Questions & explanations
1. Which statement correctly distinguishes static friction from kinetic friction?
- Static friction is always greater than kinetic friction for the same surfaces.
- Static friction acts only when the body is in motion.
- Static friction can take any value up to a maximum, while kinetic friction is constant for a given normal force.
- Kinetic friction is always greater than static friction.
Answer: Static friction can take any value up to a maximum, while kinetic friction is constant for a given normal force.
Static friction adjusts itself from zero to a maximum value (limiting friction) to prevent motion, whereas kinetic friction has a constant magnitude once motion starts, for a given normal force. The other options are incorrect because static friction is not always greater; its maximum is greater, but it can be less. Static friction acts when there is no relative motion, not only when in motion. Kinetic friction is generally less than the maximum static friction.
2. A block is placed on a rough inclined plane of inclination 30°. The coefficient of static friction between the block and the plane is 0.5. Which of the following correctly describes the motion of the block? (Take g = 10 m/s² and √3 ≈ 1.732)
- The block slides down because the component of weight down the plane is greater than the maximum static friction.
- The block does not slide because static friction is sufficient to hold it.
- The block slides down with constant velocity.
- The block remains at rest and cannot be moved even if the inclination is increased.
Answer: The block slides down because the component of weight down the plane is greater than the maximum static friction.
Weight component down the plane = mg sin30° = 0.5 mg. Maximum static friction = μs × mg cos30° = 0.5 × mg × (√3/2) ≈ 0.433 mg. Since 0.5 mg > 0.433 mg, the net force down the plane is positive, so the block slides down. Option B is false because friction is insufficient. Option C is false because friction is kinetic after motion starts, but initially it slides with acceleration. Option D is false because increasing the inclination will eventually make it slide.
3. A block of weight 100 N is suspended by two light strings making angles 30° and 60° with the horizontal as shown. Using Lami's theorem, the tension in the string making 30° with the horizontal is: (g = 10 m/s²)
- 50 N
- 86.6 N
- 100 N
- 115.5 N
Answer: 50 N
At the point where the three forces (tensions T1 and T2 in the strings and weight W) meet, Lami's theorem gives T1/sin(angle opposite T1) = T2/sin(angle opposite T2) = W/sin(angle opposite W). Here, angle between T1 (30° to horizontal) and vertical downward is 120°, between T2 (60° to horizontal) and vertical is 150°, and between T1 and T2 is 90°. So T1/sin150° = 100/sin90° ⇒ T1 = 100 × (1/2) = 50 N.
4. In a conical pendulum, a bob of mass m is whirled in a horizontal circle of radius r. The string of length L makes an angle θ with the vertical. The speed v of the bob is given by:
- A. v = √(r g tan θ)
- B. v = √(r g cot θ)
- C. v = √(L g sin θ)
- D. v = √(L g cos θ)
Answer: A. v = √(r g tan θ)
For a conical pendulum, the horizontal component of tension provides centripetal force: T sin θ = m v²/r, and vertically T cos θ = mg. Dividing gives tan θ = v²/(r g), so v = √(r g tan θ). Remove/correct this statement: substituting r=Lsinθ gives v=√(Lg sinθ tanθ), which is NOT the same as option C; option C is simply incorrect, not an equivalent rewritten form. Option A is correct as given.
5. Two blocks A and B of masses 2 kg and 3 kg respectively are connected by a light inextensible string that passes over a smooth pulley. Block A is placed on a rough horizontal surface with coefficient of kinetic friction 0.5, while block B hangs freely. The system is released from rest. What is the acceleration of the blocks? (Take g = 10 m/s²)
- 2 m/s²
- 4 m/s²
- 6 m/s²
- 8 m/s²
Answer: 4 m/s²
For block A, the equation of motion is T - f = 2a, where f = μ m_A g = 0.5 × 2 × 10 = 10 N. For block B, m_B g - T = 3a, i.e., 30 - T = 3a. Adding the two equations eliminates T: (T - 10) + (30 - T) = 2a + 3a → 20 = 5a → a = 4 m/s². The friction is overcome because the required tension (18 N) exceeds the maximum friction (10 N). Thus the acceleration is 4 m/s².
6. Which of the following is a correct law of limiting friction?
- Limiting friction is directly proportional to the normal reaction.
- Limiting friction is independent of the nature of the surfaces in contact.
- Limiting friction is less than kinetic friction for the same pair of surfaces.
- Limiting friction acts in the direction of motion of the body.
Answer: Limiting friction is directly proportional to the normal reaction.
One of the laws of limiting friction states that the limiting frictional force is directly proportional to the normal reaction, i.e., f ∝ N. The other options are false: limiting friction depends on the nature of surfaces (roughness), it is greater than kinetic friction, and it always opposes the impending motion, not acts in the direction of motion.
7. Which of the following correctly compares rolling friction with sliding friction for the same normal force and surfaces?
- Rolling friction is generally much smaller than sliding friction.
- Rolling friction is equal to sliding friction.
- Rolling friction is generally greater than sliding friction.
- Rolling friction depends on speed, while sliding friction does not.
Answer: Rolling friction is generally much smaller than sliding friction.
Rolling friction arises due to deformation at the point of contact and is much less than sliding friction because there is no interlocking of irregularities being sheared off. The other options are incorrect: rolling friction is not equal or greater; both types can depend on speed in certain ways, but the key comparison is the magnitude.
8. A rocket is propelled forward by expelling gases backward. The force that propels the rocket forward is the:
- Force exerted by the rocket on the gases
- Reaction force of the gases on the rocket
- Gravitational force acting on the rocket
- Force of atmospheric pressure on the rocket
Answer: Reaction force of the gases on the rocket
Newton's third law states that every action has an equal and opposite reaction. The rocket exerts a force on the gases (action), so the gases exert an equal and opposite force on the rocket (reaction), which propels it forward. The other options do not represent the action-reaction pair that provides thrust.
9. If the angle of friction for a pair of surfaces is λ, what is the angle of repose for the same surfaces?
- λ
- 90° – λ
- λ / 2
- 2λ
Answer: λ
The angle of repose is defined as the maximum angle of an inclined plane at which a block just begins to slide. It is equal to the angle of friction (tan θ = μ = tan λ) because both derive from the same coefficient of friction. Hence angle of repose = λ. The other options are incorrect relationships.
10. A passenger sitting inside a car feels pressed against the door when the car takes a sharp left turn. This apparent outward force experienced by the passenger, as observed from the car's frame, is called:
- A. Centripetal force
- B. Centrifugal force
- C. Gravitational force
- D. Frictional force
Answer: B. Centrifugal force
In the non‑inertial (rotating) frame of the car, a pseudo‑force – the centrifugal force – appears to act outward, pushing the passenger against the door. The actual centripetal force is provided by friction and acts inward. Options A, C, D are real forces, not the apparent outward force.
11. A bullet of mass 50 g is fired from a gun of mass 5 kg with a velocity of 200 m/s. The recoil velocity of the gun is:
- 0.5 m/s
- 1 m/s
- 2 m/s
- 4 m/s
Answer: 2 m/s
By conservation of linear momentum, initial total momentum (gun+bullet) is zero. Thus, 0 = m_bullet v_bullet + m_gun v_gun, so v_gun = - (0.05×200)/5 = -2 m/s. The magnitude of recoil velocity is 2 m/s. Other options result from incorrect mass conversions or misapplication of the law.
12. A particle is tied to a light string and whirled in a vertical circle of radius R. The minimum speed v_min that the particle must have at the highest point so that it completes the full circle is:
- A. √(g R)
- B. √(2 g R)
- C. √(3 g R)
- D. √(5 g R)
Answer: A. √(g R)
At the highest point, the minimum speed occurs when tension just becomes zero. Then mg = m v²/R gives v_min = √(g R). Higher speeds are possible but not minimal. Options B, C, D correspond to different conditions like crossing the top with finite tension or speeds at the bottom.