Mix Example - System of Particles and Rotational Motion Questions in English

(D) Given: Mass $(m) = 10 \ kg$,Radius $(r) = 0.5 \ m$,Velocity $(v) = 2 \ m/s$,Total Kinetic Energy $(E) = 32.8 \ J$.
The total kinetic energy of a body rolling without slipping is the sum of its translational and rotational kinetic energies:
$E = E_{\text{translational}} + E_{\text{rotational}}$
$E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = mk^2$ and $\omega = \frac{v}{r}$,where $k$ is the radius of gyration:
$E = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)\left(\frac{v}{r}\right)^2$
$32.8 = \frac{1}{2} \times 10 \times (2)^2 + \frac{1}{2} \times 10 \times k^2 \times \left(\frac{2}{0.5}\right)^2$
$32.8 = 20 + 5 \times k^2 \times (4)^2$
$32.8 = 20 + 80k^2$
$12.8 = 80k^2$
$k^2 = \frac{12.8}{80} = 0.16$
$k = \sqrt{0.16} = 0.4 \ m$.

(C) Case $(i)$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.
$K_{trans} = \frac{1}{2}mv^2$
Case $(ii)$: When the hollow cylinder rolls without slipping,it possesses both translational and rotational kinetic energy.
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a thin hollow cylinder,the moment of inertia $I = mR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$.
$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$
Ratio of kinetic energies:
$\frac{K_{trans}}{K_{total}} = \frac{\frac{1}{2}mv^2}{mv^2} = \frac{1}{2}$
Thus,the ratio is $1:2$.

(A) The rotational kinetic energy $(K_{rot})$ of a body is given by $K_{rot} = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular speed.
Given $I = 3 \ kg \cdot m^2$ and $\omega = 2 \ rad/s$,we have $K_{rot} = \frac{1}{2} \times 3 \times (2)^2 = \frac{1}{2} \times 3 \times 4 = 6 \ J$.
The translational kinetic energy $(K_{trans})$ of a mass $m$ moving with speed $v$ is given by $K_{trans} = \frac{1}{2} m v^2$.
Given $m = 12 \ kg$ and $K_{trans} = K_{rot} = 6 \ J$,we set up the equation: $6 = \frac{1}{2} \times 12 \times v^2$.
$6 = 6 \times v^2 \Rightarrow v^2 = 1 \Rightarrow v = 1 \ m/s$.

(D) Statement $(a)$ is incorrect because the centre of mass coincides with the centre of gravity only if the gravitational field is uniform.
Statement $(b)$ is correct because the centre of mass is defined as the point where the net gravitational torque acting on the body is zero.
Statement $(c)$ is incorrect because a couple produces only rotational motion,not translational motion,as the net force of a couple is zero.
Statement $(d)$ is correct because $\text{Mechanical Advantage} = \frac{\text{Load}}{\text{Effort}}$. If $\text{Mechanical Advantage} > 1$,then $\text{Load} > \text{Effort}$,meaning a small effort can lift a large load.
Therefore,statements $(b)$ and $(d)$ are correct.

(D) The rotational kinetic energy $(K_r)$ of a body is given by $K_r = \frac{1}{2}I\omega^2$.
Given $I = 3\ kg\cdot m^2$ and $\omega = 2\ rad/s$,we have $K_r = \frac{1}{2} \times 3 \times (2)^2 = \frac{1}{2} \times 3 \times 4 = 6\ J$.
The translational kinetic energy $(K_t)$ of a mass $m$ moving with velocity $v$ is given by $K_t = \frac{1}{2}mv^2$.
According to the problem,$K_r = K_t$.
Therefore,$6 = \frac{1}{2} \times 12 \times v^2$.
$6 = 6 \times v^2$.
$v^2 = 1$.
$v = 1\ m/s$.

(A) The total kinetic energy of a rolling body is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$.
For a disc,the radius of gyration $k$ satisfies $\frac{k^2}{R^2} = \frac{1}{2}$. Thus,$K_{disc} = \frac{1}{2}mv_d^2(1 + \frac{1}{2}) = \frac{3}{4}mv_d^2$.
For a ring,the radius of gyration $k$ satisfies $\frac{k^2}{R^2} = 1$. Thus,$K_{ring} = \frac{1}{2}mv_r^2(1 + 1) = mv_r^2$.
Given that $K_{disc} = K_{ring}$,we have $\frac{3}{4}mv_d^2 = mv_r^2$.
Rearranging the terms,we get $\frac{v_d^2}{v_r^2} = \frac{4}{3}$.
Therefore,the ratio of their velocities is $\frac{v_d}{v_r} = \sqrt{\frac{4}{3}} = \frac{\sqrt{4}}{\sqrt{3}}$.

(D) The work-energy theorem states that the work done by an external torque is equal to the change in rotational kinetic energy.
$W = \tau \cdot \theta = \Delta K$
Since the bodies are brought to rest,the work done by the torque $\tau$ must equal the initial rotational kinetic energy $K$.
$\tau \cdot \theta = K$
$\theta = \frac{K}{\tau}$
Here,$\theta$ is the angular displacement in radians,which is related to the number of rotations $n$ by $\theta = 2\pi n$.
Since both the initial kinetic energy $K$ and the applied torque $\tau$ are the same for all three objects,the angular displacement $\theta$ will be the same for all of them.
Therefore,the number of rotations $n = \frac{\theta}{2\pi} = \frac{K}{2\pi\tau}$ will be the same for the ring,the solid sphere,and the disc.

(B) The difficulty in stopping a rotating object is determined by its rotational kinetic energy,$K = \frac{1}{2} I \omega^2$. Since both objects have the same angular velocity $\omega$,the one with the smaller moment of inertia $I$ will have less rotational kinetic energy and will be easier to stop.
For a ring of mass $M$ and radius $R$ rotating about its diameter,the moment of inertia is $I_{\text{ring}} = \frac{1}{2} MR^2$.
For a solid sphere of mass $M$ and radius $R$ rotating about its diameter,the moment of inertia is $I_{\text{sphere}} = \frac{2}{5} MR^2$.
Comparing the two,$I_{\text{sphere}} = 0.4 MR^2$ and $I_{\text{ring}} = 0.5 MR^2$.
Since $I_{\text{sphere}} < I_{\text{ring}}$,the solid sphere has less rotational kinetic energy and is easier to stop.

(A) $1$. Moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5}MR^2$.
$2$. Volume of the sphere is $V = \frac{4}{3}\pi R^3$. The volume remains constant when recast into a disc,so $V = \pi r^2 t$. Thus,$t = \frac{4R^3}{3r^2}$.
$3$. The moment of inertia of a disc about its central axis perpendicular to its plane is $I_{cm} = \frac{1}{2}Mr^2$.
$4$. Using the parallel axis theorem,the moment of inertia about an axis tangent to the edge and perpendicular to the plane is $I_{tangent} = I_{cm} + Mr^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.
$5$. Given $I_{tangent} = I$,we have $\frac{3}{2}Mr^2 = \frac{2}{5}MR^2$.
$6$. Simplifying,$r^2 = \frac{4}{15}R^2$,which gives $r = \frac{2}{\sqrt{15}}R$.

(D) Given that the angular momentum $L$ is $n$ times the linear momentum $P$,we have $L = nP$.
The total kinetic energy $(KE)$ of a rigid body is the sum of its rotational kinetic energy $(KE_R)$ and translational kinetic energy $(KE_T)$.
$KE = KE_R + KE_T$
Using the formulas $KE_R = \frac{L^2}{2I}$ and $KE_T = \frac{P^2}{2m}$,where $I$ is the moment of inertia:
$KE = \frac{L^2}{2I} + \frac{P^2}{2m}$
Substituting $L = nP$ into the equation:
$KE = \frac{(nP)^2}{2I} + \frac{P^2}{2m}$
$KE = \frac{n^2P^2}{2I} + \frac{P^2}{2m}$
Factoring out $\frac{P^2}{2}$:
$KE = \frac{P^2}{2}\left(\frac{n^2}{I} + \frac{1}{m}\right)$

(A) Let the ring have mass $M = 2 \ kg$,radius $R = 0.5 \ m$,and velocity $v = 1 \ m/s$. The ball has mass $m = 0.1 \ kg$,initial velocity $u_x = -20 \ m/s$,and final velocity $v_x = 0, v_y = 10 \ m/s$. The collision point is at height $h = 0.75 \ m = 1.5R$.
Using impulse-momentum theorem for the ball in the $x$-direction: $J_x = m(v_x - u_x) = 0.1(0 - (-20)) = 2 \ Ns$. The impulse on the ring is $J_x' = -2 \ Ns$.
For the ring,the change in linear momentum is $M(v_f - v_i) = J_x'$,so $2(v_f - 1) = -2$,giving $v_f = 0$.
The impulse on the ring at height $1.5R$ creates an angular impulse about the center of mass: $\tau_{imp} = J_x' \times (1.5R - R) = -2 \times 0.25 = -0.5 \ Nms$.
Initial angular momentum $L_i = I\omega = (MR^2)(v/R) = MvR = 2 \times 1 \times 0.5 = 1 \ kgm^2/s$.
Final angular momentum $L_f = L_i + \tau_{imp} = 1 - 0.5 = 0.5 \ kgm^2/s$.
Since $L_f = I\omega_f = (MR^2)\omega_f = (2 \times 0.5^2)\omega_f = 0.5\omega_f$,we get $0.5 = 0.5\omega_f$,so $\omega_f = 1 \ rad/s$.
Since $v_f = 0$ and $\omega_f \neq 0$,the ring is in pure rotation about its center of mass.

(A) According to the principle of conservation of angular momentum,if no external torque is applied $(\tau_{ext} = 0)$,the angular momentum $L = I\omega$ remains constant.
The rotational kinetic energy is given by $K = \frac{1}{2} I\omega^2$. Substituting $\omega = \frac{L}{I}$,we get $K = \frac{L^2}{2I}$.
If the moment of inertia $I$ increases while $L$ remains constant,the kinetic energy $K$ must decrease because $K \propto \frac{1}{I}$.
Statement-$1$ claims $K$ decreases,which is possible if $I$ changes. However,the statement implies $K$ decreases without specifying a change in $I$,which is physically inconsistent with the conservation of $L$ and constant $I$. If $I$ is constant and $L$ is constant,$\omega$ must be constant,and thus $K$ must be constant. Therefore,Statement-$1$ is false.
Statement-$2$ provides the correct definitions for $L$ and $K$,so it is true.

(D) The moment of inertia of a solid sphere about its diameter is given by the formula:
$I = \frac{2}{5} M R^2$
Since the mass $M$ is kept constant,we have:
$I \propto R^2$
This relationship represents a parabola opening along the $I$-axis.
Comparing this with the given options,the graph that shows $I$ increasing quadratically with $R$ is represented by Graph $D$.

(D) The angular momentum $L$ of a particle in circular motion is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Kinetic energy $E$ is given by $E = \frac{1}{2}I\omega^2$.
From these,we can write $L = \frac{2E}{\omega}$.
Given: Frequency $f$ is doubled,so angular velocity $\omega' = 2\omega$. Kinetic energy is halved,so $E' = \frac{E}{2}$.
The new angular momentum $L'$ is:
$L' = \frac{2E'}{\omega'} = \frac{2(E/2)}{2\omega} = \frac{E}{2\omega} = \frac{1}{4} \left( \frac{2E}{\omega} \right) = \frac{L}{4}$.

(A) Let the velocity of the centre of mass of the system after the collision be $V_{CM}$.
Since there are no external horizontal forces acting on the system,the linear momentum of the system is conserved.
Let the direction of the velocity of mass $2m$ be positive ($+y$ direction) and the direction of the velocity of mass $m$ be negative ($-y$ direction).
The initial linear momentum of the system is:
$P_i = (2m)(v) + (m)(-2v) = 2mv - 2mv = 0$
After the collision,the two masses stick to the rod,so the total mass of the system is $M_{total} = 8m + 2m + m = 11m$.
By the law of conservation of linear momentum:
$P_i = P_f$
$0 = (11m) V_{CM}$
Therefore,$V_{CM} = 0$.

(C) Given: Initial angular velocity $\omega_0 = 0$,Angular acceleration $\alpha = 8 \ rad \ s^{-2}$.
$1$. For $t = 5 \ s$: The angular displacement is $\theta_5 = \omega_0 t + \frac{1}{2} \alpha t^2 = 0(5) + \frac{1}{2}(8)(5)^2 = 100 \ rad$.
The number of revolutions in $5 \ s$ is $n_5 = \frac{\theta_5}{2\pi} = \frac{100}{2\pi}$.
$2$. For $t = 6 \ s$: The angular displacement is $\theta_6 = \omega_0 t + \frac{1}{2} \alpha t^2 = 0(6) + \frac{1}{2}(8)(6)^2 = 144 \ rad$.
The number of revolutions in the $6^{th}$ second is $\Delta n = \frac{\theta_6 - \theta_5}{2\pi} = \frac{144 - 100}{2\pi} = \frac{44}{2\pi}$.
$3$. After $6 \ s$,torque is zero,so $\alpha = 0$. The angular velocity at $t = 6 \ s$ is $\omega_6 = \omega_0 + \alpha t = 0 + (8)(6) = 48 \ rad \ s^{-1}$.
In the $7^{th}$ second (i.e.,from $t=6$ to $t=7$),the angular displacement is $\theta_7 = \omega_6 \times \Delta t = 48 \times 1 = 48 \ rad$.
The number of revolutions in the $7^{th}$ second is $\frac{48}{2\pi}$.

(C) Initial angular speed of cylinder $A$ is $\omega_{0A} = 50 \text{ rev/s}$.
Initial angular speed of cylinder $B$ is $\omega_{0B} = 0 \text{ rev/s}$.
Let the magnitude of angular acceleration be $\alpha = 1 \text{ rev/s}^2$.
For cylinder $A$,the angular speed at time $t$ is $\omega_A = \omega_{0A} - \alpha t = 50 - 1t$.
For cylinder $B$,the angular speed at time $t$ is $\omega_B = \omega_{0B} + \alpha t = 0 + 1t = t$.
When the angular speeds become equal,$\omega_A = \omega_B$.
Therefore,$50 - t = t$.
$2t = 50$.
$t = 25 \text{ seconds}$.

(A) Initial angular velocity $\omega_0 = 100 \ rpm = 100 \times \frac{2\pi}{60} \ rad/s = \frac{10\pi}{3} \ rad/s$.
Final angular velocity $\omega = 0 \ rad/s$ at time $t = 15 \ s$.
Using the equation of rotational motion $\omega = \omega_0 + \alpha t$:
$0 = \frac{10\pi}{3} + \alpha(15) \implies \alpha = -\frac{10\pi}{45} = -\frac{2\pi}{9} \ rad/s^2$.
Now,the total angular displacement $\theta$ is given by $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$:
$\theta = (\frac{10\pi}{3})(15) + \frac{1}{2}(-\frac{2\pi}{9})(15)^2 = 50\pi - 25\pi = 25\pi \ rad$.
The number of rotations $N = \frac{\theta}{2\pi} = \frac{25\pi}{2\pi} = 12.5$ rotations.

(A) The angular momentum is given by $L = I\omega$.
The rotational kinetic energy is given by $E_K = \frac{1}{2} I \omega^2$.
From these two equations,we can write $I = \frac{L}{\omega}$. Substituting this into the kinetic energy equation:
$E_K = \frac{1}{2} (\frac{L}{\omega}) \omega^2 = \frac{1}{2} L \omega$.
Therefore,$L = \frac{2 E_K}{\omega}$.
Let the initial state be $(L_1, E_{K1}, \omega_1)$ and the final state be $(L_2, E_{K2}, \omega_2)$.
Given: $\omega_2 = 2\omega_1$ and $E_{K2} = \frac{E_{K1}}{2}$.
Taking the ratio:
$\frac{L_2}{L_1} = \frac{2 E_{K2} / \omega_2}{2 E_{K1} / \omega_1} = \frac{E_{K2}}{E_{K1}} \times \frac{\omega_1}{\omega_2}$.
Substituting the given values:
$\frac{L_2}{L_1} = (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{4}$.
Thus,$L_2 = \frac{L_1}{4} = \frac{L}{4}$.

(C) The rotational kinetic energy $K$ is given by $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Given that the rotational kinetic energies are equal,$K_1 = K_2$.
Therefore,$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$.
Rearranging for the ratio of angular momenta,we get $\frac{L_1^2}{L_2^2} = \frac{I_1}{I_2}$.
Substituting the given values $I_1 = I$ and $I_2 = 2I$,we have $\frac{L_1^2}{L_2^2} = \frac{I}{2I} = \frac{1}{2}$.
Taking the square root on both sides,the ratio of angular momenta is $\frac{L_1}{L_2} = \frac{1}{\sqrt{2}}$.

(D) Initial kinetic energy $(K.E.)_i = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$
$I = \frac{1}{2} M R^2 = \frac{1}{2} \times 2 \times (0.2)^2 = 0.04 \ kg \cdot m^2$
$(K.E.)_i = \frac{1}{2} \times 0.04 \times (3)^2 + \frac{1}{2} \times 0.5 \times (5)^2 = 0.18 + 6.25 = 6.43 \ J$
By conservation of angular momentum about the center of the cylinder: $L_i = L_f$
$I \omega + m v R = (I + m R^2) \omega_f$
$0.04 \times 3 + 0.5 \times 5 \times 0.2 = (0.04 + 0.5 \times (0.2)^2) \omega_f$
$0.12 + 0.5 = (0.04 + 0.02) \omega_f$
$0.62 = 0.06 \omega_f \Rightarrow \omega_f = \frac{0.62}{0.06} = 10.33 \ rad/s$
Final kinetic energy $(K.E.)_f = \frac{1}{2} (I + m R^2) \omega_f^2 = \frac{1}{2} (0.06) \times (10.33)^2 = 0.03 \times 106.7 = 3.20 \ J$
Loss in energy $\Delta E = (K.E.)_i - (K.E.)_f = 6.43 - 3.20 = 3.23 \ J \approx 3.25 \ J$ (closest option).

(B) For a thin hollow cylinder,the moment of inertia $I = MR^2$,so the radius of gyration $K$ satisfies $K^2/R^2 = 1$.
Case $1$: Sliding without rotation.
The kinetic energy is purely translational: $E_1 = \frac{1}{2} Mv^2$.
Case $2$: Rolling without slipping.
The total kinetic energy is the sum of translational and rotational kinetic energy: $E_2 = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
Since $v = R\omega$ and $I = MR^2$,we have $E_2 = \frac{1}{2} Mv^2 + \frac{1}{2} (MR^2)(v/R)^2 = \frac{1}{2} Mv^2 + \frac{1}{2} Mv^2 = Mv^2$.
Ratio of kinetic energies: $E_1 / E_2 = (\frac{1}{2} Mv^2) / (Mv^2) = 1/2$.
Thus,the ratio is $1 : 2$.

(C) The radius of the sphere at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$.
For $\Delta T = 100^{\circ}C$,$R_{100} = R_0(1 + 100\alpha)$.
The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.
Since no external torque acts on the sphere,angular momentum is conserved: $I_0 \omega_0 = I_{100} \omega_{100}$.
Substituting the values: $\frac{2}{5}MR_0^2 \omega_0 = \frac{2}{5}MR_{100}^2 \omega_{100}$.
$R_0^2 \omega_0 = [R_0(1 + 100\alpha)]^2 \omega_{100}$.
$R_0^2 \omega_0 = R_0^2(1 + 200\alpha) \omega_{100}$ (using binomial approximation $(1+x)^n \approx 1+nx$ for small $x$).
$\omega_{100} = \frac{\omega_0}{1 + 200\alpha} = \frac{\omega_0}{1 + 200(2.0 \times 10^{-5})} = \frac{\omega_0}{1 + 0.004} = \frac{\omega_0}{1.004}$.
$\omega_{100} \approx 0.996 \omega_0$.

(C) Let $x$ be the perpendicular distance from the center $O$ of the equilateral triangle to each of its sides.
Since the triangle is equilateral,the distance from the center to each side is the same.
The torque $\tau$ produced by a force $F$ about point $O$ is given by $\tau = F \times x$.
Looking at the directions of the forces $\vec F_1, \vec F_2$,and $\vec F_3$ along the sides of the triangle,we can see that $\vec F_1$ and $\vec F_2$ produce torques in the same rotational sense (e.g.,clockwise) about $O$,while $\vec F_3$ produces a torque in the opposite sense (e.g.,counter-clockwise).
For the total torque about $O$ to be zero,the sum of the torques must be zero:
$\tau_1 + \tau_2 - \tau_3 = 0$
$F_1 x + F_2 x - F_3 x = 0$
Dividing by $x$ (since $x \neq 0$):
$F_1 + F_2 - F_3 = 0$
Therefore,$F_3 = F_1 + F_2$.

(D) The moment of inertia of a thin spherical shell of mass $m$ and radius $r$ about its diameter is $I_{diam} = \frac{2}{3} mr^2$.
For the third shell,the axis $XX'$ passes through its diameter. Therefore,its moment of inertia is $I_3 = \frac{2}{3} mr^2$.
For the other two shells,the axis $XX'$ is tangent to them. According to the parallel axis theorem,the moment of inertia about a tangent is $I_{tangent} = I_{cm} + md^2$,where $d = r$. Thus,$I_1 = I_2 = \frac{2}{3} mr^2 + mr^2 = \frac{5}{3} mr^2$.
The total moment of inertia of the system about the $XX'$ axis is $I = I_1 + I_2 + I_3 = \frac{5}{3} mr^2 + \frac{5}{3} mr^2 + \frac{2}{3} mr^2 = \frac{12}{3} mr^2 = 4 mr^2$.

(A) The rotational kinetic energy of the Earth is converted into heat energy when it stops rotating.
Rotational Kinetic Energy $K = \frac{1}{2} I \omega^2$.
For a solid sphere,the moment of inertia $I = \frac{2}{5} M R^2$.
Thus,$K = \frac{1}{2} (\frac{2}{5} M R^2) \omega^2 = \frac{1}{5} M R^2 \omega^2$.
This energy is converted into heat $Q = M S \Delta \theta$,where $M$ is the mass,$S$ is the specific heat,and $\Delta \theta$ is the change in temperature.
Using the mechanical equivalent of heat $W = JQ$,we have $\frac{1}{5} M R^2 \omega^2 = J (M S \Delta \theta)$.
Solving for $\Delta \theta$,we get $\Delta \theta = \frac{R^2 \omega^2}{5Js}$.

(D) flywheel is a heavy rotating mechanical device used to store rotational energy. In a steam engine,the energy production is often cyclic or intermittent. The flywheel,due to its large moment of inertia,resists changes in rotational speed. When the engine produces excess energy,the flywheel absorbs it by increasing its angular velocity. When the engine produces less energy,the flywheel releases stored energy to maintain a constant angular velocity. Thus,it helps in keeping the speed of the engine uniform by smoothing out fluctuations in torque. Therefore,the correct option is $D$.

(B) According to the principle of conservation of angular momentum, $L = I\omega = \text{constant}$.
When the viscous liquid of mass $m$ is dropped on the platform, it spreads outwards. This increases the moment of inertia $(I)$ of the system. Since $L$ is constant, the angular velocity $(\omega)$ must decrease.
As the liquid reaches the edge and falls off the platform, the moment of inertia $(I)$ of the system decreases back towards its original value. Consequently, the angular velocity $(\omega)$ increases again.

(D) The linear momentum $P$ of a system is given by the product of the total mass $M$ and the velocity of the center of mass $v_{cm}$,i.e.,$P = M \cdot v_{cm}$.
Since the disc is rotating about a fixed axis passing through its center of mass,the center of mass of the disc remains stationary.
Therefore,the velocity of the center of mass $v_{cm} = 0$.
Consequently,the linear momentum $P = 2 \ kg \times 0 \ m/s = 0 \ kg \cdot m/s$.
Since $0 \ kg \cdot m/s$ is not listed in options $A$,$B$,or $C$,the correct option is $D$.

(D) The rotational kinetic energy of the object is given by $K_{rot} = \frac{1}{2} I \omega^2$.
Given $I = 3 \ kg \cdot m^2$ and $\omega = 2 \ rad/s$,we have $K_{rot} = \frac{1}{2} \times 3 \times (2)^2 = \frac{1}{2} \times 3 \times 4 = 6 \ J$.
The translational kinetic energy of the mass is given by $K_{trans} = \frac{1}{2} m v^2$.
Given $m = 12 \ kg$,we set $K_{trans} = K_{rot}$:
$\frac{1}{2} \times 12 \times v^2 = 6$
$6 \times v^2 = 6$
$v^2 = 1$
$v = 1 \ m/s$.

(B) According to the law of conservation of energy,the potential energy lost by the falling mass is converted into the kinetic energy of the mass and the rotational kinetic energy of the wheel.
Initial energy = Final energy
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the string does not slip,the linear velocity of the mass $v$ is related to the angular velocity of the wheel $\omega$ by $v = r\omega$.
Substituting $v = r\omega$ into the energy equation:
$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}I\omega^2$
$mgh = \frac{1}{2}(mr^2 + I)\omega^2$
Solving for $\omega$:
$\omega^2 = \frac{2mgh}{I + mr^2}$
$\omega = \sqrt{\frac{2mgh}{I + mr^2}}$

(D) The time taken for an object to roll down an inclined plane is given by $t = \sqrt{\frac{2h}{g \sin^2 \theta} (1 + \frac{I}{MR^2})}$.
Since the moment of inertia $I$ depends on the distribution of mass,and the radius of gyration $k$ is defined by $I = Mk^2$,the time taken depends on both the moment of inertia and the radius of gyration.
$A$ solid sphere has $I = \frac{2}{5}MR^2$ and a disc has $I = \frac{1}{2}MR^2$.
Because their moments of inertia and radii of gyration are different,they reach the bottom at different times.
Therefore,both $(a)$ and $(c)$ are correct.

(B) The work done to rotate a body with constant angular velocity $\omega$ is given by $W = \frac{1}{2} I \omega^2$. For the work to be minimum,the moment of inertia $I$ of the system must be minimum.
Let the axis of rotation be at a distance $x$ from the $0.3 \, kg$ mass. Then the distance from the $0.7 \, kg$ mass is $(1.4 - x)$.
The moment of inertia $I$ is given by: $I = m_1 x^2 + m_2 (L - x)^2 = 0.3 x^2 + 0.7 (1.4 - x)^2$.
To find the minimum $I$,we differentiate with respect to $x$ and set it to zero:
$\frac{dI}{dx} = 0.3(2x) + 0.7(2)(1.4 - x)(-1) = 0$
$0.6x - 1.4(1.4 - x) = 0$
$0.6x - 1.96 + 1.4x = 0$
$2.0x = 1.96$
$x = 0.98 \, m$.
Thus,the axis should pass at a distance of $0.98 \, m$ from the $0.3 \, kg$ mass.

(D) Let the length of the rod be $l$ and its mass be $m$,so $W = mg$. Initially,the rod is in equilibrium. When one man at end $B$ lets go,the rod starts rotating about the other end $A$ due to the torque produced by its weight acting at the center of mass $G$ (at distance $l/2$ from $A$).
The torque $\tau$ about point $A$ is given by $\tau = W \cdot \frac{l}{2}$.
The moment of inertia $I$ of the rod about point $A$ is $I = \frac{ml^2}{3}$.
Using $\tau = I\alpha$,we have $W \cdot \frac{l}{2} = \left(\frac{ml^2}{3}\right)\alpha$.
Substituting $W = mg$,we get $mg \cdot \frac{l}{2} = \frac{ml^2}{3} \alpha$,which simplifies to $\alpha = \frac{3g}{2l}$.
The linear acceleration $a$ of the center of mass $G$ is $a = \frac{l}{2} \alpha = \frac{l}{2} \cdot \frac{3g}{2l} = \frac{3g}{4}$.
Now,applying Newton's second law for the vertical motion of the center of mass: $W - N = ma$,where $N$ is the normal force exerted by the man at $A$.
$mg - N = m \left(\frac{3g}{4}\right)$.
$N = mg - \frac{3mg}{4} = \frac{mg}{4} = \frac{W}{4}$.
Thus,the other man feels a force of $W/4$.

(D) Given the torque $\overrightarrow{\tau} = \overrightarrow{A} \times \overrightarrow{L}$. Since $\overrightarrow{\tau} = \frac{d\overrightarrow{L}}{dt}$,we have $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{A} \times \overrightarrow{L}$.
$1$. By the definition of the cross product,$\frac{d\overrightarrow{L}}{dt}$ is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{L}$. Thus,$\frac{d\overrightarrow{L}}{dt} \perp \overrightarrow{L}$,making option $(a)$ correct.
$2$. To check the magnitude of $\overrightarrow{L}$,consider $L^2 = \overrightarrow{L} \cdot \overrightarrow{L}$. Differentiating with respect to time $t$: $\frac{d}{dt}(L^2) = 2\overrightarrow{L} \cdot \frac{d\overrightarrow{L}}{dt}$. Since $\frac{d\overrightarrow{L}}{dt} \perp \overrightarrow{L}$,the dot product $\overrightarrow{L} \cdot \frac{d\overrightarrow{L}}{dt} = 0$. Therefore,$\frac{d}{dt}(L^2) = 0$,which means the magnitude $L$ is constant. Thus,option $(c)$ is correct.
$3$. Since $\frac{d\overrightarrow{L}}{dt} = \overrightarrow{A} \times \overrightarrow{L}$,the rate of change of $\overrightarrow{L}$ is always perpendicular to $\overrightarrow{A}$. The component of $\overrightarrow{L}$ along $\overrightarrow{A}$ is given by $L_A = \overrightarrow{L} \cdot \hat{A}$. Differentiating with respect to time: $\frac{dL_A}{dt} = \frac{d\overrightarrow{L}}{dt} \cdot \hat{A} = (\overrightarrow{A} \times \overrightarrow{L}) \cdot \hat{A}$. Since the cross product $(\overrightarrow{A} \times \overrightarrow{L})$ is perpendicular to $\overrightarrow{A}$,the dot product with $\hat{A}$ is zero. Thus,$\frac{dL_A}{dt} = 0$,meaning the component of $\overrightarrow{L}$ along $\overrightarrow{A}$ is constant. Thus,option $(b)$ is correct.
Since all statements are correct,the answer is $(d)$.

(A) Before hitting the obstacle $O$,the angular momentum of the cube about $O$ is $L = Mv(a/2)$.
After hitting $O$,the cube rotates about the edge passing through $O$. The angular momentum is $L = I\omega$,where $I$ is the moment of inertia of the cube about the edge passing through $O$.
Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{Ma^2}{6}$ and $d = \sqrt{(a/2)^2 + (a/2)^2} = \frac{a}{\sqrt{2}}$.
So,$I = \frac{Ma^2}{6} + M(\frac{a^2}{2}) = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2Ma^2}{3}$.
By the conservation of angular momentum about $O$:
$Mv(a/2) = I\omega$
$Mv(a/2) = (\frac{2Ma^2}{3})\omega$
$\omega = \frac{Mv(a/2) \cdot 3}{2Ma^2} = \frac{3v}{4a}$.

(A) Let the length of the rod be $l = 2L$. The center of mass of the rod is at a distance $l/2 = L$ from the point of contact.
Initially,the height of the center of mass above the floor is $h = (l/2) \sin\alpha = L \sin\alpha$.
By the law of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:
$mgh = \frac{1}{2} I \omega^2$
Here,$I$ is the moment of inertia of the rod about the point of contact on the floor,given by $I = \frac{ml^2}{3} = \frac{m(2L)^2}{3} = \frac{4mL^2}{3}$.
Substituting the values:
$mg(L \sin\alpha) = \frac{1}{2} \left( \frac{4mL^2}{3} \right) \omega^2$
$mgL \sin\alpha = \frac{2mL^2}{3} \omega^2$
$g \sin\alpha = \frac{2L}{3} \omega^2$
$\omega^2 = \frac{3g \sin\alpha}{2L}$
$\omega = \sqrt{\frac{3g \sin\alpha}{2L}}$

(D) The total number of revolutions is equal to the area under the $\omega-t$ graph.
The graph is a trapezium with parallel sides of length $t_1 = 3.5 \text{ min}$ (from $t=0$ to $t=3.5$) and $t_2 = (3.5 - 1) = 2.5 \text{ min}$ (the flat top part).
However,looking at the graph,the base is from $t=0$ to $t=5$ minutes,and the top flat part is from $t=1$ to $t=3.5$ minutes.
The area of the trapezium is given by $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.
Here,the parallel sides are the base (length $5 \text{ min}$) and the top (length $3.5 - 1 = 2.5 \text{ min}$).
The height is the angular velocity $\omega = 3000 \text{ revolutions/min}$.
Number of revolutions $= \frac{1}{2} \times (5 + 2.5) \times 3000 = \frac{1}{2} \times 7.5 \times 3000 = 7.5 \times 1500 = 11250$ revolutions.

(B) Since there is no external torque acting on the system,the angular momentum $L = I\omega$ remains constant.
As the tortoise moves along a chord,its distance $r$ from the center of rotation changes.
The moment of inertia $I$ of the system is given by $I = I_{\text{platform}} + mr^2$,where $m$ is the mass of the tortoise and $r$ is its distance from the center.
Initially,the tortoise is at the edge $(r = R)$,so the moment of inertia is maximum. As it moves towards the center of the chord (the point closest to the center of the platform),the distance $r$ decreases,causing the moment of inertia $I$ to decrease.
Since $L = I\omega$ is constant,if $I$ decreases,the angular velocity $\omega$ must increase.
After passing the point on the chord closest to the center,the distance $r$ starts increasing again,causing the moment of inertia $I$ to increase,which in turn causes the angular velocity $\omega$ to decrease.
Therefore,the angular velocity first increases and then decreases.

(D) The moment of inertia of a solid sphere about its diameter is given by the formula: $I = \frac{2}{5}MR^2$.
Since the mass $M$ is kept constant,we can write this relationship as $I \propto R^2$.
This is the equation of a parabola of the form $y = kx^2$,where $y = I$ and $x = R$.
Therefore,the graph between $I$ and $R$ will be a parabola opening upwards,starting from the origin $(0,0)$.
Comparing this with the given options,the graph shown in option $D$ represents this parabolic relationship.

(C) The given equation is $I = I_{cm} + Mx^2$. This is of the form $y = c + ax^2$,which represents a parabola symmetric about the $I$-axis. Since $I = I_{cm}$ when $x = 0$,the vertex of the parabola is at $(0, I_{cm})$. Therefore,the graph is a parabola that is symmetric about the $I$-axis and does not pass through the origin. This corresponds to the graph shown in option $C$.

(A) The relationship between angular momentum $L$ and angular velocity $\omega$ is given by the formula $L = I\omega$,where $I$ is the moment of inertia.
Assuming the moment of inertia $I$ remains constant,the equation $L = I\omega$ represents a linear relationship of the form $y = mx$,where $I$ acts as the slope.
Therefore,the graph of $L$ versus $\omega$ is a straight line passing through the origin with a constant slope equal to $I$.
This corresponds to the graph shown in option $A$.

(B) The relationship between angular momentum $L$ and linear momentum $P$ for a particle at a distance $r$ from the axis of rotation is given by $L = rP$.
Taking the natural logarithm on both sides,we get:
$\log_e L = \log_e (rP)$
Using the logarithmic property $\log(ab) = \log a + \log b$,we have:
$\log_e L = \log_e P + \log_e r$
This equation is in the form of a straight line $y = mx + c$,where $y = \log_e L$,$x = \log_e P$,slope $m = 1$,and intercept $c = \log_e r$.
Since the intercept $c = \log_e r$ is generally non-zero,the graph is a straight line with a slope of $1$ that does not pass through the origin. This corresponds to the graph shown in option $B$.

(B) The moment of inertia of a solid sphere of mass $M$ and radius $a$ about its diameter is $I_{cm} = \frac{2}{5}Ma^2$.
Let the axis of rotation be the side $AB$ passing through spheres $1$ and $2$.
For spheres $1$ and $2$,the axis passes through their centers,so their moment of inertia about the axis is $I_1 = I_2 = \frac{2}{5}Ma^2$.
For spheres $3$ and $4$,the axis is at a perpendicular distance $b$ from their centers. Using the parallel axis theorem,$I = I_{cm} + Md^2$,we get $I_3 = I_4 = \frac{2}{5}Ma^2 + Mb^2$.
The total moment of inertia of the system about axis $AB$ is $I_{total} = I_1 + I_2 + I_3 + I_4$.
$I_{total} = \frac{2}{5}Ma^2 + \frac{2}{5}Ma^2 + (\frac{2}{5}Ma^2 + Mb^2) + (\frac{2}{5}Ma^2 + Mb^2)$.
$I_{total} = 4 \times (\frac{2}{5}Ma^2) + 2Mb^2 = \frac{8}{5}Ma^2 + 2Mb^2$.

(B) Initially,when the rod is standing vertically,its potential energy is $U_i = mg\frac{l}{2}$.
When it strikes the floor,its potential energy is converted into rotational kinetic energy.
By the law of conservation of energy: $U_i = K_f$.
$mg\left(\frac{l}{2}\right) = \frac{1}{2}I\omega^2$,where $I = \frac{ml^2}{3}$ is the moment of inertia of the rod about the hinge $A$.
Substituting $I$ and $\omega = \frac{v_B}{l}$:
$mg\left(\frac{l}{2}\right) = \frac{1}{2}\left(\frac{ml^2}{3}\right)\left(\frac{v_B}{l}\right)^2$.
$mg\frac{l}{2} = \frac{1}{6}mv_B^2$.
$v_B^2 = 3gl$.
$v_B = \sqrt{3gl}$.

(A) The angular acceleration $\alpha$ is defined as the rate of change of angular velocity with respect to time,given by $\alpha = \frac{d\omega}{dt}$.
Graphically,$\alpha$ represents the slope of the $\omega-t$ graph.
The magnitude of the angular acceleration is $|\alpha| = |\frac{d\omega}{dt}|$,which corresponds to the steepness (absolute value of the slope) of the tangent to the curve at any point.
In graph $A$,the curve starts with a steep slope and becomes flatter as $t$ increases. This means the slope $\frac{d\omega}{dt}$ is positive and its magnitude is decreasing over time.
In graph $B$,the slope is constant,so $\alpha$ is constant.
In graph $C$,the slope is constant (negative),so the magnitude of $\alpha$ is constant.
In graph $D$,the curve becomes steeper as $t$ increases,meaning the magnitude of the slope is increasing.
Therefore,the magnitude of the angular acceleration is constantly decreasing in graph $A$.

(A) Let the mass of the hemisphere be $M = 3m$. Since there are no external horizontal forces acting on the system,the horizontal position of the centre of mass remains constant.
Let $x$ be the displacement of the hemisphere to the right. The horizontal displacement of the particle relative to the table is $(R \sin \theta - x)$ to the left.
Equating the horizontal displacement of the centre of mass to zero:
$m(R \sin \theta - x) - M x = 0$
$m R \sin \theta = (M + m) x$
Substituting $M = 3m$:
$m R \sin \theta = (3m + m) x = 4m x$
$x = \frac{R \sin \theta}{4}$
Differentiating with respect to time $t$:
$\frac{dx}{dt} = \frac{R \cos \theta}{4} \frac{d\theta}{dt}$
Given that the velocity of the hemisphere is $v = \frac{dx}{dt}$ and the angular velocity of the particle relative to the centre is $\omega = \frac{d\theta}{dt}$:
$v = \frac{R \cos \theta}{4} \omega$
$\omega = \frac{4v}{R \cos \theta}$

(D) $1$. Since the surface is smooth and there are no external horizontal forces acting on the system (particle + wedge),the linear momentum of the system is conserved in the horizontal direction.
$2$. At the maximum height $h$ relative to the wedge,the particle and the wedge move with the same horizontal velocity $V$. By conservation of momentum: $mv_0 = (m + M)V$,so $V = \frac{mv_0}{m+M}$.
$3$. In the centre of mass $(COM)$ frame,the total momentum is zero. At the maximum height $h$,the relative velocity between the particle and the wedge is zero (they move together horizontally). Since the total momentum in the $COM$ frame is zero,both must be stationary with respect to the $COM$ frame.
$4$. In the $COM$ frame,the total energy is the sum of the kinetic energy of the system relative to the $COM$ and the potential energy. At the maximum height $h$,the relative kinetic energy is zero,meaning all the initial kinetic energy relative to the $COM$ has been converted into potential energy $mgh_{eff}$.
$5$. Thus,both statements $(B)$ and $(C)$ are correct.

(C) Let the mass density be $\rho \text{ kg/m}^2$.
Let the sides be $PQ = RS = a$ and $QR = PS = b$. Assume $a > b$.
The mass of the rectangle is $M = \rho ab$.
The moment of inertia of the rectangle about its center is $I_{cm} = M(a^2 + b^2)/12$.
The distance of point $P$ from the center is $\sqrt{(a/2)^2 + (b/2)^2} = \sqrt{a^2 + b^2}/2$.
Using the parallel axis theorem,the moment of inertia of the rectangle about $P$ is $I = I_{cm} + M(a^2 + b^2)/4 = M(a^2 + b^2)/3$.
The mass of the triangle $PQR$ is $M' = M/2 = \rho ab/2$.
The moment of inertia of the triangle $PQR$ about an axis through its centroid $G$ perpendicular to the plate is $I_G = M'(a^2 + b^2)/18$.
The distance of point $P$ from the centroid $G$ is $d^2 = (2a/3)^2 + (b/3)^2 = (4a^2 + b^2)/9$.
Using the parallel axis theorem,the moment of inertia of the triangle $PQR$ about $P$ is $I_P = I_G + M'd^2 = (M/2)(a^2 + b^2)/18 + (M/2)(4a^2 + b^2)/9 = (M/2)(a^2 + b^2 + 8a^2 + 2b^2)/18 = (M/2)(9a^2 + 3b^2)/18 = (M/2)(3a^2 + b^2)/6$.
Comparing $I_P$ with $I/2 = (M/2)(a^2 + b^2)/3 = (M/2)(2a^2 + 2b^2)/6$,since $a > b$,$3a^2 + b^2 > 2a^2 + 2b^2$,therefore $I_P > I/2$.