JEE Physics System Of Particles and Rotational Motion MCQs Set 40

Answer: (a) \( \dfrac g2 \)

Explanation:
1. For the hollow cylinder \( TR=mR^2\alpha \) and \( a=R\alpha \), so \( T=ma \).
2. For the mass: \( mg-T=ma \), so \( mg=2ma \).
3. \( a=\dfrac g2 \).

Teacher's Note:
1. Write one equation for the falling mass and one for the cylinder, linked by a = R alpha.

Answer: (a) \( \dfrac{4MR^2}{9\sqrt3\,\pi} \)

Explanation:
1. The cube's body diagonal equals the diameter: \( \sqrt3\,s=2R \), so \( s=\dfrac{2R}{\sqrt3} \).
2. Mass of cube: \( m=M\dfrac{s^3}{\tfrac43\pi R^3}=\dfrac{2M}{\sqrt3\,\pi} \).
3. \( I=\dfrac{ms^2}{6}=\dfrac{2M}{\sqrt3\pi}\cdot\dfrac{4R^2}{3}\cdot\dfrac16=\dfrac{4MR^2}{9\sqrt3\,\pi} \).

Teacher's Note:
1. Find the side from the diagonal, get the mass from the volume ratio, then use I = m s squared / 6 for a cube about a face axis.

Answer: (a) \( \sqrt{\dfrac32} \)

Explanation:
1. \( I=m\left(\dfrac{R^2}{4}+\dfrac{l^2}{12}\right) \) with fixed volume \( V=\pi R^2l \), so \( R^2=\dfrac{V}{\pi l} \).
2. \( I\propto \dfrac{V}{4\pi l}+\dfrac{l^2}{12} \). Setting the derivative to zero: \( -\dfrac{V}{4\pi l^2}+\dfrac l6=0\Rightarrow l^3=\dfrac{3V}{2\pi}=\dfrac32R^2l \).
3. So \( l^2=\dfrac32R^2 \) and \( \dfrac lR=\sqrt{\dfrac32} \).

Teacher's Note:
1. Eliminate R using the fixed volume and minimise the resulting function of l.

Answer: (b) \( \dfrac{3g}{2l}\sin\theta \)

Explanation:
1. Torque of the weight about the pivot: \( Mg\dfrac l2\sin\theta \).
2. \( I=\dfrac{Ml^2}{3} \).
3. \( \alpha=\dfrac{Mg(l/2)\sin\theta}{Ml^2/3}=\dfrac{3g}{2l}\sin\theta \).

Teacher's Note:
1. The lever arm of the weight about the pivot is (l/2) sin(theta) when theta is measured from the vertical.

Answer: (b) \( 4MR^2 \)

Explanation:
1. Removed mass \( =9M\cdot\dfrac{(R/3)^2}{R^2}=M \).
2. Full disc: \( \tfrac12(9M)R^2=\dfrac92MR^2 \).
3. Removed disc: \( \tfrac12M\left(\dfrac R3\right)^2+M\left(\dfrac{2R}3\right)^2=\dfrac{MR^2}{18}+\dfrac{8MR^2}{18}=\dfrac12MR^2 \).
4. Remaining: \( \dfrac92MR^2-\dfrac12MR^2=4MR^2 \).

Teacher's Note:
1. Subtract the moment of inertia of the removed disc (using the parallel axis theorem) from that of the full disc.

Answer: (a) \( \dfrac{2R}{\sqrt{15}} \)

Explanation:
1. Sphere about a diameter: \( I=\dfrac25MR^2 \).
2. Disc about its edge: \( I=\tfrac12Mr^2+Mr^2=\dfrac32Mr^2 \).
3. Equate: \( \dfrac32r^2=\dfrac25R^2\Rightarrow r^2=\dfrac{4R^2}{15} \), so \( r=\dfrac{2R}{\sqrt{15}} \).

Teacher's Note:
1. Use the parallel axis theorem for the disc about its edge and equate the two moments of inertia.

Answer: (b) 3 and 4 only

Explanation:
1. For rolling without slipping, the friction is static and does no net work (1 false).
2. Its value \( f=\tfrac13mg\sin\theta \) is constant for a given incline (2 false).
3. It acts up the plane, so it reduces the translational acceleration but provides the torque for rotation (3 true), and it decreases as \( \theta \) decreases (4 true).

Teacher's Note:
1. For a solid cylinder f = (1/3) m g sin(theta), which is constant on a fixed incline.

Answer: (d) disc

Explanation:
1. Energy: \( \tfrac12mv^2(1+k)=mg\cdot\dfrac{3v^2}{4g} \), where \( k=I/(mR^2) \).
2. So \( 1+k=\dfrac32\Rightarrow k=\dfrac12 \).
3. \( k=\tfrac12 \) corresponds to a disc (ring 1, hollow sphere 2/3, solid sphere 2/5).

Teacher's Note:
1. Compare the value of I/(mR squared) with the standard shapes.

Answer: (b) \( \dfrac{I\omega^2}{3} \)

Explanation:
1. Common speed \( \omega'=\dfrac{4\omega}{3} \).
2. Initial KE: \( \tfrac12I(4\omega^2)+\tfrac12(2I)\omega^2=3I\omega^2 \).
3. Final KE: \( \tfrac12(3I)\dfrac{16\omega^2}{9}=\dfrac83I\omega^2 \).
4. Loss \( =3I\omega^2-\dfrac83I\omega^2=\dfrac{I\omega^2}{3} \).

Teacher's Note:
1. Compare the total kinetic energy before and after reaching the common angular velocity.

Answer: (d) Statement 1 false, Statement 2 true

Explanation:
1. Acceleration \( a=\dfrac{g\sin\theta}{1+k} \), with \( k=1 \) for the hollow cylinder and \( k=\tfrac12 \) for the solid one.
2. The solid cylinder accelerates faster and reaches the bottom first, so Statement 1 is false.
3. Both start from the same height with the same mass, so the total kinetic energy at the bottom is \( mgh \) for both; Statement 2 is true.

Teacher's Note:
1. A larger I/(mR squared) means a smaller acceleration, so the solid cylinder wins the race.

Answer: (b) The block remains at rest up to a certain angle and then topples.

Explanation:
1. Sliding would begin at \( \tan\theta=\mu=\sqrt3 \), i.e. 60 degrees.
2. Toppling begins at \( \tan\theta=\dfrac{10}{15}=\dfrac23 \), i.e. about 33.7 degrees, which is smaller.
3. So the block stays at rest until about 33.7 degrees and then topples, before sliding can occur.

Teacher's Note:
1. Compare the toppling angle with the sliding angle; the smaller one decides what happens.

Answer: (a) the linear momentum of the system does not change in time

Explanation:
1. Zero net external force gives \( \dfrac{d\vec P}{dt}=0 \), so linear momentum is conserved.
2. The external forces could still form a couple, so angular momentum about a point need not stay constant, and internal forces can change kinetic and potential energies.

Teacher's Note:
1. Zero net force guarantees constant momentum only; torque and energy changes are not excluded.

Answer: (c) 13

Explanation:
1. Each sphere about its own diameter: \( \tfrac25(0.5)(0.025)^2=1.25\times10^{-4} \) kg m squared; four give \( 5\times10^{-4} \).
2. Two spheres lie on the diagonal (distance 0). The other two are \( 4/\sqrt2 \) cm from it, so \( d^2=8\times10^{-4} \) m squared and each adds \( 0.5\times8\times10^{-4} \).
3. Total: \( 5\times10^{-4}+2(0.5)(8\times10^{-4})=13\times10^{-4} \), so N = 13.

Teacher's Note:
1. Add the spheres' own moment of inertia and the parallel axis term for the two off-diagonal spheres.

Answer: (b) 2 rad/s

Explanation:
1. The perpendicular distance of each side from the centre is \( R/2=0.25 \) m.
2. Net torque: \( 3\times0.5\times0.25=0.375 \) N m.
3. \( I=\tfrac12(1.5)(0.5)^2=0.1875 \), so \( \alpha=2 \) rad/s squared and \( \omega=\alpha t=2 \) rad/s.

Teacher's Note:
1. Equal forces in the same cyclic sense give zero net force but a net torque.

Answer: (d) \( \dfrac45R \)

Explanation:
1. Angular momentum is conserved: \( MR^2\omega=\left[MR^2+\dfrac M8(r_1^2+r_2^2)\right]\dfrac{8\omega}{9} \).
2. So \( MR^2+\dfrac M8(r_1^2+r_2^2)=\dfrac98MR^2 \), giving \( r_1^2+r_2^2=R^2 \).
3. With \( r_1=\dfrac35R \): \( r_2^2=\dfrac{16}{25}R^2 \), so \( r_2=\dfrac45R \).

Teacher's Note:
1. Conserve angular momentum; the sum of squared distances then follows directly.

Answer: (d) \( \dfrac hl=\dfrac{3\sqrt3}{16},\ f=\dfrac{16\sqrt3}{3}\) N

Explanation:
1. Weight \( W=16 \) N. Let the wall reaction be N (perpendicular to the stick, so at 30 degrees above the horizontal) and the floor reaction also N.
2. Vertical: \( N+N\sin30^\circ=16\Rightarrow N=\dfrac{32}{3} \) N.
3. Horizontal: \( f=N\cos30^\circ=\dfrac{16\sqrt3}{3} \) N.
4. Moments about the bottom: \( N\cdot\dfrac{h}{\cos30^\circ}=W\cdot\dfrac l2\sin30^\circ=4l \), so \( \dfrac hl=\dfrac{4\cos30^\circ}{32/3}=\dfrac{3\sqrt3}{16} \).

Teacher's Note:
1. Balance vertical forces, horizontal forces and moments about the foot of the stick.

Answer: (b) 1, 2 and 4 only

Explanation:
1. \( \vec v=3\alpha t^2\hat i+2\beta t\hat j=(10,10) \) at t = 1 (1 true).
2. \( \vec r=(10/3,5) \), so \( L_z=m(xv_y-yv_x)=0.1\left(\dfrac{100}{3}-50\right)=-\dfrac53 \) (2 true).
3. \( \vec a=(6\alpha t,2\beta)=(20,10) \), so \( \vec F=m\vec a=(2,1) \) N, and statement 3 is false.
4. \( \tau_z=xF_y-yF_x=\dfrac{10}{3}-10=-\dfrac{20}{3} \) (4 true).

Teacher's Note:
1. Differentiate the position vector for velocity and acceleration, then use cross products for L and torque.

Answer: (d) \( h\left(\dfrac{1}{\cos(\pi/n)}-1\right) \)

Explanation:
1. When resting on a side, the centre is at the apothem height h.
2. At the top of its path the centre is directly above the pivot vertex, at the circumradius \( \dfrac{h}{\cos(\pi/n)} \).
3. Increase \( =\dfrac{h}{\cos(\pi/n)}-h \).

Teacher's Note:
1. The centre of mass rises from the apothem to the circumradius of the polygon.

Answer: (b) 2 and 3 only

Explanation:
1. The force is \( F=-\dfrac{dV}{dr}=-kr \), directed to O: \( \dfrac{mv^2}{R}=kR \).
2. So \( v=\sqrt{\dfrac km}\,R \) (2 true).
3. \( L=mvR=m\sqrt{\dfrac km}R^2=\sqrt{mk}\,R^2 \) (3 true).

Teacher's Note:
1. Equate the central force to the required centripetal force, then compute m v R.

Answer: (a) 1 and 3 only

Explanation:
1. \( \vec v=\left(\dfrac{t^2}{2},\,t\right) \), \( \vec r=\left(\dfrac{t^3}{6},\,\dfrac{t^2}{2}\right) \).
2. \( \tau_z=xF_y-yF_x=\dfrac{t^3}{6}-\dfrac{t^3}{2}=-\dfrac{t^3}{3} \), so at t = 1 the magnitude is 1/3 N m and the direction is \( -\hat k \) (1 true, 2 false).
3. \( \vec v(1)=\left(\dfrac12,1\right)=\dfrac12(\hat i+2\hat j) \) (3 true).
4. Displacement \( =\left(\dfrac16,\dfrac12\right) \), magnitude \( \sqrt{\dfrac1{36}+\dfrac14}\ne\dfrac16 \) (4 false).

Teacher's Note:
1. Integrate force for velocity and position, then take r cross F for the torque.

Answer: (b) 0.75 m

Explanation:
1. Ring: \( a=\tfrac12g\sin\theta \); disc: \( a=\tfrac23g\sin\theta \). With path length \( L=\dfrac{H}{\sin\theta} \): \( t_{ring}=2\sqrt{\dfrac{L}{g\sin\theta}} \), \( t_{disc}=\sqrt3\sqrt{\dfrac{L}{g\sin\theta}} \).
2. \( \Delta t=(2-\sqrt3)\sqrt{\dfrac{L}{g\sin\theta}}=\dfrac{2-\sqrt3}{\sqrt{10}} \Rightarrow \dfrac{L}{g\sin\theta}=\dfrac1{10} \).
3. \( L=\sin60^\circ=\dfrac{\sqrt3}{2} \) m and \( H=L\sin60^\circ=\dfrac34 \) m.

Teacher's Note:
1. Write the time to slide down the shortest path for each body and equate the difference to the given value.

Answer: (a) P: p, L, K, U, E; Q: L, E; R: L, K, U, E; S: E

Explanation:
1. P: straight line at constant velocity, no force: everything is conserved.
2. Q: an ellipse with a central force; L and E are conserved but the speed varies, so K is not.
3. R: uniform circular motion; p changes direction but L, K, U (constant radius) and E are conserved.
4. S: constant force along j; only E is conserved (torque of the force about the origin is not zero and p, K, U change).

Teacher's Note:
1. Identify the force for each path (zero, central or uniform) and check which quantities stay constant.

Answer: (a) \( \dfrac{2\sqrt3\,\pi a}{v_0} \)

Explanation:
1. The impulse is \( mv_0 \) along BC; the perpendicular distance of line BC from the centre of mass is \( \dfrac{a}{2\sqrt3} \).
2. Angular momentum about the centre: \( mv_0\dfrac{a}{2\sqrt3} \), and \( I=3m\left(\dfrac a{\sqrt3}\right)^2=ma^2 \), so \( \omega=\dfrac{v_0}{2\sqrt3\,a} \).
3. Half a revolution takes \( t=\dfrac\pi\omega=\dfrac{2\sqrt3\,\pi a}{v_0} \).

Teacher's Note:
1. The impulse gives the triangle a spin about its centre of mass; the time follows from t = pi / omega.

Answer: (a) \( \dfrac47 \)

Explanation:
1. Contact is lost when the normal force is zero: \( mg\cos\phi=\dfrac{mv^2}{R} \).
2. Energy for rolling about the edge: \( mgR(1-\cos\phi)=\tfrac12mv^2+\tfrac12\cdot\tfrac12mR^2\omega^2=\tfrac34mv^2 \).
3. So \( \cos\phi=\dfrac43(1-\cos\phi)\Rightarrow\cos\phi=\dfrac47 \).

Teacher's Note:
1. Set the normal reaction to zero for the centripetal condition and use energy conservation with the rolling kinetic energy.