A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be
a.
Yellow – Violet – Orange – Silver
Correct Answer
b.
Yellow – Green – Violet – Gold
c.
Violet – Yellow – Orange – Silver
d.
Green – Orange – Violet – Gold
Step-by-Step AI Explanation
Verified Method
47 kΩ = 47 × 10³ Ω. First digit 4 = Yellow, Second digit 7 = Violet, Multiplier 10³ = Orange, Tolerance ±10% = Silver. The tolerance is 4.7/47 × 100 = 10%, which corresponds to Silver. Hence the sequence is Yellow – Violet – Orange – Silver.
Q93mcq
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of 'n' is
a.
11
b.
20
c.
10
Correct Answer
d.
9
Step-by-Step AI Explanation
Verified Method
Series case: I = E / (nR + R) = E / [R(n+1)]. Parallel case: equivalent resistance of n resistors in parallel = R/n. Total resistance = R/n + R = R(1 + n)/n. Current = E / [R(n+1)/n] = nE / [R(n+1)] = nI₀ where I₀ = E/[R(n+1)]. So parallel current = nE/[R(n+1)] and series current I = E/[R(n+1)]. Therefore parallel current = nI. Given nI = 10I → n = 10.
Q94mcq
An em wave is propagating in a medium with a velocity V = V î. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along
a.
+ z direction
Correct Answer
b.
– y direction
c.
– z direction
d.
– x direction
Step-by-Step AI Explanation
Verified Method
For an electromagnetic wave, the relation is: direction of propagation = E × B. Here propagation is along +x (î) and E is along +y (ĵ). So: î = ĵ × B̂. This gives B̂ = k̂ (+ z direction), since ĵ × k̂ = î. Hence the magnetic field oscillates along the +z direction.
Q94mcq
An EM wave is propagating in a medium with a velocity V = V î. The instantaneous oscillating electric field of this EM wave is along +y axis. Then the direction of oscillating magnetic field of the EM wave will be along
a.
+z direction
Correct Answer
b.
–y direction
c.
–z direction
d.
–x direction
Step-by-Step AI Explanation
Verified Method
For an EM wave, the electric field (E), magnetic field (B), and propagation direction (k) are mutually perpendicular and follow the right-hand rule: k̂ × Ê = B̂. Here k̂ = î and Ê = ĵ, so B̂ = î × ĵ = k̂, i.e., +z direction.
Q95mcq
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
a.
138.88 H
b.
1.389 H
c.
0.138 H
d.
13.89 H
Correct Answer
Step-by-Step AI Explanation
Verified Method
Energy stored in inductor: U = ½LI². So L = 2U/I² = 2 × 25×10⁻³ / (60×10⁻³)² = 0.05 / (3.6×10⁻³) = 13.89 H.
Q95mcq
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
a.
138.88 H
b.
1.389 H
c.
0.138 H
d.
13.89 H
Correct Answer
Step-by-Step AI Explanation
Verified Method
Energy stored in an inductor: U = ½LI². So L = 2U/I² = 2 × 25 × 10⁻³ / (60 × 10⁻³)² = 50 × 10⁻³ / (3600 × 10⁻⁶) = 50 × 10⁻³ / 3.6 × 10⁻³ = 50/3.6 ≈ 13.89 H.
Q96mcq
The refractive index of the material of a prism is √2 and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is
a.
45°
Correct Answer
b.
30°
c.
60°
d.
zero
Step-by-Step AI Explanation
Verified Method
For the light to retrace its path after reflection from the silvered surface, it must hit that surface at 90° (normal incidence). The refracted ray inside the prism must make an angle of (90° – 30°) = 60° with the normal to the entry face, i.e., the angle of refraction r = 30°. Applying Snell's law at entry: sin i = μ sin r = √2 × sin 30° = √2 × 0.5 = 1/√2, so i = 45°.
Q97mcq
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be
a.
36 cm away from the mirror
Correct Answer
b.
30 cm towards the mirror
c.
30 cm away from the mirror
d.
36 cm towards the mirror
Step-by-Step AI Explanation
Verified Method
Using mirror formula 1/v + 1/u = 1/f with sign convention (distances measured from pole, negative for real objects/images in front of mirror). Initially u = –40 cm, f = –15 cm: 1/v = 1/f – 1/u = –1/15 + 1/40 = (–8+3)/120 = –5/120, so v = –24 cm. After displacement, u = –20 cm: 1/v = –1/15 + 1/20 = (–4+3)/60 = –1/60, so v = –60 cm. Displacement of image = 60 – 24 = 36 cm, and the image moves away from the mirror (further from –24 to –60).
Q98mcq
An electron of mass m with an initial velocity V = V₀ î (V₀ > 0) enters an electric field E = –E₀ î (E₀ = constant > 0) at t = 0. If λ₀ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is
a.
λ₀ / (1 + eE₀t / mV₀)
b.
λ₀ t
c.
λ₀ (1 + eE₀t / mV₀)
Correct Answer
d.
λ₀
Step-by-Step AI Explanation
Verified Method
The electric field E = –E₀ î decelerates the electron (force F = –eE = eE₀ î, but wait: force on electron = –e × E = –e(–E₀ î) = eE₀ î, which accelerates it). Actually force on electron (charge –e) in field –E₀ î is F = (–e)(–E₀ î) = eE₀ î, so it accelerates. Velocity at time t: v(t) = V₀ + (eE₀/m)t. de-Broglie wavelength λ = h/mv. So λ(t) = h/[m(V₀ + eE₀t/m)] = h/(mV₀) × 1/(1 + eE₀t/mV₀) = λ₀/(1 + eE₀t/mV₀). This matches option (1)/choice_a. However, re-examining: the field is –E₀ î and electron has charge –e, so force = (–e)(–E₀ î) = eE₀ î (same direction as motion). So momentum increases and wavelength decreases. Answer is choice_a: λ₀/(1 + eE₀t/mV₀).
Q99mcq
When the light of frequency 2ν₀ (where ν₀ is threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is v₁. When the frequency of the incident radiation is increased to 5ν₀, the maximum velocity of electrons emitted from the same plate is v₂. The ratio of v₁ to v₂ is
a.
1 : 4
b.
4 : 1
c.
1 : 2
Correct Answer
d.
2 : 1
Step-by-Step AI Explanation
Verified Method
Using photoelectric equation: ½mv² = h(ν – ν₀). For frequency 2ν₀: ½mv₁² = h(2ν₀ – ν₀) = hν₀. For frequency 5ν₀: ½mv₂² = h(5ν₀ – ν₀) = 4hν₀. Dividing: v₁²/v₂² = 1/4, so v₁/v₂ = 1/2, i.e., ratio is 1:2.
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For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
a.
10
b.
30
c.
20
d.
15
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Unpolarised light is incident from air on a plane surface of a material of refractive index 'μ'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
a.
Reflected light is polarised with its electric vector perpendicular to the plane of incidence
b.
i = sin⁻¹(1/μ)
c.
Reflected light is polarised with its electric vector parallel to the plane of incidence
d.
i = tan⁻¹(1/μ)
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In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to
a.
1.9 mm
b.
2.1 mm
c.
1.8 mm
d.
1.7 mm
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An inductor 20 mH, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314t. The power loss in the circuit is
a.
0.43 W
b.
2.74 W
c.
0.79 W
d.
1.13 W
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An inductor 20 mH, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf V = 10 sin 314t. The power loss in the circuit is
a.
0.43 W
b.
2.74 W
c.
0.79 W
d.
1.13 W
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A metallic rod of mass per unit length 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
a.
5.98 A
b.
14.76 A
c.
7.14 A
d.
11.32 A
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A metallic rod of mass per unit length 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
a.
5.98 A
b.
14.76 A
c.
7.14 A
d.
11.32 A
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Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
a.
25 Ω
b.
250 Ω
c.
40 Ω
d.
500 Ω
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Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
a.
25 Ω
b.
250 Ω
c.
40 Ω
d.
500 Ω
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A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from
a.
the magnetic field
b.
the lattice structure of the material of the rod
c.
the current source
d.
the induced electric field due to the changing magnetic field
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A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is
a.
339 m/s
b.
350 m/s
c.
330 m/s
d.
300 m/s
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An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is
a.
5 times greater
b.
10 times greater
c.
smaller
d.
equal
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A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s² at a distance of 5 m from the mean position. The time period of oscillation is
a.
π s
b.
2 s
c.
2π s
d.
1 s
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The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is
a.
2/3
b.
1/3
c.
2/5
d.
2/7
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At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given: Mass of oxygen molecule m = 2.76 × 10⁻²⁶ kg, Boltzmann's constant k_B = 1.38 × 10⁻²³ J K⁻¹)
a.
8.360 × 10⁴ K
b.
5.016 × 10⁴ K
c.
2.508 × 10⁴ K
d.
1.254 × 10⁴ K
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The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
a.
8 cm
b.
12.5 cm
c.
13.2 cm
d.
16 cm
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The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
a.
8 cm
b.
12.5 cm
c.
13.2 cm
d.
16 cm
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Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by l on applying a force F, how much force is needed to stretch the second wire by the same amount?
a.
6 F
b.
4 F
c.
9 F
d.
F
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Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?
a.
6 F
b.
4 F
c.
9 F
d.
F
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A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ N m⁻²) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is
a.
208.7 J
b.
42.2 J
c.
104.3 J
d.
84.5 J
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A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ Nm⁻²) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is
a.
208.7 J
b.
42.2 J
c.
104.3 J
d.
84.5 J
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The power radiated by a black body is P and it radiates maximum energy at wavelength λ₀. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength (4/3)λ₀, the power radiated by it becomes nP. The value of n is
a.
3/4
b.
81/256
c.
4/3
d.
256/81
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A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to
a.
r²
b.
r⁵
c.
r³
d.
r⁴
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The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K_A, K_B and K_C, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then
a.
K_A > K_B > K_C
b.
K_B < K_A < K_C
c.
K_A < K_B < K_C
d.
K_B > K_A > K_C
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A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K_t) as well as rotational kinetic energy (K_r) simultaneously. The ratio K_t : (K_t + K_r) for the sphere is
a.
5 : 7
b.
10 : 7
c.
7 : 10
d.
2 : 5
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If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
a.
Walking on the ground would become more difficult.
b.
Time period of a simple pendulum on the Earth would decrease.
c.
Raindrops will fall faster.
d.
'g' on the Earth will not change.
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A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
a.
Moment of inertia
b.
Rotational kinetic energy
c.
Angular velocity
d.
Angular momentum
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A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
a.
1 m/s, 3 m/s
b.
1 m/s, 3.5 m/s
c.
2 m/s, 4 m/s
d.
1.5 m/s, 3 m/s
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A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
a.
1 m/s, 3 m/s
b.
1 m/s, 3.5 m/s
c.
2 m/s, 4 m/s
d.
1.5 m/s, 3 m/s
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A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is
a.
a = g/sinθ
b.
a = g cosθ
c.
a = g/cosθ
d.
a = g tanθ
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A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is
a.
a = g/sinθ
b.
a = g cosθ
c.
a = g/cosθ
d.
a = g tanθ
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A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is
a.
0.525 cm
b.
0.053 cm
c.
0.521 cm
d.
0.529 cm
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Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation
a.
W_A > W_B > W_C
b.
W_B > W_A > W_C
c.
W_C > W_B > W_A
d.
W_A > W_C > W_B
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A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
a.
0.25
b.
0.8
c.
0.5
d.
0.4
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A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to
a.
D
b.
5D/7
c.
2D/3
d.
5D/4
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Given van der Waals constant for NH₃, H₂, O₂ and CO₂ are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
a.
H₂
b.
O₂
c.
NH₃
d.
CO₂
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Given van der Waals constant 'a' for NH₃, H₂, O₂ and CO₂ are respectively 4.17, 0.244, 1.36 and 3.59. Which one of the following gases is most easily liquefied?
a.
H₂
b.
O₂
c.
NH₃
d.
CO₂
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