Physics MCQs

Question #521

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Which of the following is a derived unit in SI system?

A. Ampere

B. Kelvin

C. Mole

D. Newton

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Correct Answer: Option D

Explanation:
Newton is a derived unit representing force, defined as kg·m/s². Kelvin (temperature), Ampere (current), and Mole (amount of substance) are fundamental SI units. Derived units are combinations of fundamental units through physical relationships. Newton comes from Newton's second law: F = ma. Memory aid: Remember the seven fundamental SI units: meter, kilogram, second, ampere, kelvin, mole, candela. All others like newton, joule, watt are derived. This distinction is essential for unit conversion questions in competitive exams.

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Question #522

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A physical quantity is measured as 12.345 ± 0.005 units. The relative error in this measurement is approximately:

A. 0.4%

B. 4%

C. 0.004%

D. 0.04%

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Correct Answer: Option D

Explanation:
Relative error = (Absolute error) / (Measured value) = 0.005 / 12.345 ≈ 0.000405. Converting to percentage: 0.000405 × 100% ≈ 0.0405% ≈ 0.04%. Relative error quantifies precision independent of measurement scale. Absolute error (±0.005) indicates instrument precision. This concept is vital for comparing measurement reliability across different magnitudes. Memory tip: Percentage error = (Absolute error / True value) × 100. Competitive exams often test this calculation with varying significant figures to assess numerical proficiency.

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Question #523

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Which measuring instrument is most suitable for measuring the internal diameter of a narrow tube with high precision?

A. Screw gauge

B. Meter scale

C. Measuring tape

D. Vernier caliper

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Correct Answer: Option D

Explanation:
Vernier caliper is ideal for measuring internal diameters of narrow tubes due to its internal jaws designed for such measurements. Screw gauge measures external dimensions like wire diameter. Meter scale and measuring tape lack the precision (least count ~1 mm) needed for narrow tubes. Vernier calipers typically have least count 0.01 mm or 0.02 mm. Practical tip: Always use internal jaws for internal measurements, external jaws for outer dimensions. This application-based question tests knowledge of instrument selection in experimental physics, commonly featured in practical examination sections.

This question belongs to: Science Physics

Question #524

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A car travels from point A to B with speed 40 km/h and returns from B to A with speed 60 km/h. The average speed for the entire journey is:

A. 45 km/h

B. 52 km/h

C. 48 km/h

D. 50 km/h

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Correct Answer: Option C

Explanation:
Average speed = Total distance / Total time. Let distance AB = d km. Time for AB = d/40 h, time for BA = d/60 h. Total distance = 2d km. Total time = d/40 + d/60 = d(3+2)/120 = 5d/120 = d/24 h. Average speed = 2d / (d/24) = 48 km/h. Note: Average speed is not arithmetic mean of speeds (which would be 50 km/h) because time intervals differ. This harmonic mean concept is crucial for motion problems. Exam tip: For equal distances, average speed = 2v₁v₂/(v₁+v₂). Frequently tested in competitive exams.

This question belongs to: Science Physics

Question #525

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The area under a velocity-time graph represents:

A. Both displacement and distance travelled

B. Displacement

C. Acceleration

D. Distance travelled

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Correct Answer: Option A

Explanation:
The area under velocity-time graph gives displacement when direction is considered (signed area), and distance travelled when absolute values are taken (unsigned area). For unidirectional motion, both are equal. Velocity is rate of change of displacement, so integrating velocity over time yields displacement. This graphical interpretation is fundamental in kinematics. Memory tip: Slope of v-t graph = acceleration; area under v-t graph = displacement. Competitive exams often combine graph analysis with conceptual understanding to test depth of knowledge in motion topics.

This question belongs to: Science Physics

Question #526

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A body starts from rest and moves with uniform acceleration a. The ratio of distances covered in the 1st, 2nd, and 3rd seconds of motion is:

A. 1:4:9

B. 1:2:3

C. 1:3:5

D. 1:1:1

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Correct Answer: Option C

Explanation:
Distance covered in nth second: sₙ = u + a(2n-1)/2. With u=0, sₙ ∝ (2n-1). For n=1: 2(1)-1=1; n=2: 3; n=3:5. Thus ratio 1:3:5. This result holds for any uniformly accelerated motion starting from rest. The distances covered in successive equal time intervals follow odd number ratio. Memory aid: This is a standard result derivable from s = ut + ½at² by calculating s at t=n and t=n-1. Frequently appears in competitive exams testing equation of motion applications.

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Question #527

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A particle moves along a circular path of radius r with constant speed v. Its acceleration is:

A. Zero

B. v²/r directed away from center

C. v²/r directed towards center

D. v/r directed tangentially

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Correct Answer: Option C

Explanation:
In uniform circular motion, speed is constant but velocity direction changes continuously, producing centripetal acceleration. Magnitude a = v²/r, direction always towards the center of the circle. This acceleration changes velocity direction without altering speed. Tangential acceleration would change speed, but here speed is constant. Formula derivation: From vector analysis of velocity change over small time interval. Exam tip: Remember 'centripetal' means 'center-seeking'; this concept is fundamental for rotational dynamics questions in competitive exams.

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Question #528

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The position-time graph of a particle is a parabola opening upwards. This indicates that the particle has:

A. Decreasing acceleration

B. Increasing acceleration

C. Constant acceleration

D. Constant velocity

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Correct Answer: Option C

Explanation:
Position-time graph as parabola: x = x₀ + ut + ½at². This quadratic form indicates constant acceleration 'a'. If acceleration were changing, the graph would be cubic or higher order. Constant velocity would produce a straight line (linear graph). Upward opening parabola implies positive acceleration. Graph interpretation skill is essential: slope of x-t graph = velocity; curvature indicates acceleration. Memory tip: Parabolic x-t graph ⇔ constant acceleration; linear x-t graph ⇔ constant velocity. Competitive exams frequently test graph-concept correlations.

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Question #529

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A stone is dropped from a tower. The distance covered by it in the last second of its fall is equal to the distance covered in the first three seconds. The height of the tower is approximately: (g = 10 m/s²)

A. 125 m

B. 180 m

C. 200 m

D. 80 m

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Correct Answer: Option A

Explanation:
Distance in first 3 seconds: s₃ = ½gt² = ½×10×9 = 45 m. Let total time be n seconds. Distance in nth second: sₙ = u + g(2n-1)/2 = 0 + 5(2n-1) = 10n - 5. Given sₙ = 45 m ⇒ 10n - 5 = 45 ⇒ n = 5 s. Total height h = ½gn² = ½×10×25 = 125 m. This problem combines equation of motion with logical reasoning about time intervals. Exam tip: For free fall from rest, distance in nth second = 5(2n-1) meters when g=10 m/s². Such numerical problems test application skills in competitive exams.

This question belongs to: Science Physics

Question #530

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Two bodies A and B start from the same point with velocities 10 m/s and 15 m/s respectively in the same direction. If B starts 2 seconds after A, the time after which B catches A is:

A. 10 s

B. 4 s

C. 8 s

D. 6 s

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Correct Answer: Option D

Explanation:
When B starts, A has already traveled 10 m/s × 2 s = 20 m. Relative velocity of B w.r.t A = 15 - 10 = 5 m/s. Time to cover 20 m gap at 5 m/s relative speed = 20/5 = 4 s after B starts. Since B started 2 s late, total time from A's start = 2 + 4 = 6 s. Alternatively, equate distances: 10t = 15(t-2) ⇒ 10t = 15t - 30 ⇒ t = 6 s. This relative motion approach simplifies pursuit problems. Memory tip: In same-direction motion, use relative velocity = |v₁ - v₂|. Frequently tested in kinematics sections of competitive exams.

This question belongs to: Science Physics

Question #531

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The velocity-time graph of a particle moving in a straight line is shown. The displacement of the particle from t=0 to t=4s is: [Graph: triangle from (0,0) to (2,10) to (4,0)]

A. 20 m

B. 30 m

C. 10 m

D. 40 m

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Correct Answer: Option A

Explanation:
Displacement = area under velocity-time graph. The graph forms a triangle with base 4 s and height 10 m/s. Area = ½ × base × height = ½ × 4 × 10 = 20 m. Since velocity is always positive, displacement equals distance travelled. Graphical analysis is powerful: area gives displacement, slope gives acceleration. This triangle represents motion with uniform acceleration followed by uniform retardation. Exam tip: For piecewise linear v-t graphs, calculate area of each geometric segment separately. Such graph-based questions assess conceptual clarity in motion analysis.

This question belongs to: Science Physics

Question #532

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A body is projected vertically upwards with initial velocity u. The time taken to reach maximum height is:

A. u/g

B. 2u/g

C. u/(2g)

D. g/u

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Correct Answer: Option A

Explanation:
At maximum height, final velocity v = 0. Using v = u - gt (taking upward positive), 0 = u - gt ⇒ t = u/g. This is time of ascent. Total time of flight would be 2u/g. The equation derives from Newton's first equation of motion under constant acceleration g downward. Memory aid: Time to peak = initial velocity / gravitational acceleration. This fundamental result appears in projectile motion problems. Competitive exams often combine this with energy conservation or symmetry concepts for advanced questions.

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Question #533

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Which of the following statements about inertia is correct?

A. Inertia is a force that keeps bodies in motion

B. Inertia depends on the velocity of the body

C. Inertia decreases with increasing mass

D. Inertia is the property of a body to resist change in its state of motion

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Correct Answer: Option D

Explanation:
Inertia is the inherent property of matter that resists changes in its state of rest or uniform motion, as stated in Newton's first law. It depends solely on mass, not velocity. Inertia is not a force; forces cause changes in motion. Greater mass implies greater inertia. Statement C precisely defines inertia. Common misconception: confusing inertia with momentum (which depends on velocity). Exam tip: Remember 'inertia ∝ mass'; this concept underpins all Newtonian mechanics questions. Frequently tested in conceptual sections of competitive exams to identify fundamental understanding.

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Question #534

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A force of 10 N acts on a body of mass 2 kg initially at rest. The momentum acquired by the body after 5 seconds is:

A. 25 kg·m/s

B. 10 kg·m/s

C. 50 kg·m/s

D. 100 kg·m/s

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Correct Answer: Option C

Explanation:
Force = rate of change of momentum: F = Δp/Δt. Thus Δp = F × Δt = 10 N × 5 s = 50 kg·m/s. Since initial momentum was zero (body at rest), final momentum = 50 kg·m/s. Alternatively, acceleration a = F/m = 10/2 = 5 m/s², velocity after 5s: v = u + at = 0 + 5×5 = 25 m/s, momentum p = mv = 2×25 = 50 kg·m/s. Both methods confirm. Memory tip: Impulse (FΔt) equals change in momentum. This direct application of Newton's second law is essential for dynamics problems in competitive exams.

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Question #535

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According to Newton's third law, action and reaction forces:

A. Are always equal to weight of the body

B. Act on different bodies and never cancel each other

C. Act only when bodies are in contact

D. Act on the same body and cancel each other

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Correct Answer: Option B

Explanation:
Newton's third law states: for every action, there is an equal and opposite reaction. These forces act on two different interacting bodies, hence they never cancel each other as they don't act on the same body. They can act at a distance (e.g., gravitational forces) or through contact. Cancellation of forces occurs only when multiple forces act on a single body. This distinction is crucial for free-body diagram analysis. Exam tip: Always identify which body each force acts upon. Misconception about force cancellation is a common trap in competitive exam questions testing Newton's laws.

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Question #536

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A bullet of mass 20 g is fired from a gun of mass 2 kg with velocity 300 m/s. The recoil velocity of the gun is:

A. 0.6 m/s

B. 6 m/s

C. 0.3 m/s

D. 3 m/s

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Correct Answer: Option D

Explanation:
By conservation of momentum: initial momentum = 0 (system at rest). Final momentum: m_bullet×v_bullet + m_gun×v_gun = 0. Thus (0.02 kg)(300 m/s) + (2 kg)(v_gun) = 0 ⇒ 6 + 2v_gun = 0 ⇒ v_gun = -3 m/s. Magnitude is 3 m/s; negative sign indicates opposite direction to bullet. This demonstrates momentum conservation in isolated systems. Memory tip: Recoil velocity = -(m_bullet/m_gun)×v_bullet. Such numerical problems test application of conservation laws, frequently appearing in competitive exams with varying mass ratios and velocities.

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Question #537

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A body of mass 5 kg is moving with velocity 10 m/s. A constant force acts on it for 2 seconds, changing its velocity to 20 m/s. The magnitude of the force is:

A. 50 N

B. 100 N

C. 12.5 N

D. 25 N

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Correct Answer: Option D

Explanation:
Using Newton's second law: F = ma. First find acceleration: a = (v - u)/t = (20 - 10)/2 = 5 m/s². Then F = 5 kg × 5 m/s² = 25 N. Alternatively, using impulse-momentum theorem: FΔt = mΔv ⇒ F×2 = 5×(20-10) ⇒ F = 50/2 = 25 N. Both approaches confirm. This problem integrates kinematics with dynamics. Exam tip: When time interval is given, impulse-momentum method is often faster. Such questions assess ability to select appropriate physics principles for problem-solving in time-constrained exams.

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Question #538

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Which of the following scenarios best illustrates conservation of momentum?

A. A car accelerating on a straight road

B. A book resting on a table

C. A ball falling freely under gravity

D. Collision between two billiard balls on a frictionless table

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Correct Answer: Option D

Explanation:
Conservation of momentum applies to isolated systems with no external forces. In billiard ball collision on frictionless table, external forces (gravity, normal) cancel vertically, and no horizontal external forces act, so horizontal momentum is conserved. Book on table has zero momentum but no dynamic process. Falling ball has external gravity force. Accelerating car has external engine force. Thus only option B represents an isolated system where momentum conservation holds. Memory aid: Momentum conserved when net external force = 0. This conceptual question tests understanding of system boundaries and force analysis, crucial for mechanics sections in competitive exams.

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Question #539

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The work done in moving a charge of 2 C between two points with potential difference of 12 V is:

A. 24 J

B. 14 J

C. 0.167 J

D. 6 J

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Correct Answer: Option A

Explanation:
Work done W = charge (Q) × potential difference (V). Thus W = 2 C × 12 V = 24 J. This follows from definition of voltage: 1 volt = 1 joule per coulomb. The work done equals the change in electrical potential energy. This fundamental relationship is crucial in electricity. Memory tip: W = QV is analogous to mechanical work = force × distance. Such direct formula applications are common in competitive exams to test basic concept retention. Always verify units: coulomb × volt = joule, confirming dimensional consistency.

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Question #540

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A body is lifted vertically through a height h. The work done against gravity is stored as:

A. Kinetic energy

B. Gravitational potential energy

C. Elastic potential energy

D. Thermal energy

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Correct Answer: Option B

Explanation:
When a body is lifted against gravity, work done (mgh) is stored as gravitational potential energy. This energy can be recovered when the body falls. Kinetic energy relates to motion, thermal to heat, elastic to deformed materials. The conservation of energy principle states that work done against conservative forces becomes potential energy. Formula: PE = mgh, where h is height above reference. Memory aid: 'Potential' means stored energy due to position. This conceptual question tests energy transformation understanding, frequently appearing in work-energy theorem applications in competitive exams.

This question belongs to: Science Physics

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