔ = 1.22λ/a = (1.22 x 500 x 10^{-9}) / (200 x 10^{-2}) = 3.05 x 10^{-7} radian = 305 x 10^{-9} radian
Topic: Wave Optics
Q6: In Young's double-slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be
Options: (a) (√3 + 1)² : 16
(b) 9 : 1
(c) 25 : 9
(d) 4 : 1
Answer: (b) 9 : 1
Solution: I1 = 4I0
I2 = I0
Imax = (√I1 + √I2)²
= (2√I0 + √I0)² = 9I0
Imin = (√I1 - √I2)²
= (2√I0 - √I0)² = I0
(Imax / Imin) = 9 / 1
Topic: Wave Optics
Q7: In Young's double-slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide are
Options: (a) 1.56 mm
(b) 7.8 mm
(c) 9.75 mm
(d) 15.6 mm
Answer: (b) 7.8 mm
Solution: Let y be the distance from the central maximum to the point where the bright fringes due to both the wavelengths coincides.
Now, for λ1, y = mλ1D/d
For λ2, y = nλ2D/d
mλ1 = nλ2
(m/n) = λ2/λ1 = 520/650 = 4/5
i.e. with respect to central maximum 4th bright fringe of λ1 coincides with 5th bright fringe of λ2
Now, y = (4 x 650 x 10^{-9} x 1.5) / (0.5 x 10^{-3}) m
 y = 7.8 x 10^{-3} m or y = 7.8 mm
Topic: Wave Optics
Q8: A single slit of width b is illuminated by coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)
Options: (a) 6.0 cm
(b) 1.5 cm
(c) 4.5 cm
(d) 3.0 cm
Answer: (b) 1.5 cm
Solution: For single slit diffraction, sinθ = nλ/b (n = 1,2,3----)
sinθ ≈ θ ≈ tanθ
b tanθ = nλ
b tanθ1 = 2λ
b(y1/D) = 2λ
tanθ2 = 2λ
b(y2/D) = 2λ
(y2 - y1) b / D = 2λ
(6 - 3) b / D = 2λ
3b / D = 2λ
λD / b = 3/2 = 1.5 cm
Topic: Wave Optics
Q9: Unpolarized light of intensity I0 is incident on the surface of a block of glass at Brewster's angle. In that case, which one of the following statements is true?
Options: (a) Transmitted light is partially polarized with intensity I0 / 2
(b) Transmitted light is completely polarized with intensity less than I0 / 2
(c) The reflected light is completely polarized with intensity less than I0 / 2
(d) The reflected light is partially polarized with intensity I0 / 2
Answer: (c) The reflected light is completely polarized with intensity less than I0 / 2
Solution: At Brewster's angle, i = tan⁻¹(μ), the reflected light is completely polarized, whereas refracted light is partially polarized. Thus, the reflected ray will have lesser intensity compared to the refracted ray.
Topic: Wave Optics
Q10: A beam of unpolarized light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is
Options: (a) I0 / 8
(b) I0
(c) I0 / 2
(d) I0 / 4
Answer: (d) I0 / 4
Solution: The intensity of light after passing polaroid A is
I1 = I0 / 2
Now this light will pass through the second polaroid B whose axis is inclined at an angle of 45° to the axis of polaroid A. So in accordance with Malus law, the intensity of light emerging from polaroid B is
I2 = I1 cos² 45 = (I0 / 2)(1/√2)² = I0 / 4
Topic: Electromagnetic Induction
Q1: When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coil is
Options: (a) 0.67 H
(b) 1.67 H
(c) 3 H
(d) 6 H
Answer: (b) 1.67 H
Solution: I1=5 A, I2=2 A, ΔI=2-5=-3 A, Δt=0.1, S=50 V, As, ε=-L(ΔI / Δt), 50=-L(-3 / 0.1), 50=30 L, L=5 / 3=1.67 H
Topic: Electromagnetic Induction
Q2: Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A= 10 cm^2 and length =20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (μ0=4 π × 10^-7 T mA^-1)
Options: (a) 2.4 π × 10^-4 H
(b) 2.4 π × 10^-5 H
(c) 4.8 π × 10^-4 H
(d) 4.8 π × 10^-5 H
Answer: (a) 2.4 π × 10^-4 H
Solution: A=π r1^2=10 cm^2, l=20 cm, N1=300, N2=400, M=μ0 N1 N2 / l, M=(4 π × 10^-7 × 300 × 400 × 10 × 10^-4) / 0.20=2.4 π × 10^-4 H
Topic: Electromagnetic Induction
Q3: The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of
Options: (a) 1 μF
(b) 2 μF
(c) 4 μF
(d) 8 μF
Answer: (a) 1 μF
Solution: For maximum power, L ω=1 / C ω, C=1 / L ω^2, C=1 /[10 ×(2 π × 50)^2]=1 /(10 × 10^4 ×(π)^2)=10^-6 F=1 μF (Taking π^2=10)
Topic: Electromagnetic Induction
Q4: A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2 V. The current reaches half of its steady state value in
Options: (a) 0.15 s
(b) 0.3 s
(c) 0.05 s
(d) 0.1 s
Answer: (d) 0.1 s
Solution: During growth of charge in an inductance, I=I0(1-e^(-Rt / L)), or I0 / 2=I0(1-e^(-Rt / L)), e^(-Rt / L)=1 / 2=2^-1, or Rt / L=ln 2 ⇒ t=(L / R) ln 2, t=[(300 × 10^-3) / 2] ×(0.693), t=0.1 sec
Topic: Electromagnetic Induction
Q5: In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to
Options: (a) 4 L
(b) 2 L
(c) L / 2
(d) L / 4
Answer: (c) L / 2
Solution: At resonance, ω=1 / √(L C) when ω is constant, (1 / L1 C1)=(1 / L2 C2)=(1 / LC)=1 / L2(2 C)=1 / 2 L2 C, L2=L / 2
Topic: Electromagnetic Induction
Q6: Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon
Options: (a) the rates at which currents are changing in the two coils
(b) relative position and orientation of the two coils
(c) the materials of the wires of the coils
(d) the currents in the two coils
Answer: (c) the materials of the wires of the coils
Solution: Mutual inductance between two coils depends on the materials of the wires of the coils
Topic: Electromagnetic Induction
Q7: The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is
Options: (a) R / ω L
(b) R / (R^2+ω^2 L^2)^(1 / 2)
(c) ω L / R
(d) R / (R^2-ω^2 L^2)^(1 / 2)
Answer: (b) R / (R^2+ω^2 L^2)^(1 / 2)
Solution: Power Factor, cos φ=1 / (1+tan^2 φ)^(1 / 2), cos φ=1 / (1+(ω L / R)^2)^(1 / 2)(because tan φ=ω L / R), cos φ=R / (R^2+ω^2 L^2)^(1 / 2)
Topic: Electromagnetic Induction
Q8: An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to
Options: (a) 80 H
(b) 0.08 H
(c) 0.044 H
(d) 0.065 H
Answer: (d) 0.065 H
Solution: Given I=10 A, V=80 V, R=V / I=80 / 10=8 Ω and ω=50 Hz, I=V / (8^2+X_L^2)^(1 / 2), 10=220 / (64+X_L^2)^(1 / 2), (64+X_L^2)^(1 / 2)=22, Squaring on both sides, we get 64+X_L^2=484, X_L^2=484-64=420, X_L=(420)^(1 / 2) ⇒ 2 π × ω L=(420)^(1 / 2), Series inductor on an arc lamp, L=(420)^(1 / 2) / (2 π × 50)=0.065 H
Topic: Electromagnetic Induction
Q9: A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90 %, the output current would be
Options: (a) 50 A
(b) 25 A
(c) 45 A
(d) 20 A
Answer: (c) 45 A
Solution: Efficiency η=0.9=Ps / Pp, Vs Is=0.9 × Vp Ip, Is=(0.9 × 2300 × 5) / 230=45 A
Topic: Electromagnetic Induction
Q10: A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be
Options: (a) (3 / 2) nBA ω
(b) nBA
(c) 3 nBA ω
(d) (1 / 2) nBA ω
Answer: (b) nBA
Solution: Emf induced in the coil is given by ε=BAn ω sin ω t, εmax=BA (The induced emf is maximum when sin ω t=1)
Topic: Electromagnetic Induction
Q11: A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth's magnetic field of 0.3 × 10^-4 Wb/m^2. The value of the induced emf in wire is
Options: (a) 0.3 × 10^-3 V
(b) 2.5 × 10^-3 V
(c) 1.5 × 10^-3 V
(d) 1.1 × 10^-3 V
Answer: (c) 1.5 × 10^-3 V
Solution: The motional emf is given as ε=|v(l × B)| =5 ×(0.3 × 10^-4)(10) sin 90°=1.5 × 10^-3 V
Topic: Electromagnetic Induction
Q12: A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil
Options: (a) increases by a factor of 27
(b) decreases by a factor of 3
(c) increases by a factor of 3
(d) decreases by a factor of 9
Answer: (c) increases by a factor of 3
Topic: Electromagnetic Induction
Q13: The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is
Options: (a) 637.5 J
(b) 540 J
(c) 437.5 J
(d) 740 J
Answer: (c) 437.5 J
Solution: ε=25 V, I1=10 A, I2=25 A, t=1 s, ΔE=?, ε=L(ΔI / Δt), 25=L[(25-10) / 1], L=25 / 15=(5 / 3) H, ΔE=1 / 2 L(I2^2-I1^2)=(1 / 2) ×(5 / 3) ×(25^2-10^2)=(5 / 3) × 525=437.5 J