Physics MCQs

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Physics MCQs

Practice the latest Physics MCQs with answers and detailed explanations. This section includes chapter-wise multiple-choice questions covering mechanics, motion, force, work and energy, heat, light, electricity, magnetism, modern physics, waves, optics, and other important topics. These exam-oriented MCQs are ideal for Class 9–12, NEET, JEE, CUET, SSC, Banking, Railway, JKSSB, Defence, Police, and other competitive exams.

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Question #161
Which of the following physical quantities has the dimensional formula [M^1 L^2 T^-3 A^-2]?
A. Electrical Resistance
B. Electrical Capacitance
C. Electrical Conductance
D. Magnetic Flux
Discuss
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Correct Answer: Option A

Explanation:
Electrical resistance (R) is Voltage/Current. Voltage (V) is Work/Charge. Work is [ML^2T^-2] and Charge is [AT]. So, V = [ML^2T^-3A^-1]. Therefore, R = V/A = [ML^2T^-3A^-2]. Memorizing the dimensional formula for resistance is highly recommended for competitive exams.
This question belongs to: Science Physics
Question #162
If the error in the measurement of the momentum of a particle is +100%, what will be the percentage error in the measurement of its kinetic energy?
A. 300%
B. 100%
C. 400%
D. 200%
Discuss
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Correct Answer: Option A

Explanation:
Kinetic Energy (K) is given by K = p^2 / 2m. If momentum (p) increases by 100%, the new momentum is p' = p + 1.0p = 2p. The new kinetic energy is K' = (2p)^2 / 2m = 4(p^2 / 2m) = 4K. The increase is 4K - K = 3K. Percentage increase = (3K/K) * 100 = 300%.
This question belongs to: Science Physics
Question #163
One Parsec is a unit of astronomical distance. It is defined as the distance at which a radius of one astronomical unit (AU) subtends an angle of one second of arc. One parsec is approximately equal to:
A. 9.46 x 10^15 meters
B. 3.0 x 10^8 meters
C. 3.26 light years
D. 1.5 x 10^11 meters
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Correct Answer: Option C

Explanation:
The parsec (parallax second) is the largest standard unit of distance used in astronomy. 1 Astronomical Unit (AU) is approx 1.5 x 10^11 m. 1 Light Year (ly) is approx 9.46 x 10^15 m. 1 Parsec is approximately 3.08 x 10^16 m. Converting parsecs to light years: (3.08 x 10^16) / (9.46 x 10^15) ≈ 3.26 light years.
This question belongs to: Science Physics
Question #164
A particle moves such that its position coordinates (x, y) are given by x = 2t and y = t^2. The equation of its trajectory is:
A. y = x
B. y = x^2 / 2
C. y = x^2 / 4
D. y = 2x^2
Discuss
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Correct Answer: Option C

Explanation:
The equation of trajectory represents the path of the particle in the x-y plane, eliminating the time variable 't'. Given x = 2t, we can write t = x/2. Substitute this into the y equation: y = t^2 = (x/2)^2 = x^2 / 4. This confirms the particle moves in a parabolic path.
This question belongs to: Science Physics
Question #165
A body covers 12 m in the 2nd second and 20 m in the 4th second of its motion. Assuming uniform acceleration, what is its initial velocity?
A. 4 m/s
B. 6 m/s
C. 2 m/s
D. 8 m/s
Discuss
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Correct Answer: Option B

Explanation:
The distance covered in the nth second is Sn = u + a/2 * (2n - 1). For n=2: 12 = u + a/2 * (3) => 2u + 3a = 24. For n=4: 20 = u + a/2 * (7) => 2u + 7a = 40. Subtracting equations: 4a = 16 => a = 4 m/s^2. Substitute 'a' back: 2u + 3(4) = 24 => 2u = 12 => u = 6 m/s.
This question belongs to: Science Physics
Question #166
In non-uniform circular motion, the net acceleration of the particle is directed:
A. Strictly towards the center
B. Outwards, away from the center
C. Strictly along the tangent to the circle
D. At an angle to both the radius and the tangent
Discuss
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Correct Answer: Option D

Explanation:
In non-uniform circular motion, the speed of the particle changes. This requires a tangential acceleration (at) to change the speed, and a centripetal/radial acceleration (ac) to change the direction. The net acceleration is the vector sum of these two mutually perpendicular components, so it points inward at an angle to the radius.
This question belongs to: Science Physics
Question #167
If a stone is dropped from a balloon rising upwards with a velocity of 10 m/s at a height of 75 m, what is the initial velocity of the stone relative to the ground?
A. 9.8 m/s downwards
B. 10 m/s upwards
C. 0 m/s
D. 10 m/s downwards
Discuss
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Correct Answer: Option B

Explanation:
Due to the inertia of motion, when an object is dropped from a moving vehicle (or balloon), it inherits the instantaneous velocity of that vehicle. Since the balloon was moving upwards at 10 m/s, the stone's initial velocity relative to the ground is 10 m/s upwards. It will rise slightly before falling.
This question belongs to: Science Physics
Question #168
Assertion (A): The coefficient of kinetic friction is always strictly less than the coefficient of limiting static friction. Reason (R): Once motion starts, the inertia of rest is broken and interlocking of surface irregularities is less effective.
A. Both A and R are true and R is the correct explanation of A.
B. A is true but R is false.
C. A is false but R is true.
D. Both A and R are true but R is NOT the correct explanation of A.
Discuss
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Correct Answer: Option A

Explanation:
Static friction opposes impending motion, reaching a maximum called limiting friction. Once the object moves, it doesn't have enough time for the microscopic irregularities (asperities) of the two surfaces to interlock strongly. Thus, the force required to keep it moving (kinetic friction) is slightly less than the force required to start the motion.
This question belongs to: Science Physics
Question #169
Two masses, M1 = 3 kg and M2 = 2 kg, are connected by a light inextensible string passing over a frictionless pulley (Atwood machine). The acceleration of the system is (Take g = 10 m/s^2):
A. 10 m/s^2
B. 4 m/s^2
C. 5 m/s^2
D. 2 m/s^2
Discuss
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Correct Answer: Option D

Explanation:
For an Atwood machine, the common acceleration of both masses is given by the formula a = [(M1 - M2) / (M1 + M2)] * g. Substituting the given values: a = [(3 - 2) / (3 + 2)] * 10 = (1 / 5) * 10 = 2 m/s^2. The heavier mass accelerates downwards, the lighter one upwards.
This question belongs to: Science Physics
Question #170
A particle moves in a conservative force field where its potential energy U is given by U(x) = ax^2 - bx. The particle will be in stable equilibrium at position x equal to:
A. b / 2a
B. 2b / a
C. b / a
D. a / 2b
Discuss
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Correct Answer: Option A

Explanation:
Equilibrium occurs where the net force is zero, meaning dU/dx = 0. Given U(x) = ax^2 - bx, taking the derivative: dU/dx = 2ax - b = 0. Therefore, x = b / 2a. To confirm it's stable, the second derivative d^2U/dx^2 must be positive. d^2U/dx^2 = 2a. Assuming 'a' is positive, it represents a stable equilibrium point.
This question belongs to: Science Physics
Question #171
A variable force F = (3x^2 + 2x) N acts on a particle. The work done by this force in moving the particle from x = 1 m to x = 2 m is:
A. 10 J
B. 14 J
C. 16 J
D. 12 J
Discuss
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Correct Answer: Option A

Explanation:
Work done by a variable force is the definite integral of F dx. W = Integral(3x^2 + 2x) dx from 1 to 2. W = [x^3 + x^2] evaluated from 1 to 2. Upper limit: (2^3 + 2^2) = 8 + 4 = 12. Lower limit: (1^3 + 1^2) = 1 + 1 = 2. Work = 12 - 2 = 10 Joules.
This question belongs to: Science Physics
Question #172
An engine pumps water continuously through a hose. Water leaves the hose with a velocity 'v' and mass per unit length is 'm'. What is the power imparted to the water?
A. mv^2
B. 1/2 mv^3
C. 1/2 m^2v
D. mv^3
Discuss
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Correct Answer: Option B

Explanation:
Power is the rate of doing work or imparting kinetic energy (dK/dt). Mass exiting per second = m * v (mass/length * length/time). Kinetic energy imparted per second = 1/2 * (mass per second) * v^2 = 1/2 * (mv) * v^2 = 1/2 * m * v^3. This is a very common advanced mechanics question.
This question belongs to: Science Physics
Question #173
Kepler's Second Law of planetary motion (Law of Areas) is a direct consequence of the conservation of:
A. Kinetic Energy
B. Angular Momentum
C. Total Mechanical Energy
D. Linear Momentum
Discuss
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Correct Answer: Option B

Explanation:
Kepler's Second Law states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time (areal velocity is constant). Because the gravitational force is a central force (acting strictly along the radius vector), it exerts zero torque on the planet. When net torque is zero, angular momentum is strictly conserved.
This question belongs to: Science Physics
Question #174
Due to the Earth's rotation, the acceleration due to gravity (g) is minimum at:
A. The North Pole
B. The Equator
C. A latitude of 45 degrees
D. The South Pole
Discuss
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Correct Answer: Option B

Explanation:
The effective acceleration due to gravity at latitude lambda is g' = g - R * w^2 * cos^2(lambda), where w is angular velocity. At the equator, lambda = 0, so cos(0) = 1. This subtracts the maximum outward centrifugal acceleration (Rw^2) from g. At the poles, lambda = 90, centrifugal effect is zero. Thus, g is minimum exactly at the equator.
This question belongs to: Science Physics
Question #175
The gravitational potential strictly at the center of a solid, uniform Earth of mass M and radius R is:
A. -GM / R
B. Zero
C. -3GM / 2R
D. -GM / 2R
Discuss
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Correct Answer: Option C

Explanation:
The gravitational potential V inside a uniform solid sphere at a distance r from the center is V = -GM * (3R^2 - r^2) / (2R^3). At the exact center of the Earth, r = 0. Substituting r = 0 gives V = -GM(3R^2) / 2R^3 = -3GM / 2R. The potential is strictly non-zero and highly negative.
This question belongs to: Science Physics
Question #176
What is the mathematical relationship between the escape velocity (Ve) and the orbital velocity (Vo) of a satellite revolving very close to the surface of the Earth?
A. Ve = sqrt(2) * Vo
B. Ve = Vo
C. Ve = 2 * Vo
D. Vo = sqrt(2) * Ve
Discuss
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Correct Answer: Option A

Explanation:
The orbital velocity of a satellite very close to Earth is Vo = sqrt(gR). The escape velocity from Earth's surface is Ve = sqrt(2gR). Therefore, substituting Vo into the escape velocity equation yields Ve = sqrt(2) * Vo. A satellite requires a 41.4% increase in speed to escape orbit completely.
This question belongs to: Science Physics
Question #177
A geostationary satellite orbits the Earth from:
A. West to East
B. North to South
C. South to North
D. East to West
Discuss
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Correct Answer: Option A

Explanation:
A geostationary satellite must appear stationary to an observer on the Earth's surface. To achieve this, it must orbit in the exact equatorial plane and strictly follow the Earth's rotation. Since the Earth rotates on its axis from West to East, the geostationary satellite must also orbit from West to East.
This question belongs to: Science Physics
Question #178
According to the Equation of Continuity in fluid dynamics, when water flows through a pipe of varying cross-section, the velocity of the fluid is:
A. Directly proportional to the cross-sectional area.
B. Independent of the cross-sectional area.
C. Directly proportional to the square of the cross-sectional area.
D. Inversely proportional to the cross-sectional area.
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Correct Answer: Option D

Explanation:
The Equation of Continuity states that for an incompressible, non-viscous fluid in streamlined flow, the mass flow rate is constant. Therefore, A1V1 = A2V2, where A is the area of cross-section and V is the velocity. This means V is inversely proportional to A (V ∝ 1/A). As the pipe narrows, the fluid speeds up.
This question belongs to: Science Physics
Question #179
Torricelli's Law states that the velocity of efflux of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth 'h' is identical to:
A. The velocity acquired by a freely falling body dropped from height 'h'.
B. The critical velocity for turbulent flow.
C. The terminal velocity of the fluid.
D. The velocity of a sound wave in that fluid.
Discuss
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Correct Answer: Option A

Explanation:
Torricelli's Law is a direct application of Bernoulli's principle. It proves mathematically that the velocity of fluid exiting a small hole (efflux velocity v) is v = sqrt(2gh). This is exactly the same velocity a solid body would acquire if it fell freely under gravity from rest over the same vertical height 'h'.
This question belongs to: Science Physics
Question #180
What is the excess pressure strictly inside a soap bubble of radius R and surface tension T?
A. T / 2R
B. T / R
C. 4T / R
D. 2T / R
Discuss
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Correct Answer: Option C

Explanation:
A liquid drop has only one free surface, so its excess pressure is P = 2T/R. However, a soap bubble in the air has two completely free surfaces (an inner one and an outer one). Because both surfaces exert surface tension forces inward, the total excess pressure inside a soap bubble is exactly double: P = 4T/R.
This question belongs to: Science Physics

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