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Question 1Physics

A body starts from rest and moves with uniform acceleration. The ratio of the distance covered in the first 2 s to that covered in the first 4 s is

  1. 1 : 2
  2. 1 : 4 Correct answer.
  3. 1 : 3
  4. 1 : 8

Explanation

For motion from rest with uniform acceleration, distance is proportional to the square of the time, s = ½at². So the ratio is (2²) : (4²) = 4 : 16 = 1 : 4. The result depends only on the squared time ratio, not on the value of the acceleration.

Question 2Physics

A stone is thrown vertically upward and returns to the thrower's hand. Taking upward as positive and neglecting air resistance, the acceleration of the stone at its highest point is

  1. 9.8 m/s² upward
  2. zero
  3. equal to its initial velocity
  4. 9.8 m/s² downward Correct answer.

Explanation

At the highest point the velocity is momentarily zero, but the acceleration is still that due to gravity, directed downward. Gravity acts throughout the flight, so the acceleration remains constant at about 9.8 m/s² downward. Zero velocity does not imply zero acceleration.

Question 3Physics

A car travels the first half of a straight road at 40 km/h and the second half at 60 km/h. Its average speed for the whole journey is

  1. 52 km/h
  2. 50 km/h
  3. 45 km/h
  4. 48 km/h Correct answer.

Explanation

For equal distances at speeds u and v, the average speed is the harmonic mean 2uv/(u+v). Substituting gives 2×40×60/100 = 48 km/h. The arithmetic mean 50 km/h is wrong because more time is spent on the slower half.

Question 4Physics

A body moves in a circle of radius 7 m at a constant speed of 11 m/s. Its centripetal acceleration is approximately

  1. 77 m/s²
  2. 1.57 m/s²
  3. 17.3 m/s² Correct answer.
  4. 121 m/s²

Explanation

Centripetal acceleration is a = v²/r. Here v²/r = 121/7 ≈ 17.3 m/s². The acceleration points toward the centre and changes only the direction of velocity, not its magnitude.

Question 5Physics

A block of mass 2 kg rests on a horizontal surface with coefficient of static friction 0.4 (g = 10 m/s²). The minimum horizontal force needed to just start sliding the block is

  1. 20 N
  2. 4 N
  3. 2 N
  4. 8 N Correct answer.

Explanation

The limiting static friction is f = μN = μmg for a horizontal surface. Here 0.4 × 2 × 10 = 8 N, so a force just above 8 N is needed to start motion. The normal force equals the weight because there is no vertical acceleration.

Question 6Physics

A person of mass 60 kg stands in a lift that accelerates upward at 2 m/s² (g = 10 m/s²). The reading of a weighing scale under the person is

  1. 720 N Correct answer.
  2. 120 N
  3. 600 N
  4. 480 N

Explanation

The scale reads the normal force N, found from N − mg = ma. Thus N = m(g + a) = 60 × 12 = 720 N. Upward acceleration makes the apparent weight larger than the true weight.

Question 7Physics

Two bodies of masses 4 kg and 6 kg attract each other with a gravitational force F. If the distance between them is doubled, the force becomes

  1. 2F
  2. 4F
  3. F/4 Correct answer.
  4. F/2

Explanation

Newton's law of gravitation gives F ∝ 1/r². Doubling the separation multiplies the force by 1/2² = 1/4. The masses are unchanged, so only the distance factor matters.

Question 8Physics

A satellite orbits Earth in a circular orbit of radius r. If the orbital radius is increased to 4r, the orbital speed becomes

  1. unchanged
  2. one-fourth the original speed
  3. twice the original speed
  4. half the original speed Correct answer.

Explanation

For a circular orbit, gravity supplies the centripetal force, giving v = √(GM/r). Since v ∝ 1/√r, increasing r fourfold divides the speed by √4 = 2. So the new speed is half the original.

Question 9Physics

A force of 20 N acts on a body and displaces it by 5 m. If the force makes an angle of 60° with the displacement, the work done is

  1. 25 J
  2. 0 J
  3. 100 J
  4. 50 J Correct answer.

Explanation

Work is W = Fs cos θ. Here 20 × 5 × cos 60° = 100 × 0.5 = 50 J. Only the component of force along the displacement does work.

Question 10Physics

A body of mass 2 kg moving at 3 m/s collides head-on and sticks to a stationary body of mass 4 kg. The speed of the combined mass after collision is

  1. 1 m/s Correct answer.
  2. 1.5 m/s
  3. 3 m/s
  4. 0.5 m/s

Explanation

In a perfectly inelastic collision, momentum is conserved: m₁u₁ = (m₁+m₂)v. Thus 2×3 = 6×v, giving v = 1 m/s. Kinetic energy is lost, but momentum is not.

Question 11Physics

A ball dropped from a height h rebounds to a height h/4. The coefficient of restitution between the ball and the floor is

  1. 0.75
  2. 0.5 Correct answer.
  3. 2.0
  4. 0.25

Explanation

The coefficient of restitution equals the square root of the ratio of rebound height to drop height. Here e = √(h/4 ÷ h) = √(1/4) = 0.5. It is a dimensionless number between 0 and 1.

Question 12Physics

A machine lifts a 50 kg load through 4 m in 2 s at constant speed (g = 10 m/s²). The power output of the machine is

  1. 200 W
  2. 500 W
  3. 2000 W
  4. 1000 W Correct answer.

Explanation

Work done against gravity is mgh = 50 × 10 × 4 = 2000 J. Power is work divided by time, 2000/2 = 1000 W. Constant speed means no kinetic energy is gained.

Question 13Physics

A solid sphere and a hollow cylinder of the same mass and radius roll down an incline from rest without slipping. Which reaches the bottom first and why?

  1. The hollow cylinder, because its larger moment of inertia speeds it up
  2. Both reach together, because their masses and radii are equal
  3. The solid sphere, because its smaller moment of inertia leaves more energy for translation Correct answer.
  4. The solid sphere, because it has greater gravitational force on it

Explanation

Rolling bodies accelerate at g sin θ/(1 + I/mR²). The solid sphere has I/mR² = 0.4, smaller than the hollow cylinder's 1.0, so a larger fraction of the released energy goes into translation and it descends faster. The result is independent of mass and radius.

Question 14Physics

A flywheel of moment of inertia 2 kg·m² is acted on by a constant torque of 6 N·m starting from rest. Its angular velocity after 4 s is

  1. 24 rad/s
  2. 3 rad/s
  3. 12 rad/s Correct answer.
  4. 6 rad/s

Explanation

Angular acceleration is α = τ/I = 6/2 = 3 rad/s². Using ω = αt gives ω = 3 × 4 = 12 rad/s. The torque produces uniform angular acceleration from rest.

Question 15Physics

A body floats with 3/4 of its volume submerged in a liquid of density 800 kg/m³. The density of the body is

  1. 200 kg/m³
  2. 600 kg/m³ Correct answer.
  3. 800 kg/m³
  4. 1067 kg/m³

Explanation

For a floating body, the fraction submerged equals the ratio of body density to liquid density. Thus ρ_body = (3/4) × 800 = 600 kg/m³. The buoyant force balances the weight.

Question 16Physics

Water flows steadily through a horizontal pipe whose cross-section narrows from 8 cm² to 2 cm². If the speed in the wider section is 1 m/s, the speed in the narrow section is

  1. 16 m/s
  2. 4 m/s Correct answer.
  3. 2 m/s
  4. 0.25 m/s

Explanation

The equation of continuity gives A₁v₁ = A₂v₂ for an incompressible fluid. Thus v₂ = (8/2) × 1 = 4 m/s. The speed rises where the pipe narrows.

Question 17Physics

In a simple pendulum, the time period is measured for small oscillations. If the length is made four times larger, the time period becomes

  1. unchanged
  2. four times
  3. half
  4. twice Correct answer.

Explanation

The period of a simple pendulum is T = 2π√(L/g), so T ∝ √L. Quadrupling the length multiplies the period by √4 = 2. The period is independent of the mass of the bob.

Question 18Physics

A particle executes simple harmonic motion with amplitude A. At the instant its displacement is A/2 from the mean position, the ratio of its kinetic energy to its potential energy is

  1. 4 : 1
  2. 1 : 3
  3. 3 : 1 Correct answer.
  4. 1 : 1

Explanation

In SHM, kinetic energy is proportional to A² − x² and potential energy to x². At x = A/2, the ratio is (A² − A²/4) : (A²/4) = (3/4) : (1/4) = 3 : 1. Total energy is constant.

Question 19Physics

A wave of frequency 500 Hz travels in a medium with speed 350 m/s. Its wavelength is

  1. 1.4 m
  2. 0.7 m Correct answer.
  3. 0.35 m
  4. 700 m

Explanation

The wave relation is v = fλ, so λ = v/f = 350/500 = 0.7 m. Wavelength is the distance the wave advances in one period.

Question 20Physics

A source of sound moves toward a stationary observer. The observer hears a sound whose

  1. frequency is higher and wavelength is shorter Correct answer.
  2. frequency is lower and wavelength is longer
  3. frequency is higher and wavelength is longer
  4. frequency and wavelength are both unchanged

Explanation

As the source approaches, successive wavefronts are emitted from positions closer to the observer, so the observed wavelength is compressed. Since speed is fixed by the medium, a shorter wavelength means a higher observed frequency.

Question 21Physics

Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The beat frequency heard is

  1. 4 Hz Correct answer.
  2. 8 Hz
  3. 2 Hz
  4. 516 Hz

Explanation

Beat frequency equals the difference of the two frequencies, |260 − 256| = 4 Hz. The loudness rises and falls four times per second as the waves alternately reinforce and cancel.

Question 22Physics

The speed of sound in air is about 340 m/s. An echo is heard 0.6 s after a clap. The distance of the reflecting wall from the clap is

  1. 204 m
  2. 102 m Correct answer.
  3. 510 m
  4. 340 m

Explanation

The sound travels to the wall and back, so the total path is 2d = vt = 340 × 0.6 = 204 m. Hence d = 102 m. The factor of two accounts for the round trip.

Question 23Physics

A concave mirror of focal length 10 cm forms an image of an object placed 15 cm in front of it. The image distance is

  1. 10 cm behind the mirror
  2. 30 cm behind the mirror
  3. 30 cm in front of the mirror Correct answer.
  4. 6 cm in front of the mirror

Explanation

Using 1/f = 1/v + 1/u with f = −10 cm and u = −15 cm (sign convention) gives 1/v = −1/10 + 1/15 = −1/30, so v = −30 cm. The negative sign places a real image 30 cm in front of the mirror.

Question 24Physics

Light passes from water (refractive index 1.33) into air. The critical angle for total internal reflection is closest to

  1. 90°
  2. 61°
  3. 41°
  4. 49° Correct answer.

Explanation

The critical angle satisfies sin C = n₂/n₁ = 1/1.33 ≈ 0.752, giving C ≈ 49°. Beyond this angle, light travelling from the denser to the rarer medium is totally internally reflected.

Question 25Physics

A convex lens of focal length 20 cm forms a real image at 60 cm from the lens. The object distance from the lens is

  1. 60 cm
  2. 15 cm
  3. 40 cm
  4. 30 cm Correct answer.

Explanation

Using 1/f = 1/v − 1/u with a real image at v = 60 cm and f = 20 cm, 1/u = 1/60 − 1/20 = −1/30, so u = −30 cm. The object lies 30 cm in front of the lens.

Question 26Physics

A stone thrown vertically upward returns to the thrower's hand after 4 s (g = 10 m/s², air resistance neglected). The maximum height reached above the hand is

  1. 80 m
  2. 40 m
  3. 20 m Correct answer.
  4. 10 m

Explanation

The time to reach the top is half the total flight time, 2 s, and at the top the velocity is zero. Using h = ½gt² with t = 2 s gives h = ½×10×4 = 20 m.

Question 27Physics

A body starts from rest and moves with uniform acceleration. The ratio of the distance covered in the first 2 s to that covered in the next 2 s is

  1. 1 : 4
  2. 1 : 1
  3. 1 : 3 Correct answer.
  4. 1 : 2

Explanation

From rest with uniform acceleration, distances in successive equal time intervals are in the ratio 1 : 3 : 5 : …, the odd-number ratio. So the first 2 s and the next 2 s give 1 : 3.

Question 28Physics

A vector of magnitude 5 units makes an angle of 60° with the x-axis. The magnitude of its component along the y-axis is closest to

  1. 5.0 units
  2. 4.3 units Correct answer.
  3. 2.5 units
  4. 0 units

Explanation

The y-component is 5 sin 60° = 5 × 0.866 ≈ 4.3 units. The x-component would be 5 cos 60° = 2.5 units.

Question 29Physics

A block of mass 5 kg rests on a rough horizontal floor with coefficient of static friction 0.4 (g = 10 m/s²). A horizontal force of 15 N is applied to the block. The frictional force acting on the block is

  1. 5 N
  2. 20 N
  3. 0 N
  4. 15 N Correct answer.

Explanation

The limiting static friction is μmg = 0.4×5×10 = 20 N. Since the applied 15 N is below this limit, the block stays at rest and static friction adjusts to equal the applied force, 15 N.

Question 30Physics

Two forces of 6 N and 8 N act at a point at right angles to each other. The magnitude of their resultant is

  1. 14 N
  2. 7 N
  3. 2 N
  4. 10 N Correct answer.

Explanation

For perpendicular forces the resultant magnitude is √(6² + 8²) = √100 = 10 N. This is the Pythagorean combination of the two force vectors.

Question 31Physics

A particle moves in a circle of radius 2 m at a constant speed of 4 m/s. Its centripetal acceleration is

  1. 8 m/s² Correct answer.
  2. 2 m/s²
  3. 16 m/s²
  4. 4 m/s²

Explanation

Centripetal acceleration is v²/r = 16/2 = 8 m/s², directed toward the centre. The speed is constant, so there is no tangential acceleration.

Question 32Physics

A body of weight W is pulled up a rough inclined plane of angle θ at constant speed by a force parallel to the plane. If friction is neglected, the pulling force needed is

  1. W
  2. W tan θ
  3. W cos θ
  4. W sin θ Correct answer.

Explanation

Resolving the weight along the incline gives the component W sin θ acting down the slope. At constant speed with no friction, the applied force must exactly balance this component.

Question 33Physics

A body of mass 2 kg moves in a straight line with velocity changing from 4 m/s to 10 m/s in 3 s. The net force acting on it is

  1. 6 N
  2. 12 N
  3. 4 N Correct answer.
  4. 2 N

Explanation

Acceleration is (10 − 4)/3 = 2 m/s², and Newton's second law gives F = ma = 2×2 = 4 N. The force equals mass times the rate of change of velocity.

Question 34Physics

A man of mass 70 kg stands on the floor of a lift moving downward with a constant velocity of 3 m/s (g = 10 m/s²). The normal force exerted by the floor on him is

  1. 210 N
  2. 700 N Correct answer.
  3. 490 N
  4. 910 N

Explanation

Constant velocity means zero acceleration, so the net force on the man is zero. The normal force therefore equals his weight, mg = 70×10 = 700 N.

Question 35Physics

A rocket in deep space expels gas backward at high speed. The rocket accelerates forward because

  1. the gas pushes against the vacuum of space
  2. the momentum of the expelled gas is balanced by the forward momentum of the rocket Correct answer.
  3. gravity from nearby stars pulls the rocket forward
  4. the rocket's mass decreases, so it speeds up at constant momentum

Explanation

By conservation of momentum of the rocket–gas system, the backward momentum of the ejected gas is equal and opposite to the forward momentum gained by the rocket. No external medium is needed.

Question 36Physics

A body of mass 2 kg moving at 6 m/s collides head-on with a stationary body of mass 4 kg and sticks to it. The kinetic energy lost in the collision is

  1. 0 J
  2. 12 J
  3. 36 J
  4. 24 J Correct answer.

Explanation

Momentum conservation gives v = (2×6)/6 = 2 m/s. Initial KE = ½×2×36 = 36 J, final KE = ½×6×4 = 12 J, so the loss is 24 J.

Question 37Physics

A body of mass 1 kg is at rest at a height of 5 m above the ground (g = 10 m/s²). It falls freely. Its kinetic energy just before hitting the ground is

  1. 5 J
  2. 10 J
  3. 50 J Correct answer.
  4. 100 J

Explanation

By conservation of energy, the potential energy mgh = 1×10×5 = 50 J converts fully into kinetic energy at ground level. So KE just before impact is 50 J.

Question 38Physics

A uniform metre rule is pivoted at its centre. A 20 g mass hangs 20 cm to the left of the pivot. To balance the rule, the mass that must hang 40 cm to the right of the pivot is

  1. 40 g
  2. 10 g Correct answer.
  3. 20 g
  4. 80 g

Explanation

For rotational balance the moments must be equal: 20 g × 20 cm = m × 40 cm, giving m = 10 g. The longer lever arm needs a smaller mass.

Question 39Physics

A body of mass 2 kg is tied to a string and whirled in a vertical circle of radius 1 m. At the lowest point its speed is 6 m/s (g = 10 m/s²). The tension in the string at that point is

  1. 20 N
  2. 72 N
  3. 52 N
  4. 92 N Correct answer.

Explanation

At the lowest point the tension acts upward and must supply the centripetal force plus balance the weight: T = mv²/r + mg = 2×36/1 + 20 = 92 N.

Question 40Physics

A body of mass 2 kg is displaced 3 m along the x-direction by a force of 10 N directed at 120° to the x-axis. The work done by the force is

  1. 30 J
  2. 15 J
  3. −15 J Correct answer.
  4. −30 J

Explanation

Work is Fd cos θ = 10×3×cos 120° = 30×(−0.5) = −15 J. The negative sign shows the force opposes the displacement.

Question 41Physics

A solid cylinder of mass 2 kg and radius 0.5 m rotates about its central axis. Its moment of inertia is

  1. 0.125 kg·m²
  2. 0.25 kg·m² Correct answer.
  3. 1.00 kg·m²
  4. 0.50 kg·m²

Explanation

For a solid cylinder about its central axis, I = ½MR² = ½×2×0.25 = 0.25 kg·m².

Question 42Physics

A wheel rotating at 20 rad/s is brought to rest uniformly in 5 s. The angular displacement during this time is

  1. 100 rad
  2. 50 rad Correct answer.
  3. 4 rad
  4. 20 rad

Explanation

With uniform angular deceleration, the average angular velocity is (20 + 0)/2 = 10 rad/s. The angular displacement is 10×5 = 50 rad.

Question 43Physics

A body of density 900 kg/m³ floats in a liquid of density 1200 kg/m³. The fraction of the body's volume that remains above the liquid surface is

  1. 0.50
  2. 0.75
  3. 0.33
  4. 0.25 Correct answer.

Explanation

The submerged fraction equals the density ratio 900/1200 = 0.75, so the fraction above the surface is 1 − 0.75 = 0.25.

Question 44Physics

A hydraulic press has a small piston of area 5 cm² and a large piston of area 200 cm². A force of 50 N on the small piston produces a force on the large piston of

  1. 2000 N Correct answer.
  2. 250 N
  3. 50 N
  4. 500 N

Explanation

By Pascal's law the pressure is transmitted undiminished, so F₂ = F₁(A₂/A₁) = 50×(200/5) = 2000 N.

Question 45Physics

A body floats in water with 60% of its volume submerged. When placed in another liquid it floats with 40% submerged. The density of the second liquid is

  1. 667 kg/m³
  2. 800 kg/m³
  3. 1500 kg/m³ Correct answer.
  4. 1000 kg/m³

Explanation

The body's density is 0.6×1000 = 600 kg/m³. In the second liquid, 0.4ρ = 600, so ρ = 1500 kg/m³. A less-submerged body indicates a denser liquid.

Question 46Physics

A steel ball sinks in water but a steel ship floats. The ship floats because

  1. water pressure on the bottom is greater than on the top of the hull
  2. its large hollow shape displaces water whose weight equals the ship's weight Correct answer.
  3. the ship's engines push it upward against gravity
  4. steel becomes less dense than water when shaped into a ship

Explanation

A hollow hull encloses air, so the ship's average density is less than that of water. It floats by displacing a volume of water whose weight equals its own weight.

Question 47Physics

A particle executes simple harmonic motion with amplitude 5 cm and angular frequency 2 rad/s. Its maximum speed is

  1. 5 cm/s
  2. 20 cm/s
  3. 10 cm/s Correct answer.
  4. 2.5 cm/s

Explanation

The maximum speed in SHM is v_max = Aω = 5×2 = 10 cm/s, attained as the particle passes through the mean position.

Question 48Physics

A simple pendulum has a time period of 2 s on Earth. Taken to a planet where acceleration due to gravity is four times that on Earth, its time period becomes

  1. 0.5 s
  2. 2 s
  3. 4 s
  4. 1 s Correct answer.

Explanation

The period T = 2π√(L/g) varies as 1/√g. Quadrupling g halves the period, so T = 2/2 = 1 s.

Question 49Physics

Water flows through a horizontal pipe that narrows from a wider section to a narrower one. In the narrower section, compared with the wider section, the flow speed is

  1. higher and the pressure lower Correct answer.
  2. higher and the pressure higher
  3. lower and the pressure lower
  4. the same and the pressure the same

Explanation

By the equation of continuity, a smaller cross-section gives a higher flow speed. By Bernoulli's principle for a horizontal pipe, higher speed corresponds to lower pressure.

Question 50Physics

The dimensional formula of the universal gravitational constant G is

  1. [M⁻²L³T⁻²]
  2. [ML³T⁻²]
  3. [M⁻¹L³T⁻²] Correct answer.
  4. [M⁻¹L²T⁻²]

Explanation

From Newton's law F = Gm₁m₂/r², we get G = Fr²/(m₁m₂). Substituting the dimensions of force [MLT⁻²], length squared [L²] and mass squared [M²] gives [ML³T⁻²]/[M²] = [M⁻¹L³T⁻²].