Fifth Chapter Solved Exercise MCQs Of FSc First Year Physics

Fifth Chapter Solved Exercise MCQs of FSc First Year Physics is a helpful resource for students preparing for exams, featuring solved multiple-choice questions. Each MCQ comes with a clear explanation, making it easier for you to understand the correct answers. Use this resource to enhance your studies and feel ready for your exams.

Fifth chapter solved MCQs with explanation

1. The right-hand rule for rotating bodies is used to find the direction of:

(a) Angular velocity

(b) Torque

(c) Angular acceleration

(d) All

Explanation: Option (d) is correct: The right-hand rule is used to determine the direction of angular velocity, torque, and angular acceleration.

2. The mud files off the tire of a fast-moving car in the direction of:

(a) Parallel to the moving tire

(b) Antiparallel to the tire            

(c) Tangent to the moving tire

(d) None of these

Explanation: Option (a) is correct: The mud is flung off the tire due to the centrifugal force, which acts tangentially to the tire’s motion.

3. The correct SI unit of angular momentum is:

(a) kgm²s⁻¹

(b) kgms⁻²

(c) kgm⁻¹s⁻¹

(d) kgm²s⁻²

Explanation: Option (a) is correct: Angular momentum is the product of the moment of inertia and angular velocity. Its SI unit is kgm²s⁻¹.

4. A 30kg flywheel is moving with uniform angular acceleration. If the radius of the flywheel is 2m, then its moment of inertia is:

(a) 120 kgm²

(b) 30 kgm²

(c) 2 kgm²

(d) 60 kgm²

Explanation: Option (a) is correct: The moment of inertia for a flywheel is calculated using the formula I = (1/2) * m * r^2. Given a mass of 30 kg and a radius of 2 m, we get: I = (1/2) * 30 kg * (2 m)^2 = 60 kg·m² The uniform angular acceleration doesn’t affect this calculation, as moment of inertia depends only on mass distribution and axis of rotation.

5. The moment of inertia of a 100kg sphere having a radius of 5cm is:

(a) 0.1 kgm²

(b) 5 kgm²

(c) 500 kgm²

(d) 2.5 kgm²

Explanation: The moment of inertia of a solid sphere is (2/5)MR². Plugging in the values, we get (2/5)(100 kg)(0.05 m)² = 2.5 kgm².

6. The hoop and disc have the same mass and radius. Their rotational KE are related by the equation:

(a) KE_hoop = KE_disc

(b) KE_hoop = 2KE_disc

(c) KE_hoop = (1/2)KE_disc

(d) None

Explanation: The rotational kinetic energy of a rotating object is given by KE = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. For a hoop and a disc of the same mass and radius, the hoop has a larger moment of inertia, resulting in a higher rotational kinetic energy.

7. The critical velocity of an artificial satellite is:

(a) 7.9 miles s⁻¹

(b) 7.9 km s⁻¹

(c) 7.9 kmh⁻¹

(d) 790 ms⁻¹

Explanation: Option (b) is correct: Critical velocity, also known as orbital velocity, is the minimum speed required for a satellite to maintain a stable circular orbit around a celestial body. For Earth, the approximate critical velocity is 7.9 kilometers per second.  

8. In angular motion, Newton’s 2nd law of motion is:

(a) F = ma

(b) F = ΔP/Δt

(c) τ = Iα

(d) All of above

Explanation: Option (c) is correct: Newton’s second law of motion for rotational motion states that the torque acting on an object is equal to the product of its moment of inertia and its angular acceleration.  

9. A man of weight W is standing in an elevator which is ascending with uniform acceleration a. Then its apparent weight is:

(a) mg

(b) mg – ma

(c) mg + ma

(d) ma – mg

Explanation: Option (c) is correct: When an elevator is accelerating upwards, the apparent weight of a person inside is greater than their actual weight due to the additional force exerted by the elevator floor.

10. If a body of mass 10kg is allowed to fall freely, its apparent weight becomes:

(a) Zero

(b) 89N

(c) 9.8N

(d) 10N

Explanation: Option (a) is correct: When a body is in free fall, it experiences weightlessness, and its apparent weight becomes zero.

11. If external torque is zero, then which of these quantities is constant:

(a) Angular momentum

(b) Force

(c) Linear momentum

(d) None of these

Explanation: Option (a) is correct: According to the law of conservation of angular momentum, if no external torque acts on a system, its total angular momentum remains constant.

12. The acceleration due to gravity on the moon is 1/6 of that on Earth. What will be the mass of the body on the moon if its mass on Earth is m?

(a) m/6

(b) 6m

(c) m

(d) m/3

Explanation: Option (c) is correct: Mass is a scalar quantity that does not change with location. Therefore, the mass of a body remains the same regardless of its location.

13. The value of angular momentum of a body is maximum if θ is equal to:

(a) 0°

(b) 45°

(c) 90°

(d) 180°

Explanation: Option (c) is correct: Angular momentum is maximum when the angle between the position vector and the velocity vector is 90°, as this maximizes the sine function in the formula for angular momentum.

14. The diver spins faster when the moment of inertia becomes:

(a) Smaller

(b) Greater

(c) Remains the same

(d) None of these

Explanation: Option (a) is correct: According to the law of conservation of angular momentum, when the moment of inertia decreases, the angular velocity increases to maintain constant angular momentum.

15. When a body moves in a circle, then its linear and angular velocity are:

(a) Parallel

(b) Perpendicular

(c) Antiparallel

(d) None

Explanation: Option (b) is correct: In circular motion, the linear velocity is tangent to the circle, while the angular velocity is perpendicular to the plane of the circle.

16. Torque per unit moment of inertia is equivalent to:

(a) Angular velocity

(b) Angular acceleration

(c) Inertia

(d) Radius of gyration

Explanation: Option (b) is correct: Torque divided by moment of inertia is equal to angular acceleration.

17. The moment of linear momentum is called:

(a) Torque

(b) Couple

(c) Impulse

(d) Angular momentum

Explanation: Option (d) is correct: The moment of linear momentum, also known as angular momentum, is defined as the product of the linear momentum of an object and the perpendicular distance from the axis of rotation to the line of action of the momentum. Mathematically, it’s expressed as \( L = r \times p \), where \( L \) is angular momentum, \( r \) is the position vector, and \( p \) is linear momentum. Therefore, when referring to the “moment of linear momentum,” we are discussing angular momentum.

18. The ratio of angular velocities of the hour hand and minute hand of a watch is:

(a) 1:1

(b) 720:1

(c) 43200:1

(d) 1:12

Explanation: Option (d) is correct: The ratio of angular velocities of the hour hand and minute hand is 1:12 because the minute hand completes one full revolution in 60 minutes, while the hour hand does so in 12 hours. Thus, in the same hour, the minute hand makes 12 times the rotation of the hour hand, leading to the ratio of 1:12.

19. If the Earth shrinks to half of its radius without change in mass, the duration of the day will:

(a) 6hrs

(b) 12hrs

(c) 24hrs

(d) 48hrs

Explanation: Option (a) is correct: If the Earth shrinks to half its radius without changing its mass, the duration of a day will become 6 hours. 

20. The ratio of angular frequency and linear frequency is:

(a) 2πf

(b) π

(c) 1/(2π)

(d) π/2

Explanation: Option (a) is correct: In the case of frequency, it is the number of repeated events per unit time. So, the formula for frequency will be, f= 1 / T, where T is time. Thus, by establishing the link between the formula of angular frequency and frequency, we get ω =2πf.