JEE Physics System Of Particles and Rotational Motion MCQs Set 69

Answer: (c) \( L\frac{M}{m+M} \)

Explanation:
1. Take \( m \) at the origin and \( M \) at \( x=L \).
2. \( x_{cm}=\dfrac{m\cdot0+M\cdot L}{m+M}=\dfrac{ML}{m+M} \).

Teacher's Note:
1. The centre of mass is closer to the heavier mass.
2. Its distance from a mass is proportional to the other mass.

Answer: (a) in all cases

Explanation:
1. There is no external horizontal force on the system (water resistance is neglected).
2. So the centre of mass cannot accelerate; being at rest initially, it stays at rest however the man moves.

Teacher's Note:
1. Internal forces cannot change the motion of the centre of mass.
2. The result does not depend on masses or how the man walks.

Answer: (a) in all cases

Explanation:
1. The only external force is gravity (taking air resistance as negligible), which acts as if at the centre of mass.
2. So the centre of mass follows a parabola irrespective of the rotation or shape of the stick.

Teacher's Note:
1. The centre of mass of a projectile system moves as a point projectile.
2. Uniformity and rotation affect the spin, not the path of the centre of mass.

Answer: (a) remain stationary

Explanation:
1. The system is stationary, so the net external force is zero (weight balanced by buoyancy).
2. Internal motion cannot move the centre of mass, so it remains stationary while the man and balloon move towards each other.

Teacher's Note:
1. The man and balloon shift in opposite directions to keep the centre of mass fixed.
2. This is the same idea as the man walking on a boat.

Answer: (d) any point on the system or outside it

Explanation:
1. If the net force is zero and the torque about one point is zero, the torque about every point is zero.
2. So the torque condition can be applied about any point, inside or outside the system.

Teacher's Note:
1. Choose the point that eliminates unknown forces to simplify the algebra.
2. This works because the net force is also zero in equilibrium.

Answer: (a) 1:2

Explanation:
1. Let the strings be at 0 and 100 cm. The weights are \( mg \) at 50 cm and \( 2mg \) at 75 cm; total \( 3mg \).
2. Torques about the 0 end: \( T_2\times100=mg\times50+2mg\times75=200mg \), so \( T_2=2mg \).
3. \( T_1=3mg-2mg=mg \), so \( T_1:T_2=1:2 \).

Teacher's Note:
1. Take torques about one support to eliminate one tension.
2. The load nearer a support makes that support take more of the weight.

Answer: (d) 3:5

Explanation:
1. Let the forces be \( F_1,F_2 \) at separation \( d \). The resultant acts at distance \( \dfrac{F_2d}{F_1+F_2} \) from \( F_1 \).
2. After interchange it is \( \dfrac{F_1d}{F_1+F_2} \) from the same point, so the shift is \( \dfrac{(F_1-F_2)d}{F_1+F_2}=\dfrac d4 \).
3. \( 4(F_1-F_2)=F_1+F_2 \) gives \( 3F_1=5F_2 \), so the ratio of the forces is \( 3:5 \).

Teacher's Note:
1. The resultant of like parallel forces divides the separation in the inverse ratio of the forces.
2. Order the ratio consistently as smaller : larger or larger : smaller; here it is 3:5.

Answer: (c) 70 g

Explanation:
1. Take torques about the 40-cm pivot. Anticlockwise: \( 10\times30+20\times20=700 \) g cm from the weights on the left.
2. The scale's weight \( W \) acts at the 50-cm mark, 10 cm to the right, giving \( W\times10 \).
3. \( 10W=700 \), so \( W=70 \) g.

Teacher's Note:
1. The centre of a uniform scale is at the 50-cm mark.
2. Torque is weight times perpendicular distance from the pivot.

Answer: (d) it will not topple for any value of \( v \)

Explanation:
1. On a smooth surface there is no friction, so no horizontal force acts at the base.
2. Without a torque about the centre of mass, the block keeps moving uniformly and never topples.

Teacher's Note:
1. Toppling needs a force at the base (friction or an obstacle) that produces a torque.
2. Speed alone cannot topple a block on a frictionless surface.

Answer: (b) \( \mu>\frac ah \)

Explanation:
1. Friction \( f=\mu mg \) acts at the base, a distance \( h/2 \) below the centre of mass. The normal reaction \( N=mg \) shifts forward by at most \( a/2 \) from the centre.
2. Torque balance about the centre of mass at the point of toppling: \( N\cdot\dfrac a2=f\cdot\dfrac h2 \).
3. Toppling occurs when \( \mu mg\dfrac h2>mg\dfrac a2 \), i.e. \( \mu>\dfrac ah \).

Teacher's Note:
1. Compare the toppling torque of friction with the restoring torque of the weight.
2. A tall, narrow block topples more easily.

Answer: (d) 150

Explanation:
1. Average angular velocity \( =\dfrac{20\pi+40\pi}{2}=30\pi \) rad/s.
2. Angle \( =30\pi\times10=300\pi \) rad, which is \( \dfrac{300\pi}{2\pi}=150 \) rotations.

Teacher's Note:
1. For uniform acceleration the average velocity is the mean of the two.
2. Divide the angle by \( 2\pi \) to get rotations.

Answer: (c) 30

Explanation:
1. Starting from rest, \( \theta=\tfrac12\alpha t^2 \), so the angle is proportional to \( t^2 \).
2. In 6 s the fan makes \( 4\times10=40 \) rotations.
3. In the next 3 s it makes \( 40-10=30 \) rotations.

Teacher's Note:
1. Distances covered in equal successive intervals from rest are in the ratio 1:3:5.
2. The second 3 seconds therefore give three times the first.

Answer: (d) 12

Explanation:
1. \( \omega^2-\left(\dfrac\omega2\right)^2=2\alpha\,(36) \) gives \( \dfrac{3\omega^2}{4}=72\alpha \) (in rotations units).
2. Further: \( \left(\dfrac\omega2\right)^2=2\alpha\,N \), so \( N=\dfrac{\omega^2/4}{3\omega^2/4}\times36=12 \).

Teacher's Note:
1. Use \( \omega^2=\omega_0^2-2\alpha\theta \) with the angle measured in rotations throughout.
2. Equal steps of speed reduction need unequal angles.

Answer: (b) \( n \)

Explanation:
1. \( \alpha=-k\omega \) gives \( \omega\dfrac{d\omega}{d\theta}=-k\omega \), so \( \dfrac{d\omega}{d\theta}=-k \).
2. The angular velocity falls linearly with angle. Falling from \( \omega_0 \) to \( \omega_0/2 \) takes \( n \) rotations, so the remaining half takes another \( n \) rotations.

Teacher's Note:
1. Retardation proportional to speed means speed is a linear function of angle turned.
2. Compare with constant retardation, where the answer would be \( n/3 \).

Answer: (a) \( \Delta E\propto\theta^2 \)

Explanation:
1. Retarding torque \( \tau=-k\theta \) (proportional to the angle).
2. Work done \( =\displaystyle\int_0^\theta k\theta\,d\theta=\tfrac12k\theta^2 \).
3. So the loss of kinetic energy \( \Delta E\propto\theta^2 \).

Teacher's Note:
1. Loss of kinetic energy equals the work done by the retarding torque.
2. A torque growing linearly with angle gives a quadratic energy loss.

Answer: (b) \( \omega\propto\frac1\tau \)

Explanation:
1. Power \( P=\tau\omega=\text{constant} \).
2. So \( \omega=\dfrac P\tau\propto\dfrac1\tau \).

Teacher's Note:
1. Constant power means torque and angular velocity vary inversely.
2. This is the rotational form of \( P=Fv \).

Answer: (d) independent of \( \theta \)

Explanation:
1. Each rod is straight with its midpoint at the intersection, so each rod has \( I=\dfrac{ml^2}{12} \) about the axis through its centre perpendicular to the rod.
2. The total is \( \dfrac{ml^2}{6} \), which does not involve the angle.

Teacher's Note:
1. The moment of inertia of a rod about a perpendicular axis through its centre is independent of its orientation in the plane.
2. Moments of inertia about the same axis add.

Answer: (c) \( \frac{ml^2}{3}\sin^2\theta \)

Explanation:
1. Only the component of distance perpendicular to the axis matters: an element at distance \( x \) along the rod is at \( x\sin\theta \) from the axis.
2. \( I=\displaystyle\int_0^l\dfrac ml x^2\sin^2\theta\,dx=\dfrac{ml^2}{3}\sin^2\theta \).

Teacher's Note:
1. At \( \theta=90^\circ \) this reduces to \( ml^2/3 \).
2. At \( \theta=0 \) the rod lies along the axis and \( I=0 \).

Answer: (d) \( \frac12m(R^2+r^2) \)

Explanation:
1. With surface density \( \sigma \), \( m=\sigma\pi(R^2-r^2) \).
2. \( I=\tfrac12\sigma\pi(R^4-r^4)=\tfrac12\sigma\pi(R^2-r^2)(R^2+r^2) \).
3. So \( I=\tfrac12m(R^2+r^2) \), where \( m \) is the mass of the remaining disc.

Teacher's Note:
1. Subtract the moment of inertia of the removed part, then express the result in the remaining mass.
2. Check: \( r=0 \) gives \( \tfrac12mR^2 \), the solid disc.