All Free Physics MCQs with Answers

51. A car of mass 1000 kg takes a bend of radius 50 m at 10 m per second. The centripetal force needed is

• A. 200 N
• B. 2000 N
• C. 5000 N
• D. 10000 N

Explanation: The force is mv squared over r, which is 1000 times 100 divided by 50, giving 2000 N. On a level road this force must come from friction between the tyres and the surface, so if the required force exceeds the maximum friction available the car skids outwards. Doubling the speed would quadruple the force, which is why bends are so much more dangerous at speed.

52. The so-called centrifugal force experienced by a passenger on a turning bus is

• A. a real force acting outwards on the passenger
• B. the reaction to the passenger's weight
• C. an apparent force arising because the passenger is in a rotating, non inertial frame
• D. always larger than the centripetal force

Explanation: In the ground frame the passenger simply continues in a straight line by inertia while the bus turns beneath them, so no outward force is needed to explain the motion. The outward push is felt only because the observer is accelerating with the bus, which makes it a fictitious or pseudo force. The genuine force in the situation is the inward one supplied by the seat and the floor.

53. For a car to take a banked curve safely without relying on friction, the correct banking angle depends on

• A. the mass of the car only
• B. the speed and the radius of the curve
• C. the weight of the passengers
• D. the width of the road

Explanation: Setting the horizontal component of the normal force equal to mv squared over r gives tan theta equal to v squared over rg, and the mass cancels completely. That is why a banked track works equally well for a light car and a loaded truck at the same design speed. The angle is correct for one speed only, with friction covering the difference at others.

54. The time period T and the angular speed omega of circular motion are related by

• A. T equals omega over 2 pi
• B. T equals 2 pi omega
• C. T equals 2 pi over omega
• D. T equals omega squared

Explanation: One complete revolution covers 2 pi radians, so the time taken is 2 pi divided by the angular speed. A larger omega therefore means a shorter period, which is the inverse relationship the formula expresses. Frequency, being the reciprocal of the period, is omega divided by 2 pi.

55. The moment of inertia of a body depends on

• A. its mass only
• B. its mass and how that mass is distributed about the axis of rotation
• C. its angular speed
• D. the torque applied to it

Explanation: Moment of inertia is the sum of each mass element multiplied by the square of its distance from the axis, so moving mass further out increases it sharply even when the total mass is unchanged. This is why a hollow cylinder has a larger moment of inertia than a solid one of the same mass. It plays the same role in rotation that mass plays in linear motion.

56. A skater spinning with arms outstretched pulls the arms inwards. The result is that

• A. the angular speed increases, because the moment of inertia decreases and angular momentum is conserved
• B. the angular speed decreases
• C. the angular momentum increases
• D. nothing changes

Explanation: With no external torque the angular momentum, the product of moment of inertia and angular speed, stays constant, so reducing the moment of inertia must raise the angular speed. The skater's rotational kinetic energy actually increases, and the extra energy comes from the muscular work done pulling the arms in against the outward tendency. The same principle explains the spin of a collapsing star.

57. The angular momentum of a rotating rigid body is given by

• A. I alpha
• B. I omega
• C. m v r squared
• D. half I omega squared

Explanation: Angular momentum is the moment of inertia multiplied by the angular speed, the rotational analogue of linear momentum mv. The product I alpha is torque, the analogue of force, and half I omega squared is rotational kinetic energy. Keeping these three analogues apart is what most rotational questions actually test.

58. A satellite in a circular orbit around the Earth is held in orbit by

• A. the vacuum of space
• B. its own centrifugal force
• C. the gravitational pull of the Earth acting as the centripetal force
• D. the thrust of its engines

Explanation: Gravity supplies exactly the inward force needed to bend the satellite's straight line motion into a circle, so no engine is required once the orbit is established. Setting the gravitational force equal to mv squared over r gives the orbital speed, which depends on the radius but not on the satellite's mass. Astronauts feel weightless because they and the craft are falling towards the Earth together, not because gravity is absent.

59. A geostationary satellite must have an orbital period of

• A. 1 hour
• B. 12 hours
• C. 24 hours
• D. 365 days

Explanation: To stay above the same point on the equator the satellite must complete one orbit in exactly the time the Earth takes to rotate once, which is 24 hours. This fixes the orbital radius at about 42,300 km from the centre of the Earth. Communication and television satellites use this orbit so that a dish on the ground can be aimed once and left fixed.

60. The rotational kinetic energy of a body is

• A. half m v squared
• B. I omega
• C. half I omega squared
• D. I alpha

Explanation: Replacing mass with moment of inertia and linear speed with angular speed in the familiar expression gives half I omega squared. A rolling body has both this and the translational half m v squared, which is why a ball rolling down a slope reaches the bottom more slowly than one that slides without friction. The share of the energy going into rotation depends on the shape of the body.