JEE Physics Laws Of Motion MCQs Set 05

Answer: (c) 200 cm s^-1

Explanation:
1. Acceleration \( a = F/m = 100/5 = 20 \) cm s^-2.
2. Starting from rest, \( v = at = 20 \times 10 = 200 \) cm s^-1.

Teacher's Note:
1. Dyne and gram go together in the CGS system.
2. Do not convert units unless the options need it.

Answer: (a) \( 12\hat\jmath \) N s

Explanation:
1. Velocity \( \vec v = d\vec r/dt = 3\hat\imath + 4\sin t\,\hat\jmath \).
2. At \( t = 0 \): \( \vec v = 3\hat\imath \). At \( t = \pi/2 \): \( \vec v = 3\hat\imath + 4\hat\jmath \).
3. Impulse = change in momentum \( = 3\,(4\hat\jmath) = 12\hat\jmath \) N s.

Teacher's Note:
1. Impulse equals the change in linear momentum.
2. Differentiate the position vector once to get the velocity.

Answer: (b) 20 N

Explanation:
1. Thrust \( F = u\,\dfrac{dm}{dt} \).
2. \( F = 400 \times 0.05 = 20 \) N.

Teacher's Note:
1. Thrust equals exhaust speed times the rate of mass ejection.
2. Check the unit: kg/s times m/s gives newton.

Answer: (c) 2.5 m s^-2

Explanation:
1. In the lift frame the body decelerates at \( g + a \).
2. Time of flight \( t = \dfrac{2u}{g + a} \Rightarrow g + a = \dfrac{2\times5}{0.8} = 12.5 \).
3. \( a = 2.5 \) m s^-2.

Teacher's Note:
1. For an upward accelerating lift the effective gravity is \( g + a \).
2. Use the relative velocity and relative acceleration.

Answer: (c) 3w

Explanation:
1. Mass \( m = w/g \), upward deceleration \( 2g \).
2. \( N - w = m(2g) = 2w \).
3. \( N = 3w \), which is the force exerted on the ground.

Teacher's Note:
1. A retarding motion needs an upward net force.
2. The ground force must support the weight and also stop the packet.

Answer: (a) \( \rho u^2 A \)

Explanation:
1. Mass striking the wall per second \( = \rho A u \).
2. Each kg loses its horizontal speed u, so \( F = \dfrac{dp}{dt} = (\rho A u)\,u = \rho A u^2 \).

Teacher's Note:
1. The water leaves sideways, so all of its horizontal momentum is destroyed.
2. Force equals rate of change of momentum.

Answer: (b) \( \dfrac{F}{\sin\theta} \)

Explanation:
1. At B the forces are the tension T along BA, the horizontal pull F and the tension Mg in the vertical rope.
2. Horizontal: \( T\sin\theta = F \). Vertical: \( T\cos\theta = Mg \).
3. So \( T = \dfrac{F}{\sin\theta} \).

Teacher's Note:
1. Resolve the tension along horizontal and vertical directions.
2. The angle is measured from the vertical, so the horizontal component has the sine.

Answer: (a) 40 N

Explanation:
1. \( F = \dfrac{\Delta p}{\Delta t} = \dfrac{0.2\times20}{0.1} \).
2. \( F = 40 \) N.

Teacher's Note:
1. Use the change in momentum divided by the time.
2. The ball ends at rest, so the change equals the initial momentum.

Answer: (d) 90000 N

Explanation:
1. \( T - mg = ma \).
2. \( T = m(g + a) = 6000\times15 = 90000 \) N.

Teacher's Note:
1. Weighing 6000 kg means the mass is 6000 kg.
2. Upward acceleration increases the tension above the weight.

Answer: (b) is downwards

Explanation:
1. For the system A + B, the wall is smooth, so the only vertical forces are the weights and the upward component \( F\sin\theta \). Hence \( F\sin\theta = (m_A + m_B)g \).
2. For B alone: \( F\sin\theta + f = m_Bg \Rightarrow f = m_Bg - (m_A + m_B)g = -m_Ag \).
3. The friction on B is negative of the upward direction, with magnitude \( m_Ag \), so it acts downwards.

Teacher's Note:
1. Check the whole system first, then each block.
2. Block A pulls block B down through friction, so B experiences a downward friction.

Answer: (a) 100g N

Explanation:
1. At the junction, string A (vertical) carries the weight \( Mg \).
2. Vertical: \( T_C\sin45^\circ = Mg \Rightarrow T_C = \sqrt2\,Mg \).
3. Horizontal: \( T_B = T_C\cos45^\circ = Mg = 100g \) N.

Teacher's Note:
1. Use the junction of the three strings as the point of equilibrium.
2. At 45 degrees the horizontal and vertical components are equal.

Answer: (b) less than 6 kg

Explanation:
1. The acceleration is \( a = \dfrac{5-1}{5+1}g = \dfrac{2g}{3} \).
2. Tension \( T = \dfrac{2\times1\times5}{6}g = \dfrac{5g}{3} \).
3. The balance supports the pulley with a force \( 2T = \dfrac{10g}{3} \), which reads \( \dfrac{10}{3} = 3.33 \) kg, less than 6 kg.

Teacher's Note:
1. The heavier mass accelerates down, so the system is not in equilibrium.
2. The balance reads twice the string tension, not the total weight.

Answer: (c) \( \dfrac{v}{\sin\theta} \)

Explanation:
1. The rope is inextensible, so the components of velocity along the rope are equal for the block and the rope.
2. The rope makes an angle \( 90^\circ - \theta \) with the horizontal, so \( v_{block}\cos(90^\circ - \theta) = v \).
3. \( v_{block} = \dfrac{v}{\sin\theta} \).

Teacher's Note:
1. Use the constraint that the rope length is constant.
2. The angle theta is measured from the vertical, so the horizontal component brings in the sine.

Answer: (a) 1 and 2

Explanation:
1. Perpendicular to the incline the force F presses the block into the surface, so \( N = mg\cos\theta + F\sin\theta \) (statement 1 is correct).
2. Along the incline, F pushes up the slope with \( F\cos\theta \) and gravity pulls down with \( mg\sin\theta \), so friction is zero when they are equal (statement 2 is correct).
3. Limiting friction is \( \mu N = \mu(mg\cos\theta + F\sin\theta) \), so statements 3 and 4 are wrong.
4. The correct pair is 1 and 2.

Teacher's Note:
1. Resolve every force along and perpendicular to the plane.
2. Limiting friction is always mu times the normal reaction.

Answer: (c) \( \dfrac{\mu l}{1+\mu} \)

Explanation:
1. Let the hanging mass fraction be \( l_1/l \). The weight of the hanging part is balanced by the maximum friction on the part on the table.
2. \( \dfrac{l_1}{l}Mg = \mu\,\dfrac{l - l_1}{l}Mg \).
3. \( l_1 = \mu(l - l_1) \Rightarrow l_1 = \dfrac{\mu l}{1+\mu} \).

Teacher's Note:
1. Friction acts only on the part of the rope lying on the table.
2. Compare weights using lengths because the rope is uniform.

Answer: (b) 0.6 m s^-2

Explanation:
1. Treat the two blocks as one system of mass 5 kg.
2. Net horizontal force \( = 5 - 2 = 3 \) N.
3. \( a = 3/5 = 0.6 \) m s^-2.

Teacher's Note:
1. Contact forces between the blocks cancel for the combined system.
2. Only external forces enter the equation.

Answer: (b) 36 N

Explanation:
1. A is dragged only by friction, so its maximum acceleration is \( \mu_1 g = 0.2\times10 = 2 \) m s^-2.
2. For the whole system just before slipping: \( F - \mu_2(m_A + m_B)g = (m_A + m_B)a \).
3. \( F = 0.4\times60 + 6\times2 = 24 + 12 = 36 \) N.

Teacher's Note:
1. Use the maximum acceleration that friction can give to the upper block.
2. Do not forget the ground friction on the lower block.

Answer: (b) 0.74 s

Explanation:
1. In the frame of the plane there is a pseudo force \( mg \) to the left.
2. Along the incline, taking up the slope as positive: \( a = g\cos30^\circ - g\sin30^\circ = g(0.866 - 0.5) = 0.366g = 3.66 \) m s^-2, so the block moves up the slope.
3. \( 1 = \tfrac12(3.66)t^2 \Rightarrow t = \sqrt{0.546} = 0.74 \) s.

Teacher's Note:
1. In an accelerating frame add a pseudo force opposite to the acceleration of the frame.
2. Both the pseudo force and gravity press the block into the plane, so it stays in contact.

Answer: (c) 3 m s^-2

Explanation:
1. The engine force is constant, \( F = ma = 6m \).
2. The mass to be accelerated is now \( 2m \).
3. \( a' = F/2m = 3 \) m s^-2.

Teacher's Note:
1. Same force on twice the mass gives half the acceleration.
2. Assume the pulling force stays the same.

Answer: (a) 2 kg

Explanation:
1. Motion just begins when the weight of B equals the limiting friction on A.
2. \( m_Bg = \mu m_Ag \Rightarrow m_B = 0.2\times10 = 2 \) kg.

Teacher's Note:
1. Limiting friction on A is mu times its weight.
2. Static and kinetic friction are equal here, so the block just starts to move at this value.

Answer: (c) 0.5

Explanation:
1. The upper block B (3 kg) is accelerated only by friction, so \( a_{max} = \mu g \).
2. For the combined system of 10 kg: \( 49 = 10\,a_{max} \Rightarrow a_{max} = 4.9 \) m s^-2.
3. \( \mu = 4.9/9.8 = 0.5 \).

Teacher's Note:
1. The ground is smooth, so only the applied force accelerates the pair.
2. The upper block can be accelerated at most by \( \mu g \).

Answer: (c) \( 30^\circ \)

Explanation:
1. Minimum force to move up: \( mg(\sin\theta + \mu\cos\theta) \). Minimum force to prevent sliding down: \( mg(\sin\theta - \mu\cos\theta) \).
2. \( \sin\theta + \mu\cos\theta = 3(\sin\theta - \mu\cos\theta) \Rightarrow 2\mu\cos\theta = \sin\theta \).
3. \( \tan\theta = 2\mu = \dfrac{1}{\sqrt3} \Rightarrow \theta = 30^\circ \).

Teacher's Note:
1. Friction acts down the slope when the body is pushed up and up the slope when it is about to slide down.
2. This gives tan theta = 2 mu.

Answer: (d) \( 3F \)

Explanation:
1. The two blocks move together. Net force \( = 2F - F = F \), so \( a = F/2m \) towards the left.
2. The contact surface is inclined at \( 30^\circ \) to the horizontal, so the normal reaction N has a horizontal component \( N\sin30^\circ \) on A, directed to the left.
3. For A: \( N\sin30^\circ - F = m\cdot\dfrac{F}{2m} \Rightarrow \dfrac N2 = \dfrac{3F}{2} \).
4. \( N = 3F \).

Teacher's Note:
1. Treat the pair as one body to get the common acceleration.
2. The normal reaction is perpendicular to the contact surface.

Answer: (a) the relative velocity of the two surfaces in contact

Explanation:
1. Kinetic friction is \( f_k = \mu_kN \).
2. It depends on the nature of the surfaces (through \( \mu_k \)) and on the normal reaction.
3. It is nearly independent of the relative velocity.

Teacher's Note:
1. Kinetic friction is practically constant for small speeds.
2. It also does not depend on the area of contact.

Answer: (d) All three are possible.

Explanation:
1. A block just slides when \( \tan\theta = \mu \), which depends only on the surfaces, not on the mass.
2. So \( \mu_A = \tan30^\circ \) and \( \mu_B = \tan40^\circ \) tell us nothing about the masses.
3. Any of the three relations between the masses is possible.

Teacher's Note:
1. Mass cancels in the condition for sliding on an incline.
2. The angle of repose depends only on the pair of surfaces.