JEE Physics Laws Of Motion MCQs Set 08

Answer: (c) 10 m/s²

Explanation:
1. Thrust \( = u\,\dfrac{dm}{dt} = 300 \times 2 = 600 \) N.
2. \( a = 600/60 = 10 \) m/s².

Teacher's Note:
1. Initial acceleration uses the initial mass.
2. No gravity is acting here.

Answer: (c) 7 kg/s

Explanation:
1. Net upward force: \( u\,\dfrac{dm}{dt} - mg = ma \).
2. \( \dfrac{dm}{dt} = \dfrac{m(g+a)}{u} = \dfrac{800 \times 14}{1600} = 7 \) kg/s.

Teacher's Note:
1. Thrust must beat the weight and also provide the acceleration.
2. Convert km/s into m/s.

Answer: (c) 200 kg/s

Explanation:
1. To just lift the rocket, thrust equals weight: \( u\,\dfrac{dm}{dt} = mg \).
2. \( \dfrac{dm}{dt} = \dfrac{800 \times 10}{40} = 200 \) kg/s.

Teacher's Note:
1. Just lifting means zero acceleration, so thrust equals weight.
2. Use the exhaust speed relative to the rocket.

Answer: (a) along the wall

Explanation:
1. The wall exerts a force on the ball only along the normal to the wall (it is smooth).
2. No force acts along the wall, so the momentum component parallel to the wall is conserved.
3. The normal component reverses, so it is not conserved.

Teacher's Note:
1. Momentum is conserved only in a direction in which no external force acts.
2. Smooth surfaces give no tangential force.

Answer: (b) \( 2av^2d\sin\theta \)

Explanation:
1. Mass striking per second \( = a\,v\,d \).
2. The component of velocity normal to the wall is \( v\sin\theta \), and it reverses on elastic rebound.
3. Force \( = avd \times 2v\sin\theta = 2av^2d\sin\theta \).

Teacher's Note:
1. Only the normal component of velocity changes.
2. The angle is measured with the wall, so the normal component is v sin θ.

Answer: (c) Newton's third law

Explanation:
1. Gases are pushed backward by the rocket (action), so the gases push the rocket forward (reaction).
2. This is the principle of Newton's third law, together with momentum conservation.

Teacher's Note:
1. Rocket propulsion is an action-reaction pair.
2. No outside medium is needed.

Answer: (c) the ground exerts on him

Explanation:
1. The horse pushes the ground backward with its hooves.
2. By Newton's third law the ground pushes the horse forward with friction.
3. This external forward force moves the horse.

Teacher's Note:
1. Forces within the horse-wagon pair cannot start the motion.
2. The ground provides the external forward force.

Answer: (c) 2a

Explanation:
1. Force on B \( = 2m \cdot a \).
2. By the third law the force on A has the same magnitude \( 2ma \).
3. Acceleration of A \( = 2ma/m = 2a \).

Teacher's Note:
1. Action and reaction are equal in magnitude.
2. The lighter body gets the greater acceleration.

Answer: (b) the weight of all bodies will increase

Explanation:
1. Mass is the quantity of matter and does not depend on gravity.
2. Weight \( = mg \) increases when g increases.

Teacher's Note:
1. Mass is constant everywhere.
2. Weight depends on the local value of g.

Answer: (c) inertia of rest keeps the upper part of the body at rest while the lower part of the body moves forward with the horse

Explanation:
1. The lower part of the body is in contact with the horse and moves with it.
2. The upper part tends to stay at rest by inertia of rest.
3. So the rider appears to fall backward.

Teacher's Note:
1. This is a standard illustration of the law of inertia.
2. The body resists a sudden change of its state of rest.

Answer: (c) both A and B are true

Explanation:
1. The floor force exceeds the weight when the acceleration is upward, since \( N - mg = ma \).
2. In A the speed increases while moving up, so the acceleration is upward.
3. In B the speed decreases while moving down, so the acceleration is again upward.
4. Both are true.

Teacher's Note:
1. Look at the direction of acceleration, not of velocity.
2. Upward acceleration makes the apparent weight larger.

Answer: (d) zero

Explanation:
1. The cork floats at rest.
2. Its weight is balanced by the buoyant force, so the net force is zero.

Teacher's Note:
1. Floating means equilibrium.
2. Weight and upthrust are equal and opposite.

Answer: (b) The force on the 6th coin due to the 7th coin is 4mg (downwards).

Explanation:
1. Above the 3rd coin there are 7 coins, so the force is 7mg, which makes (a) false.
2. The 7th coin presses on the 6th with the weight of coins 7 to 10, which is 4mg downward, so (b) is true.
3. The reaction of the 6th on the 7th is 4mg upward, so (c) is false.
4. The net force on the 10th coin is zero, so (d) is false.

Teacher's Note:
1. The force at any contact equals the weight of everything above it.
2. Count the coins above the contact.

Answer: (a) \( N_A = \dfrac{1000}{\sqrt{3}} \) N, \( N_B = \dfrac{500}{\sqrt{3}} \) N

Explanation:
1. The normal at A makes 30° with the vertical. Vertical balance: \( N_A\cos30° = mg = 500 \), so \( N_A = 1000/\sqrt{3} \) N.
2. Horizontal balance: \( N_B = N_A\sin30° = 500/\sqrt{3} \) N.

Teacher's Note:
1. Resolve the reaction at A along the vertical and horizontal.
2. The wall supplies the horizontal balance.

Answer: (c) \( mg\cos\theta + F\sin\theta \)

Explanation:
1. Take components perpendicular to the plane.
2. The weight contributes \( mg\cos\theta \) into the plane.
3. The horizontal force pushes the block into the plane with component \( F\sin\theta \).
4. So \( N = mg\cos\theta + F\sin\theta \).

Teacher's Note:
1. Resolve every force along and perpendicular to the incline.
2. A horizontal force pressing on the incline adds to the normal reaction.

Answer: (c) 4 N

Explanation:
1. Acceleration of the system \( = (6-3)/(2+1) = 1 \) m/s².
2. For the 2 kg block: \( 6 - N = 2 \times 1 \).
3. So \( N = 4 \) N.

Teacher's Note:
1. Find the common acceleration first.
2. Then apply Newton's law to one block alone.

Answer: (d) when the elevator moves downward with decreasing velocity

Explanation:
1. Apparent weight is more than real weight only when the acceleration is upward.
2. Moving downward with decreasing velocity means the acceleration is upward.
3. In the other cases the acceleration is zero or downward.

Teacher's Note:
1. Decreasing speed while moving down means upward acceleration.
2. Apparent weight is \( m(g + a) \) for upward a.

Answer: (b) 400 N

Explanation:
1. For downward acceleration \( N = m(g - a) \).
2. \( N = 80 \times (10 - 5) = 400 \) N.

Teacher's Note:
1. Downward acceleration reduces the reading.
2. The scale shows the normal force.

Answer: (c) 3 : 2

Explanation:
1. Upward acceleration: \( N_1 = 60(10+2) = 720 \) N.
2. Downward acceleration: \( N_2 = 60(10-2) = 480 \) N.
3. Ratio \( = 720 : 480 = 3 : 2 \).

Teacher's Note:
1. Use \( g + a \) for upward and \( g - a \) for downward acceleration.
2. The mass cancels in the ratio.

Answer: (d) infinitely large

Explanation:
1. The vertical component of the two tensions must balance the weight: \( 2T\sin\alpha = 15 \), where α is the angle of the rope below the horizontal.
2. For a perfectly horizontal rope \( \alpha = 0 \), so \( T = 15/(2\sin\alpha) \to \infty \).
3. So the rope can never be made completely straight.

Teacher's Note:
1. A horizontal rope has no vertical component of tension.
2. That is why a loaded rope always sags.

Answer: (c) 9 : 4

Explanation:
1. \( T_3 \) supports only the 4 kg block, so \( T_3 = 4g \).
2. \( T_1 \) supports all three blocks, so \( T_1 = (2+3+4)g = 9g \).
3. Ratio \( T_1 : T_3 = 9 : 4 \).

Teacher's Note:
1. Each string supports everything hanging below it.
2. Equilibrium means zero acceleration.

Answer: (d) 45°

Explanation:
1. The lower half of the rope carries the weight \( 6 \times 10 = 60 \) N.
2. At the mid point, the upper tension T satisfies \( T\cos\theta = 60 \) (vertical) and \( T\sin\theta = 60 \) (horizontal).
3. So \( \tan\theta = 1 \) and \( \theta = 45° \).

Teacher's Note:
1. Balance the forces at the junction in two directions.
2. Equal vertical and horizontal pulls give 45 degrees.

Answer: (c) \( T = R + W \)

Explanation:
1. The sphere is in equilibrium, so the vector sum of the forces is zero, and (a) is true.
2. \( T\cos\theta = W \) and \( T\sin\theta = R \), so \( T^2 = R^2 + W^2 \) and \( R = W\tan\theta \), making (b) and (d) true.
3. T equals the vector resultant of R and W, not their scalar sum, so (c) is false.

Teacher's Note:
1. Forces at different angles add as vectors.
2. T is greater than R and W individually but less than R + W.

Answer: (c) 25 N

Explanation:
1. Acceleration \( a = (40 - 10)/(10+6+4) = 1.5 \) m/s² towards the 40 N force.
2. For \( m_1 \): \( T_1 - 10 = 10 \times 1.5 \).
3. So \( T_1 = 25 \) N.

Teacher's Note:
1. Apply Newton's second law to the block at the end.
2. Take the direction of the larger force as positive.

Answer: (c) 120 N

Explanation:
1. The string \( T_1 \) pulls the 3 kg and 12 kg blocks, so \( a = 30/15 = 2 \) m/s².
2. The horizontal component of F accelerates all 30 kg: \( F\cos60° = 30 \times 2 = 60 \) N.
3. So \( F = 120 \) N.

Teacher's Note:
1. Only the horizontal component of F produces acceleration here.
2. The vertical component just reduces the normal reaction.