JEE Physics System Of Particles and Rotational Motion MCQs Set 44

Answer: (b) \( \dfrac{F}{2MR} \)

Explanation:
1. From the previous analysis \( f=F/4 \).
2. \( \alpha=\dfrac{2f}{MR}=\dfrac{F}{2MR} \).

Teacher's Note:
1. Friction at the bottom of the disc gives the torque about the centre.
2. The disc's moment of inertia is half M R squared.

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

Explanation:
1. Sphere: \( I=\tfrac25MR^2 \). Disc about an axis through the edge, perpendicular to its plane: \( I=\tfrac12Mr^2+Mr^2=\tfrac32Mr^2 \).
2. \( \tfrac32r^2=\tfrac25R^2\Rightarrow r^2=\dfrac{4R^2}{15} \Rightarrow r=\dfrac{2R}{\sqrt{15}} \).

Teacher's Note:
1. Mass does not change on melting.
2. Use the parallel axes theorem for the edge axis.

Answer: (b) \( \sqrt2 \)

Explanation:
1. \( \tfrac12kx_1^2=\tfrac12I(2\omega)^2=2I\omega^2 \) and \( \tfrac12kx_2^2=\tfrac12(2I)\omega^2=I\omega^2 \).
2. \( \dfrac{x_1^2}{x_2^2}=2\Rightarrow\dfrac{x_1}{x_2}=\sqrt2 \).

Teacher's Note:
1. The spring energy is converted fully into rotational kinetic energy.
2. The ratio of squares of the compressions equals the ratio of energies.

Answer: (a) \( \dfrac{2I\omega}{3t} \)

Explanation:
1. Angular momentum is conserved: \( I(2\omega)+2I\omega=3I\omega_c\Rightarrow\omega_c=\dfrac{4\omega}{3} \).
2. For disc A: \( \tau t=I\left(2\omega-\dfrac{4\omega}3\right)=\dfrac{2I\omega}{3} \).
3. \( \tau=\dfrac{2I\omega}{3t} \).

Teacher's Note:
1. The torque on each disc equals its change of angular momentum over t.
2. The two torques are equal and opposite.

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

Explanation:
1. Initial KE: \( \tfrac12I(4\omega^2)+\tfrac12(2I)\omega^2=3I\omega^2 \).
2. Final: \( \omega_c=\dfrac{4\omega}3 \): \( \tfrac12(3I)\dfrac{16\omega^2}{9}=\dfrac{8I\omega^2}{3} \).
3. Loss \( =3-\dfrac83=\dfrac{I\omega^2}{3} \).

Teacher's Note:
1. Compute the energy before and after.
2. The common angular velocity comes from conservation of angular momentum.

Answer: (d) Assertion is false but Reason is true

Explanation:
1. The velocity of the centre of mass is governed by the net external force, not by the torque, so a body free of torque can still accelerate (Assertion false).
2. The linear momentum of an isolated system (no external force) is conserved (Reason true).

Teacher's Note:
1. Torque affects rotation, force affects the motion of the centre of mass.
2. A false assertion with a true reason gives the fourth type of option.

Answer: (c) Assertion is false but Reason is true

Explanation:
1. Acceleration \( a=\dfrac{g\sin\theta}{1+k^2/R^2} \); the hollow cylinder (\( k^2/R^2=1 \)) is slower than the solid one (\( \tfrac12 \)), so it reaches the bottom later: Assertion false.
2. Both lose the same potential energy, so their total kinetic energies are the same: Reason true.

Teacher's Note:
1. The body with more of its mass far from the axis accelerates less.
2. The reason states a true fact that does not support the assertion.

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

Explanation:
1. Net external force zero means \( d\vec P/dt=0 \), so linear momentum is constant.
2. Torque may be non-zero (for example a couple), so angular momentum can change; kinetic and potential energies may also change.

Teacher's Note:
1. A couple has zero resultant force but a non-zero torque.
2. Internal forces can change energies.

Answer: (b) (B) and (C)

Explanation:
1. With the centre speed v: \( V_A=0 \), \( V_B=v \), \( V_C=2v \) (all in the forward direction).
2. (A): \( V_C-V_A=2v \) but \( 2(V_B-V_C)=-2v \): false. (B): \( 2v-v=v=v-0 \): true. (C): \( 2v=2\times v \): true. (D): \( 2v\ne4v \): false.

Teacher's Note:
1. The contact point is at rest, the centre moves at v and the top at 2v.
2. Compare vectors, not just magnitudes.

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

Explanation:
1. Sliding would need \( \tan\theta=\mu=\sqrt3\Rightarrow\theta=60^\circ \).
2. Toppling occurs when \( \tan\theta=\dfrac{10}{15}=\dfrac23 \), i.e. \( \theta\approx33.7^\circ \), which is smaller. So the block topples before it ever slides.

Teacher's Note:
1. Compare the two critical angles; the smaller one acts first.
2. The topple condition uses the ratio of base to height.

Answer: (c) 9

Explanation:
1. For each sphere \( I_{cm}=\tfrac25mR^2=\tfrac25(0.5)\left(\dfrac{\sqrt5}{2}\times10^{-2}\right)^2=2.5\times10^{-5} \) kg m². Four spheres: \( 1.0\times10^{-4} \).
2. Two spheres lie on the diagonal; the other two are at distance \( 2\sqrt2 \) cm from it: \( 2\times0.5\times(2\sqrt2\times10^{-2})^2=8\times10^{-4} \).
3. Total \( =9\times10^{-4} \) kg m², so N = 9.

Teacher's Note:
1. Use the parallel axes theorem for the two off-diagonal spheres.
2. The spheres on the diagonal contribute only their own moment of inertia.

Answer: (b) 3

Explanation:
1. Let the small disc have mass m; the big disc then has mass 4m.
2. \( I_O=\tfrac12(4m)(2R)^2-\left[\tfrac12mR^2+mR^2\right]=8mR^2-1.5mR^2=6.5mR^2 \).
3. \( I_P=\left[8mR^2+4m(2R)^2\right]-\left[\tfrac12mR^2+m(R^2+4R^2)\right]=24mR^2-5.5mR^2=18.5mR^2 \).
4. \( I_P/I_O=\dfrac{18.5}{6.5}=2.85\approx3 \).

Teacher's Note:
1. Treat the lamina as the full disc minus the removed disc, using the parallel axes theorem.
2. The distance of the removed disc's centre from P is the root of R squared plus 4R squared.

Answer: (d) (A) and (B)

Explanation:
1. Rolling: \( v_O=\omega(3R)\hat i=3R\omega\hat i \): (A) true.
2. The disc turns anticlockwise at \( \omega/2 \): its point P has an extra velocity \( \dfrac\omega2R(-\sin30^\circ\hat i+\cos30^\circ\hat k)=R\omega\left(-\tfrac14\hat i+\tfrac{\sqrt3}{4}\hat k\right) \).
3. Total \( \vec v_P=\tfrac{11}{4}R\omega\hat i+\tfrac{\sqrt3}{4}R\omega\hat k \): (B) true, (C) and (D) false.

Teacher's Note:
1. Add the velocity of the centre and the velocity due to the spin about the centre.
2. The anticlockwise sense in the x-z plane gives the components used above.

Answer: (d) 8

Explanation:
1. Disc: \( I_d=\tfrac12(50)(0.4)^2=4 \) kg m².
2. Each ring about the disc's axis: \( mr^2+md^2=6.25(0.04+0.04)=0.5 \) kg m²; two rings: 1.0.
3. \( 4\times10=(4+1)\omega'\Rightarrow\omega'=8 \) rad/s.

Teacher's Note:
1. Angular momentum about the axis is conserved.
2. Use the parallel axes theorem for each ring.

Answer: (c) (C) and (D)

Explanation:
1. If both surfaces are rough: floor friction \( \mu_2N_2=N_1 \) and \( N_2+\mu_1N_1=mg \), so \( N_2(1+\mu_1\mu_2)=mg \): (C) true.
2. If the wall is smooth: \( N_2=mg \) and, from torques about the foot, \( N_1L\sin\theta=mg\dfrac L2\cos\theta\Rightarrow N_1\tan\theta=\dfrac{mg}{2} \): (D) true, (A) false.
3. With a smooth floor \( (\mu_2=0) \) there is no horizontal force to balance \( N_1 \), so the ladder cannot be in equilibrium: (B) false.

Teacher's Note:
1. Write the force balance in both directions and torque about the foot.
2. Take care to put friction in the right direction at each contact.

Answer: (a) 2

Explanation:
1. The net force is zero, so the disc only rotates. Each force acts along a side at distance \( R/2 \) from O (the inradius of the inscribed triangle).
2. Total torque \( =3\times0.5\times0.25=0.375 \) N m; \( I=\tfrac12(1.5)(0.5)^2=0.1875 \).
3. \( \alpha=2 \) rad/s², so after 1 s \( \omega=2 \) rad/s.

Teacher's Note:
1. The three forces form a closed triangle, so their vector sum is zero.
2. The apothem of an equilateral triangle in a circle of radius R is R over 2.

Answer: (c) 6

Explanation:
1. A thin shell of radius r and thickness dr has \( dI=\tfrac23r^2\,dm \), with \( dm=\rho\,4\pi r^2dr \).
2. \( I_A=\tfrac{8\pi k}{3R}\int_0^Rr^5dr=\dfrac{8\pi kR^5}{18} \) and \( I_B=\tfrac{8\pi k}{3R^5}\int_0^Rr^9dr=\dfrac{8\pi kR^5}{30} \).
3. \( \dfrac{I_B}{I_A}=\dfrac{18}{30}=\dfrac{6}{10} \), so n = 6.

Teacher's Note:
1. Split the sphere into thin spherical shells.
2. A thin shell has moment of inertia two thirds m r squared about a diameter.

Answer: (b) (A), (B) and (D)

Explanation:
1. \( \vec v=3\alpha t^2\hat i+2\beta t\hat j=10\hat i+10\hat j \): (A) true.
2. \( \vec r=\tfrac{10}{3}\hat i+5\hat j \); \( \vec L=m(\vec r\times\vec v)=0.1\left(\tfrac{10}{3}\cdot10-5\cdot10\right)\hat k=-\tfrac53\hat k \): (B) true.
3. \( \vec a=6\alpha t\hat i+2\beta\hat j=20\hat i+10\hat j \), \( \vec F=m\vec a=(2\hat i+\hat j) \) N, not \( \hat i+2\hat j \): (C) false.
4. \( \vec\tau=\vec r\times\vec F=\left(\tfrac{10}{3}(1)-5(2)\right)\hat k=-\tfrac{20}{3}\hat k \): (D) true.

Teacher's Note:
1. Differentiate the position vector once for velocity and twice for acceleration.
2. Torque is r cross F.

Answer: (c) \( \Delta=h\left(\dfrac1{\cos(\pi/n)}-1\right) \)

Explanation:
1. The centre of mass is at the circumradius \( R_c \) from every vertex. When it rests on a side, its height is the apothem \( h=R_c\cos(\pi/n) \).
2. At the highest point of the roll, the vertex is the pivot and the centre of mass is vertically above it at height \( R_c=\dfrac{h}{\cos(\pi/n)} \).
3. \( \Delta=R_c-h=h\left(\dfrac1{\cos(\pi/n)}-1\right) \).

Teacher's Note:
1. The height varies between the apothem and the circumradius.
2. Check with a square: apothem a over 2, circumradius a over root 2.

Answer: (c) (B), (C) and (D)

Explanation:
1. The floor is smooth, so no horizontal force acts and the midpoint falls vertically (D). Its height is \( \tfrac L2\cos\theta \), so the displacement \( \tfrac L2(1-\cos\theta) \) (C).
2. Torque of the weight about the contact point is \( mg\,\tfrac L2\sin\theta \) (B).
3. The end A moves on \( x=\tfrac L2\sin\theta,\ y=L\cos\theta \), an ellipse, not a parabola (A false).

Teacher's Note:
1. The centre of mass of a body free of horizontal force keeps its x-coordinate.
2. Eliminate theta to get the path of the end A.

Answer: (b) (A) and (C)

Explanation:
1. \( \vec a=(t,1) \), \( \vec v=\left(\tfrac{t^2}2,t\right) \), \( \vec r=\left(\tfrac{t^3}{6},\tfrac{t^2}{2}\right) \). At t = 1: \( \vec v=(0.5,1)=\tfrac12(\hat i+2\hat j) \) (C true).
2. \( \vec\tau=\vec r\times\vec F=\left(\tfrac16(1)-\tfrac12(1)\right)\hat k=-\tfrac13\hat k \): magnitude 1/3 (A true), direction \( -\hat k \) (B false).
3. Displacement \( =\sqrt{\tfrac1{36}+\tfrac14}=\dfrac{\sqrt{10}}{6} \) m, not 1/6 (D false).

Teacher's Note:
1. Integrate the acceleration twice from rest.
2. The z-component of r cross F is x F_y minus y F_x.

Answer: (a) 0.75

Explanation:
1. Rolling: \( a=\dfrac{g\sin\theta}{1+k^2/R^2} \) with \( L=\dfrac h{\sin\theta} \), so \( t=\sqrt{\dfrac{2L}a}=\sqrt{\dfrac{2h(1+k^2/R^2)}{g\sin^2\theta}} \).
2. Ring: \( k^2/R^2=1 \), disc: \( \tfrac12 \): \( \Delta t=\sqrt{\dfrac{2h}{g\sin^2\theta}}\left(\sqrt2-\sqrt{\tfrac32}\right)=\sqrt{\dfrac h{3.75}}\,\dfrac{2-\sqrt3}{\sqrt2} \).
3. Setting \( \Delta t=\dfrac{2-\sqrt3}{\sqrt{10}} \): \( \sqrt{\dfrac{h}{3.75}}=\dfrac{\sqrt2}{\sqrt{10}} \Rightarrow \dfrac h{3.75}=0.2\Rightarrow h=0.75 \) m.

Teacher's Note:
1. The time to roll down is proportional to the square root of (1 + k squared over R squared).
2. Write the time in terms of the height and the angle.