JEE Physics Laws Of Motion MCQs Set 21

Answer: (b) \( \frac g3 \)

Explanation:
1. Weight in the stationary lift is \( mg \); in the descending lift the reading is \( N=m(g-a) \).
2. Given \( \frac{mg}{m(g-a)}=\frac32 \): \( 2g=3g-3a \).
3. So \( a=\frac g3 \).

Teacher's Note:
1. The apparent weight is what the machine reads, which is the normal reaction.
2. For a downward acceleration the reading falls below mg.

Answer: (a) increasing at 20 m and decreasing at 50 m

Explanation:
1. The reading is the tension \( T \) in the spring: \( mg-T=ma \) with downward as positive.
2. At 20 m the reading 30 is below the true 32, so \( a>0 \) downward, and the speed is increasing.
3. At 50 m the reading 36 is above 32, so the net acceleration is upward while the lift moves down, and the speed is decreasing.

Teacher's Note:
1. Compare the reading with the true weight: smaller means acceleration downward, larger means upward.
2. Speed increases when acceleration and velocity point the same way.

Answer: (d) 4 m/s²

Explanation:
1. Take both masses as a system; the spring force is internal.
2. External force: \( F=m_1a_1+m_2a_2 \), so \( 200=10\times12+20a_2 \).
3. Hence \( a_2=4 \) m/s² (spring force on the 20 kg mass is \( 200-80=120 \) N).

Teacher's Note:
1. For a system with different accelerations, add the masses times their accelerations.
2. Internal spring forces never appear in the system equation.

Answer: (c) The normal reaction on the block is 11 N

Explanation:
1. In the wedge frame the pseudo force on the block is \( 1\times5=5 \) N to the left, pressing it into the face.
2. Along the face: \( mg\sin37^\circ-ma\cos37^\circ=6-4=2 \) N, so the block slides down at 2 m/s² (not stationary, not 1 m/s²).
3. Perpendicular to the face: \( N=mg\cos37^\circ+ma\sin37^\circ=8+3=11 \) N.
4. Net force on the wedge is \( 2\times5=10 \) N, not 2 N.

Teacher's Note:
1. Work in the wedge frame: real forces plus a pseudo force opposite to the frame's acceleration.
2. Resolve along and perpendicular to the face, and check every option.

Answer: (b) \( \frac{1000}{\sqrt3} \) N and \( \frac{500}{\sqrt3} \) N

Explanation:
1. The incline pushes with \( N_A \) perpendicular to the incline, at 30° to the vertical; the wall pushes horizontally with \( N_B \).
2. Vertical: \( N_A\cos30^\circ=500 \), so \( N_A=\frac{1000}{\sqrt3} \) N.
3. Horizontal: \( N_B=N_A\sin30^\circ=\frac{500}{\sqrt3} \) N.

Teacher's Note:
1. Only the incline can carry the weight; the wall only provides the horizontal balance.
2. The perpendicular to a 30° incline is 30° from the vertical.

Answer: (a) \( \frac{20}{3} \) m

Explanation:
1. \( a=\frac{(6-3)g}{6+3}=\frac{10}{3} \) m/s².
2. Distance \( s=\frac12at^2=\frac12\cdot\frac{10}{3}\cdot4=\frac{20}{3} \) m (about 6.7 m).

Teacher's Note:
1. Atwood's machine gives the acceleration as the difference of masses over their sum times g.
2. Use the constant-acceleration formula from rest.

Answer: (d) 80 N

Explanation:
1. Tension: \( T=\frac{2m_1m_2}{m_1+m_2}g=\frac{2\cdot3\cdot6}{9}\times10=40 \) N.
2. The string pulls the pulley down with \( T \) on each side, so the clamp must supply \( 2T=80 \) N upward.

Teacher's Note:
1. The clamp force equals the downward pull of both string segments, not the total weight.
2. The pulley is massless, so no weight is added.

Answer: (c) 112 N

Explanation:
1. Differentiate twice: \( a_x=10 \) m/s² and \( a_y=18t=36 \) m/s² at \( t=2 \) s.
2. Force components: \( F_x=3\times10=30 \) N and \( F_y=3\times36=108 \) N.
3. Magnitude \( \sqrt{30^2+108^2}\approx112 \) N.

Teacher's Note:
1. Find the acceleration components by differentiating the coordinates twice.
2. Combine the force components with the Pythagoras theorem.

Answer: (b) 6.0 N

Explanation:
1. The agent supports the whole chain of 0.5 kg: \( F-0.5g=0.5\times2 \).
2. So \( F=0.5(10+2)=6.0 \) N.
3. (The force between adjacent links from the bottom up is 1.2 N, 2.4 N, 3.6 N and 4.8 N.)

Teacher's Note:
1. The force on the top link must carry the full weight plus the mass times acceleration.
2. Treat the whole chain as one system.

Answer: (a) 720 N

Explanation:
1. Only the acceleration matters, not the velocity.
2. Taking upward as positive: \( N-mg=ma \).
3. So \( N=60(10+2)=720 \) N (the lift is slowing down while moving down).

Teacher's Note:
1. The velocity direction does not decide the reading, the acceleration does.
2. Upward acceleration gives a reading larger than the true weight.

Answer: (d) \( \frac{15}{4} \) m/s², opposite to B's

Explanation:
1. The push on B is \( 75\times3=225 \) N.
2. By the third law B pushes A with 225 N in the opposite direction.
3. So \( a_A=\frac{225}{60}=3.75=\frac{15}{4} \) m/s², opposite to B's.

Teacher's Note:
1. Action and reaction are equal in magnitude, so accelerations are inversely proportional to masses.
2. The lighter body always accelerates more.

Answer: (c) 322 N

Explanation:
1. The rope is massless, so its tension at D equals the sum of the forces the monkeys exert on it.
2. A (accelerating down): \( m(g-a)=10\times8=80 \) N. B (constant speed): \( 15\times10=150 \) N. C (accelerating up): \( 8(10+1.5)=92 \) N.
3. Total \( 80+150+92=322 \) N.

Teacher's Note:
1. A monkey exerts on the rope the force the rope exerts on it, which depends on its own acceleration.
2. Constant speed means no extra force beyond the weight.

Answer: (b) \( (g+a)\sin\theta \)

Explanation:
1. In the elevator frame the block feels an extra pseudo force \( ma \) downward, so the effective weight is \( m(g+a) \).
2. Along the smooth face only the component along the slope acts: \( m(g+a)\sin\theta=ma_{rel} \).
3. So \( a_{rel}=(g+a)\sin\theta \).

Teacher's Note:
1. An upward accelerating frame behaves like a stronger gravity.
2. Replace g by g + a in any problem in that frame.

Answer: (a) \( \frac{4F}{M+m}-g \)

Explanation:
1. Take the painter and platform as one system. Each rope pulls up on his hands with \( F \) and on the platform with \( F \), so each rope contributes \( 2F \).
2. Two ropes give a total upward force \( 4F \): \( 4F-(M+m)g=(M+m)a \).
3. So \( a=\frac{4F}{M+m}-g \).

Teacher's Note:
1. Count the rope segments that act on the system, here four.
2. Choose the system so the internal rope pulls appear as external forces.

Answer: (d) 70 kg

Explanation:
1. The maximum reading occurs during upward acceleration, the minimum during deceleration while moving up.
2. Half the sum of the readings is the true value, because \( m(g+a)+m(g-a)=2mg \): \( \frac{80.5+59.5}{2}=70 \) kg.

Teacher's Note:
1. The extra and the deficit in the reading are equal, so the true mass lies exactly halfway.
2. The readings in kg are the forces divided by g.

Answer: (c) 1.5 m/s²

Explanation:
1. The true mass is 70 kg and the maximum reading is \( m(g+a)=80.5g \).
2. Hence \( 70(10+a)=805 \): \( 10+a=11.5 \) so \( a=1.5 \) m/s².

Teacher's Note:
1. Subtract the true weight from the maximum reading to isolate \( ma \).
2. Keep the readings in kg consistent with g.

Answer: (b) 165 N

Explanation:
1. The rope pulls the monkey up with tension \( T \), equal to the force the monkey applies on it.
2. Newton's second law: \( T-mg=ma \).
3. So \( T=15(10+1)=165 \) N.

Teacher's Note:
1. The force needed is weight plus mass times acceleration.
2. Same as the reading of a weighing machine in an upward-accelerating lift.

Answer: (a) 36 m

Explanation:
1. Height equals the area under the speed-time graph (a trapezium).
2. Phase 1: \( \frac12\times2\times3.6=3.6 \) m. Phase 2: \( 8\times3.6=28.8 \) m. Phase 3: \( \frac12\times2\times3.6=3.6 \) m.
3. Total \( 3.6+28.8+3.6=36 \) m, or \( \frac12(12+8)\times3.6=36 \) m.

Teacher's Note:
1. Areas under the speed-time graph add up to distance.
2. The trapezium rule with parallel sides 12 and 8 gives a quick check.

Answer: (d) \( \frac W2\csc\theta \)

Explanation:
1. By symmetry each support carries half the weight, so the vertical component of the end tension is \( \frac W2 \).
2. The tension acts along the chain at angle \( \theta \) above the horizontal: \( 2T\sin\theta=W \).
3. So \( T=\frac{W}{2\sin\theta}=\frac W2\csc\theta \).

Teacher's Note:
1. Isolate the whole chain and balance vertical forces.
2. The tension is smallest when the chain is steep, since then the vertical component does most of the work.

Answer: (c) zero

Explanation:
1. Let the lengths on the two faces be \( L_1 \) and \( L_2 \), with the ends at the same level: \( L_1\sin\alpha=L_2\sin\beta \).
2. The driving force along the rope is \( \frac ML(L_1\sin\alpha-L_2\sin\beta)g=0 \).
3. So the initial acceleration is zero.

Teacher's Note:
1. Both ends are at equal heights, so the weight components along the two faces are equal.
2. Equal and opposite driving forces leave no net acceleration.

Answer: (b) \( w\left(1-\frac ag\right) \)

Explanation:
1. Mass is \( \frac wg \). Taking downward as positive: \( w-f=\frac wga \).
2. So \( f=w-\frac{wa}{g}=w\left(1-\frac ag\right) \).

Teacher's Note:
1. Resistance is the part of the weight that does not go into acceleration.
2. If a = g the resistance is zero.

Answer: (a) \( \sqrt{40} \) m/s

Explanation:
1. The groove is smooth, so only gravity does work, and the height lost is the vertical drop from O to B: \( 4\sin30^\circ=2 \) m (A and B are at the same level).
2. Energy: \( v^2=2gh=2\times10\times2=40 \).
3. So \( v=\sqrt{40} \) m/s. (Check: groove length 5 m, acceleration along it \( 4 \) m/s², \( v^2=2\times4\times5=40 \).)

Teacher's Note:
1. Speed gained depends only on the vertical drop when there is no friction.
2. The shape or length of the path does not matter.

Answer: (d) \( \frac{a}{2gk} \)

Explanation:
1. In the wire's frame the bead feels \( mg \) down and the pseudo force \( ma \) horizontally opposite to the acceleration.
2. For equilibrium on a smooth wire, the resultant of these must be normal to the wire, so the tangent slope satisfies \( \tan\phi=\frac ag \).
3. The slope of the parabola is \( \frac{dy}{dx}=2kx \), so \( 2kx=\frac ag \) and \( x=\frac{a}{2gk} \).

Teacher's Note:
1. On a smooth wire the effective gravity must be perpendicular to the wire.
2. The slope of the curve at the equilibrium point equals a/g.

Answer: (c) 4.9 m/s² in the vertical direction

Explanation:
1. On a smooth incline of angle \( \varphi \) with the horizontal, the acceleration along the plane is \( g\sin\varphi \) and its vertical component is \( g\sin^2\varphi \).
2. A: \( g\sin^260^\circ=\frac{3g}{4} \). B: \( g\sin^230^\circ=\frac g4 \).
3. Relative vertical acceleration \( \frac{3g}{4}-\frac g4=\frac g2=4.9 \) m/s², vertical.

Teacher's Note:
1. The vertical component of acceleration on a smooth incline is g sin² of the incline angle.
2. Subtract the two vertical components because both are downward.