JEE Physics Laws Of Motion MCQs Set 09

Answer: (c) 4 m/s²

Explanation:
1. Total weight \( = 15 \times 10 = 150 \) N.
2. Net upward force \( = 210 - 150 = 60 \) N.
3. \( a = 60/15 = 4 \) m/s².

Teacher's Note:
1. Treat the two blocks as one system for the acceleration.
2. Subtract the total weight from the applied force.

Answer: (a) 80 N

Explanation:
1. The inclined rope is the only one with a vertical component.
2. Vertical balance: \( T\sin30° = 40 \).
3. So \( T = 80 \) N.

Teacher's Note:
1. The horizontal rope gives no vertical support.
2. The angle is measured from the ceiling, so use sine.

Answer: (b) 10 m/s²

Explanation:
1. The maximum tension is \( 30 \times 10 = 300 \) N.
2. For the monkey: \( T - mg = ma \), so \( 300 - 150 = 15a \).
3. \( a = 10 \) m/s².

Teacher's Note:
1. Maximum tension equals the weight of the maximum mass.
2. Upward acceleration needs tension above the weight.

Answer: (b) \( Fx/L \)

Explanation:
1. Acceleration of the string \( a = F/M \).
2. The length x of string beyond the point has mass \( Mx/L \), and the tension must accelerate it.
3. \( T = \dfrac{Mx}{L}\cdot\dfrac{F}{M} = \dfrac{Fx}{L} \).

Teacher's Note:
1. Tension at a point accelerates the portion of the string beyond that point.
2. Measure x from the free end.

Answer: (b) g/2

Explanation:
1. Mass on the heavier side \( = 10 + 5 = 15 \) kg and on the lighter side \( = 5 \) kg.
2. \( a = \dfrac{15 - 5}{15 + 5}g = \dfrac{g}{2} \).

Teacher's Note:
1. Blocks hanging together on one side act as one mass.
2. Use the Atwood machine formula.

Answer: (a) 5.67 mg

Explanation:
1. Acceleration of the masses \( a = \dfrac{2m - m}{3m}g = \dfrac{g}{3} \).
2. Tension in the string \( T = m(g + a) = \dfrac{4mg}{3} \).
3. Rope force \( = 3mg + 2T = 3mg + \dfrac{8mg}{3} = 5.67\,mg \).

Teacher's Note:
1. The string pulls the pulley down on both sides with force T.
2. Add the weight of the pulley.

Answer: (c) less than that in case (b)

Explanation:
1. In case (a): \( a = \dfrac{2m - m}{2m + m}g = \dfrac{g}{3} \).
2. In case (b): \( a = \dfrac{(mg + mg) - mg}{2m} = \dfrac{g}{2} \).
3. So the acceleration in (a) is less than in (b).

Teacher's Note:
1. Add the extra force to the driving side in case (b).
2. Compare the two accelerations.

Answer: (c) \( mg/(m + m') \)

Explanation:
1. The only driving force is the weight \( mg \) of the hanging mass.
2. The total mass moved is \( m + m' \).
3. \( a = mg/(m + m') \).

Teacher's Note:
1. A frictionless horizontal surface exerts no resistance.
2. Net force divided by total mass gives acceleration.

Answer: (c) \( \dfrac{m_1 g}{2(m_1 + m_2)} \)

Explanation:
1. The hanging weight \( m_1 g \) pulls the system one way and F opposes it.
2. Net force \( = m_1 g - m_1 g/2 = m_1 g/2 \).
3. \( a = \dfrac{m_1 g/2}{m_1 + m_2} = \dfrac{m_1 g}{2(m_1 + m_2)} \).

Teacher's Note:
1. F acts against the weight of the hanging block.
2. Use net force divided by total mass.

Answer: (a) 60 N

Explanation:
1. The string pulls only the body A (the one behind), which has the acceleration 3 m/s².
2. \( T = 20 \times 3 = 60 \) N.

Teacher's Note:
1. Tension equals the mass of the body behind times the acceleration.
2. F itself would be 150 N.

Answer: (c) 20 kg

Explanation:
1. The hanging block pulls with tension \( T = mg \).
2. For equilibrium of M on the smooth plane, \( Mg\sin30° = T = mg \).
3. \( M = m/\sin30° = 20 \) kg.

Teacher's Note:
1. Balance the component of weight along the plane with the tension.
2. sin 30 degrees is one half.

Answer: (d) neither compressed nor stretched regardless of the values of \( m_1 \) and \( m_2 \)

Explanation:
1. In free fall both masses have the same acceleration g.
2. The spring force on each mass would have to be zero for this, so the spring keeps its natural length.

Teacher's Note:
1. In free fall every body has the same acceleration, so there is no relative effect.
2. Weightlessness removes the stretching.

Answer: (b) 5 m/s²

Explanation:
1. The spring force on the 1 kg mass is \( 1 \times 10 = 10 \) N towards the right.
2. By the third law the spring pulls the 2 kg mass back with 10 N.
3. Net force on the 2 kg mass \( = 20 - 10 = 10 \) N, so \( a = 5 \) m/s².

Teacher's Note:
1. The spring exerts equal and opposite forces on its two ends.
2. Use the 1 kg mass to find the spring force.

Answer: (b) 5.3 m/s²

Explanation:
1. Spring force on the 20 kg mass \( = 20 \times 2 = 40 \) N.
2. The spring pulls back on the 30 kg mass with 40 N.
3. \( a = (200 - 40)/30 = 5.33 \) m/s².

Teacher's Note:
1. Springs give equal and opposite forces at both ends.
2. Instantaneous accelerations of the two masses can differ.

Answer: (a) first increases then decreases to zero

Explanation:
1. When the man jumps off he pushes down harder on the platform, so the reading first increases.
2. After he leaves the platform the reading drops to zero.

Teacher's Note:
1. To jump, the man must push the platform with a force more than his weight.
2. That extra push raises the reading briefly.

Answer: (b) any inertial observer

Explanation:
1. Newton's laws hold in inertial frames.
2. All inertial observers measure the same acceleration, so \( \vec{F} = m\vec{a} \) is valid for any inertial observer.

Teacher's Note:
1. Inertial frames move with constant velocity relative to each other.
2. Non-inertial frames need pseudo forces.

Answer: (a) linear velocity

Explanation:
1. A frame moving with constant linear velocity relative to an inertial frame has no acceleration.
2. So Newton's laws hold in it.
3. A frame with constant angular velocity or constant acceleration is accelerated, so it is non-inertial.

Teacher's Note:
1. Constant velocity means no acceleration.
2. Rotating frames are non-inertial.

Answer: (c) \( \dfrac{\sqrt{5}}{2}mg \)

Explanation:
1. In the car frame the sphere is in equilibrium under mg (down), pseudo force \( m \cdot g/2 \) (horizontal) and tension T.
2. \( T = \sqrt{(mg)^2 + (mg/2)^2} = mg\sqrt{1 + 1/4} \).
3. \( T = \dfrac{\sqrt{5}}{2}mg \).

Teacher's Note:
1. Combine the weight and the pseudo force as perpendicular vectors.
2. The string tilts back from the vertical.

Answer: (c) a, d

Explanation:
1. The force on the floor is less than the weight only when the acceleration is downward.
2. Going up and slowing down means downward acceleration (a).
3. Going down and speeding up also means downward acceleration (d).

Teacher's Note:
1. Decide the direction of acceleration from the change in speed.
2. Downward acceleration reduces the apparent weight.

Answer: (a) the pans will remain at the same level

Explanation:
1. Both pans and masses experience the same effective gravity \( g + a \).
2. The torques about the pivot are equal, so the balance stays level.

Teacher's Note:
1. Equal masses give equal apparent weights in the same frame.
2. The beam stays horizontal.

Answer: (b) \( g/2 \)

Explanation:
1. \( \dfrac{g + a}{g - a} = 3 \), so \( g + a = 3g - 3a \).
2. \( 4a = 2g \), giving \( a = g/2 \).

Teacher's Note:
1. Use g + a for upward and g - a for downward acceleration.
2. Form the ratio and solve.

Answer: (a) zero

Explanation:
1. During the jump both fall freely with acceleration g.
2. The load exerts no force on the head, so it has no apparent weight.

Teacher's Note:
1. Free fall means weightlessness.
2. The normal force between the man and the load is zero.

Answer: (c) \( mg/\cos\theta \)

Explanation:
1. The block moves horizontally with the wedge, so the vertical forces balance.
2. The force from the smooth wedge is the normal N, with \( N\cos\theta = mg \).
3. \( N = mg/\cos\theta \).

Teacher's Note:
1. The block has no vertical acceleration.
2. The horizontal component of N gives the acceleration.

Answer: (a) 1

Explanation:
1. In a single fixed pulley the effort needed equals the load, \( F = W \).
2. Mechanical advantage \( = W/F = 1 \).

Teacher's Note:
1. A fixed pulley only changes the direction of the force.
2. A movable pulley has a mechanical advantage of 2.

Answer: (c) 5

Explanation:
1. Mechanical advantage \( = \dfrac{\text{load}}{\text{effort}} = \dfrac{400}{80} \).
2. So it equals 5.

Teacher's Note:
1. Mechanical advantage is the ratio of load to effort.
2. It has no unit.