JEE Physics System Of Particles and Rotational Motion MCQs Set 66

Answer: (a) \( \sqrt{\dfrac{a^2+b^2+ab}{3}} \)

Explanation:
1. Mass of a thin ring: \( dm=\dfrac{\sigma_0}{r}\,2\pi r\,dr=2\pi\sigma_0\,dr \), so \( M=2\pi\sigma_0(b-a) \).
2. \( I=\int_a^b r^2\,2\pi\sigma_0\,dr=\dfrac{2\pi\sigma_0}{3}(b^3-a^3) \).
3. \( k^2=\dfrac IM=\dfrac{b^2+ab+a^2}{3} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) 140

Explanation:
1. Disc: \( I_1=\tfrac12\cdot\dfrac{7M}8(2R)^2=\dfrac{7MR^2}{4} \).
2. Second part has mass M/8, hence one-eighth of the volume, so its radius is R/2: \( I_2=\tfrac25\cdot\dfrac M8\cdot\dfrac{R^2}{4}=\dfrac{MR^2}{80} \).
3. \( I_1/I_2=\dfrac74\times80=140 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) \( \dfrac{2MR^2}{3} \)

Explanation:
1. \( M=\int_0^R kr^2\,2\pi r\,dr=\dfrac{\pi kR^4}{2} \).
2. \( I=\int_0^R kr^2\,2\pi r\cdot r^2\,dr=\dfrac{\pi kR^6}{3} \).
3. \( I=\dfrac{\pi kR^6/3}{\pi kR^4/2}M=\dfrac{2MR^2}{3} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (d) \( \dfrac{2k}{I} \) \( \theta \)

Explanation:
1. Work-energy: \( \tau\,d\theta=d(KE) \), so \( \tau=\dfrac{d}{d\theta}(k\theta^2)=2k\theta \).
2. \( \alpha=\dfrac\tau I=\dfrac{2k}{I}\theta \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (a) \( \dfrac{8}{5} \)

Explanation:
1. \( M=\int_0^R\rho_0 r\,2\pi r\,dr=\dfrac{2\pi\rho_0R^3}{3} \).
2. \( I_{cm}=\int_0^R\rho_0r\,2\pi r\cdot r^2dr=\dfrac{2\pi\rho_0R^5}{5}=\dfrac35MR^2 \).
3. Parallel axis: \( I=\dfrac35MR^2+MR^2=\dfrac85MR^2 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) A parabola (quadratic in x) that starts from a positive value on the vertical axis

Explanation:
1. Parallel-axis theorem: \( I(x)=I_{cm}+Mx^2=\tfrac25MR^2+Mx^2 \).
2. This is quadratic in x with a non-zero value \( \tfrac25MR^2 \) at x = 0.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) 16 cm

Explanation:
1. Hollow cylinder: \( I=\tfrac12M(R_1^2+R_2^2)=\tfrac12M(100+400)=250M \) (cm² units).
2. Thin shell: \( MR^2=250M\Rightarrow R=\sqrt{250}\approx15.8\approx16 \) cm.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (d) \( \dfrac{15}{16} \) \( I_0 \)

Explanation:
1. Triangle DEF has half the side, so one-quarter of the area and mass, and the same centroid G.
2. \( I\propto ML^2 \), so \( I_{DEF}=\dfrac14\cdot\dfrac14I_0=\dfrac{I_0}{16} \).
3. \( I'=I_0-\dfrac{I_0}{16}=\dfrac{15}{16}I_0 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (a) \( \dfrac{137}{15} \) \( MR^2 \)

Explanation:
1. Each ball: \( \tfrac25MR^2+M(2R)^2=\tfrac{22}5MR^2 \); two balls give \( \tfrac{44}5MR^2 \).
2. Rod: \( \dfrac{M(2R)^2}{12}=\dfrac{MR^2}3 \).
3. Total \( =\dfrac{44}5+\dfrac13=\dfrac{137}{15}MR^2 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) \( \dfrac{181}{2} \) \( MR^2 \)

Explanation:
1. About O: \( \tfrac12MR^2+6\left[\tfrac12MR^2+M(2R)^2\right]=\dfrac{55}2MR^2 \).
2. Shift to P with total mass 7M and OP = 3R: \( I_P=\dfrac{55}2MR^2+7M(3R)^2=\dfrac{181}2MR^2 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

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

Explanation:
1. The removed disc has mass \( 9M\times\dfrac19=M \).
2. Full disc: \( \tfrac12(9M)R^2=\dfrac92MR^2 \). Removed disc about O: \( \tfrac12M\left(\dfrac R3\right)^2+M\left(\dfrac{2R}3\right)^2=\dfrac{MR^2}{2} \).
3. Remaining: \( \dfrac92MR^2-\dfrac12MR^2=4MR^2 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

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

Explanation:
1. With \( V=\pi R^2l \) fixed, \( R^2=\dfrac V{\pi l} \): \( f(l)=\dfrac V{4\pi l}+\dfrac{l^2}{12} \).
2. \( f'(l)=-\dfrac V{4\pi l^2}+\dfrac l6=0\Rightarrow l^3=\dfrac{3V}{2\pi}=\dfrac32R^2l \).
3. \( l^2=\dfrac32R^2\Rightarrow \dfrac lR=\sqrt{\dfrac32} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

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

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

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

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

Explanation:
1. The body diagonal of the largest cube equals 2R, so its side is \( a=\dfrac{2R}{\sqrt3} \).
2. Mass of cube \( m=M\dfrac{a^3}{\tfrac43\pi R^3}=\dfrac{2M}{\sqrt3\,\pi} \).
3. \( I=\dfrac{ma^2}6=\dfrac16\cdot\dfrac{2M}{\sqrt3\pi}\cdot\dfrac{4R^2}3=\dfrac{4MR^2}{9\sqrt3\pi} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

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

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

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (d) \( \dfrac{1}{2} \) \( MR^2 \)

Explanation:
1. The full disc has mass 4M, so \( I_{full}=\tfrac12(4M)R^2=2MR^2 \).
2. By symmetry the quarter has one-fourth: \( I=\dfrac{2MR^2}4=\dfrac12MR^2 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (a) \( \dfrac{3\rho L^3}{8\pi^2} \)

Explanation:
1. Radius \( r=\dfrac L{2\pi} \), mass \( m=\rho L \).
2. About a tangent in the plane: \( I=\tfrac12mr^2+mr^2=\tfrac32mr^2=\dfrac32\rho L\cdot\dfrac{L^2}{4\pi^2}=\dfrac{3\rho L^3}{8\pi^2} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) I

Explanation:
1. For a square, the moment of inertia tensor in the plane is isotropic: \( I_x=I_y \).
2. For an axis at angle \( \theta \): \( I_\theta=I_x\cos^2\theta+I_y\sin^2\theta=I \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) 9

Explanation:
1. Two spheres lie on the diagonal; the other two are at distance \( 2\sqrt2 \) cm from it.
2. Own moments: \( 4\times\tfrac25(0.5)\left(\tfrac{\sqrt5}{2}\times10^{-2}\right)^2=1\times10^{-4} \).
3. Offset: \( 2\times0.5\times(2\sqrt2\times10^{-2})^2=8\times10^{-4} \). Total \( =9\times10^{-4} \), so N = 9.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (d) \( -48\,\hat k \)

Explanation:
1. \( \vec v=2\hat i-6t\hat j \). At t = 2: \( \vec r=4\hat i-12\hat j \), \( \vec v=2\hat i-12\hat j \).
2. \( \vec L=m(\vec r\times\vec v)=2\left[4(-12)-(-12)(2)\right]\hat k=2(-24)\hat k=-48\hat k \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (a) \( \dfrac{M\omega_0}{M+6m} \)

Explanation:
1. No external torque: \( \dfrac{ML^2}{12}\omega_0=\left(\dfrac{ML^2}{12}+2m\dfrac{L^2}4\right)\omega \).
2. \( \omega=\dfrac{M\omega_0}{M+6m} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) 6 kg m²/s

Explanation:
1. Energy: \( v_B^2=5^2+2(10)(10)=225\Rightarrow v_B=15 \) m/s.
2. At B the velocity is horizontal and O is 20 m directly above, so \( L=mvr_\perp=0.02\times15\times20=6 \) kg m²/s.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) \( \dfrac{4}{5} \) \( R \)

Explanation:
1. Conservation of angular momentum: \( MR^2\omega=\left[MR^2+\dfrac M8(r_1^2+r_2^2)\right]\dfrac89\omega \).
2. \( MR^2+\dfrac M8(r_1^2+r_2^2)=\dfrac98MR^2\Rightarrow r_1^2+r_2^2=R^2 \).
3. With \( r_1=\dfrac35R \): \( r_2^2=R^2\left(1-\dfrac9{25}\right)=\dfrac{16}{25}R^2\Rightarrow r_2=\dfrac45R \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (d) changes in direction but not in magnitude

Explanation:
1. The vector from the point of suspension to the bob sweeps a cone, so \( \vec L=\vec r\times m\vec v \) has constant magnitude but a component perpendicular to the axis that rotates.
2. The tension and gravity together give a non-zero torque about the suspension point, so L is not conserved.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (a) \( \vec L_O \) remains constant while \( \vec L_P \) varies with time

Explanation:
1. About O the position vector is radial in the plane and \( \vec v \) is tangential, so \( \vec L_O=mR^2\omega\hat k \) is constant.
2. About P the position vector has a vertical component, so \( \vec L_P \) has a horizontal component that rotates with the particle: it varies in direction with time.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) K/2

Explanation:
1. No external torque, so L is constant: \( K=\dfrac{L^2}{2I} \).
2. If I doubles, K becomes K/2.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) The centre of the circle

Explanation:
1. About the centre the position vector is perpendicular to the velocity at all times, and the force (centripetal) passes through the centre, so the torque is zero.
2. About any other point the torque of the centripetal force is not zero, so L changes.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (d) \( \dfrac{3}{2} \) \( MR^2\omega \)

Explanation:
1. \( L=I_{cm}\omega+Mv_{cm}r_\perp \) with \( v_{cm}=\omega R \) and perpendicular distance R of the line of motion from O.
2. \( L=\tfrac12MR^2\omega+MR^2\omega=\tfrac32MR^2\omega \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (a) remains constant

Explanation:
1. \( L=mvd \), where d is the perpendicular distance of the line of motion from the origin.
2. Both v and d are constant, so L is constant and non-zero.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) \( \dfrac{mv^3}{4\sqrt2\,g} \)

Explanation:
1. At the top the velocity is horizontal, \( v_x=\dfrac v{\sqrt2} \), and the height is \( h=\dfrac{v^2\sin^2 45^\circ}{2g}=\dfrac{v^2}{4g} \).
2. \( L=mv_xh=m\dfrac v{\sqrt2}\cdot\dfrac{v^2}{4g}=\dfrac{mv^3}{4\sqrt2\,g} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) \( \dfrac{\omega M}{M+2m} \)

Explanation:
1. \( Mr^2\omega=(M+2m)r^2\omega' \).
2. \( \omega'=\dfrac{M\omega}{M+2m} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.