Rolling On Inclined Plane Questions in English

(C) For a sphere rolling up an incline without slipping,the acceleration $a$ is given by $a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$.
For a solid sphere,the moment of inertia $I = \frac{2}{5} mR^2$.
Substituting this,$a = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{7/5} = \frac{5}{7} g \sin \theta$.
Given $g = 9.8 \, m/s^2$ (or $10 \, m/s^2$ for approximation),$\theta = 30^{\circ}$,and $v_0 = 1 \, m/s$.
Using $g = 9.8 \, m/s^2$: $a = \frac{5}{7} \times 9.8 \times \sin(30^{\circ}) = \frac{5}{7} \times 9.8 \times 0.5 = 3.5 \, m/s^2$.
The time taken to reach the highest point is $t_{up} = \frac{v_0}{a} = \frac{1}{3.5} = \frac{2}{7} \, s$.
The total time to return to point $A$ is $T = 2 \times t_{up} = 2 \times \frac{2}{7} = \frac{4}{7} \approx 0.57 \, s$.

(C) Let the solid cylinder be in equilibrium.
The forces acting along the inclined plane are the tension $T$ and the friction $f$. The component of gravity along the plane is $mg \sin 60^{\circ}$.
For translational equilibrium: $T + f = mg \sin 60^{\circ} \quad ......(i)$
For rotational equilibrium about the center of the cylinder: $TR - fR = 0 \implies T = f \quad ......(ii)$
Substituting $(ii)$ into $(i)$: $2f = mg \sin 60^{\circ} \implies f = \frac{mg \sin 60^{\circ}}{2} = \frac{mg \sqrt{3}}{4} \approx 0.433 mg$.
The limiting friction is $f_{L} = \mu_{s} N = \mu_{s} mg \cos 60^{\circ} = 0.4 \times mg \times 0.5 = 0.2 mg$.
Since the required friction $(0.433 mg)$ is greater than the limiting friction $(0.2 mg)$,the cylinder will not remain in static equilibrium and will roll down.
The friction acting will be kinetic friction: $f_{k} = \mu_{k} N$. Assuming $\mu_{k} = \mu_{s} = 0.4$,we get $f_{k} = 0.4 \times mg \times 0.5 = 0.2 mg = \frac{mg}{5}$.

(A) The total initial energy of the sphere is $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$.
Since it rolls without slipping,$\omega_0 = \frac{v_0}{a}$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}ma^2$.
Substituting these into the energy equation: $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{2}{5}ma^2)(\frac{v_0}{a})^2 = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$.
At the maximum height $h$,the sphere momentarily comes to rest,so its final kinetic energy is zero. By the law of conservation of energy,$E_i = E_f = mgh$.
Therefore,$mgh = \frac{7}{10}mv_0^2$,which gives $h = \frac{7v_0^2}{10g}$.
The distance $d$ traveled along the incline is related to height $h$ by $h = d \sin \theta$.
Thus,$d \sin \theta = \frac{7v_0^2}{10g}$,which implies $d = \frac{7v_0^2}{10g \sin \theta}$.
This matches option $A$.

(D) If the disk slips on an inclined plane,its acceleration is $a_{1} = g \sin \theta$.
Using the equation of motion $L = \frac{1}{2} a_{1} t_{1}^{2}$,we get $t_{1} = \sqrt{\frac{2L}{a_{1}}} \quad \dots (i)$.
If the disk rolls on an inclined plane,its acceleration is $a_{2} = \frac{g \sin \theta}{1 + \frac{I}{mR^{2}}}$.
For a circular disk,the moment of inertia $I = \frac{1}{2} mR^{2}$,so $a_{2} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{2}{3} g \sin \theta$.
Using the equation of motion $L = \frac{1}{2} a_{2} t_{2}^{2}$,we get $t_{2} = \sqrt{\frac{2L}{a_{2}}} \quad \dots (ii)$.
Taking the ratio $\frac{t_{2}}{t_{1}} = \sqrt{\frac{a_{1}}{a_{2}}} = \sqrt{\frac{g \sin \theta}{\frac{2}{3} g \sin \theta}} = \sqrt{\frac{3}{2}}$.
Comparing this with $\sqrt{\frac{3}{x}}$,we get $x = 2$.

(B) Let $K_r$ be the rotational kinetic energy and $K_t$ be the translational kinetic energy.
Given $K_r = 0.5 K_t$,where $K_r = \frac{1}{2} I \omega^2$ and $K_t = \frac{1}{2} m v^2$.
Since the body rolls without slipping,$v = \omega R$,so $\omega = \frac{v}{R}$.
Substituting this into the rotational energy formula: $K_r = \frac{1}{2} I (\frac{v}{R})^2 = \frac{1}{2} (\frac{I}{R^2}) v^2$.
According to the problem,$\frac{1}{2} (\frac{I}{R^2}) v^2 = 0.5 \times (\frac{1}{2} m v^2)$.
This simplifies to $\frac{I}{R^2} = 0.5 m$,which means $I = 0.5 m R^2 = \frac{1}{2} m R^2$.
The moment of inertia $I = \frac{1}{2} m R^2$ corresponds to a solid cylinder or a disk.

(B) For a body rolling down an inclined plane of height $h$,the conservation of energy gives $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since $v = R\omega$,we have $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{I}{mR^2})$.
For a ring,$I = mR^2$,so $mgh = \frac{1}{2}mv_R^2(1 + 1) = mv_R^2$. Thus,$v_R = \sqrt{gh}$.
For a solid cylinder,$I = \frac{1}{2}mR^2$,so $mgh = \frac{1}{2}mv_c^2(1 + \frac{1}{2}) = \frac{3}{4}mv_c^2$. Thus,$v_c = \sqrt{\frac{4gh}{3}}$.
The ratio of velocities is $\frac{v_R}{v_c} = \frac{\sqrt{gh}}{\sqrt{4gh/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Comparing this to $\frac{\sqrt{x}}{2}$,we get $x = 3$.

(A) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$.
The moment of inertia $I$ for the objects are: $I_{\text{ring}} = mR^2$,$I_{\text{cylinder}} = \frac{1}{2}mR^2$,and $I_{\text{sphere}} = \frac{2}{5}mR^2$.
Comparing the moments of inertia: $I_{\text{ring}} > I_{\text{cylinder}} > I_{\text{sphere}}$.
Since $a$ is inversely proportional to the factor $(1 + \frac{I}{mR^2})$,the acceleration follows the order: $a_{\text{ring}} < a_{\text{cylinder}} < a_{\text{sphere}}$.
Using the kinematic equation $v^2 = u^2 + 2as$ with $u = 0$,the final velocity $v = \sqrt{2as}$ is directly proportional to the square root of acceleration.
Therefore,the final velocities at the bottom of the plane follow the order: $v_{\text{ring}} < v_{\text{cylinder}} < v_{\text{sphere}}$.
Thus,the sphere has the greatest velocity and the ring has the least velocity.

(D) Let $M = 12 \, kg$ be the mass of the wheel and $m = 3 \, kg$ be the hanging mass. When the wheel moves down by a vertical height $h$,the mass $m$ moves up by a vertical height $h$.
Applying the principle of conservation of mechanical energy:
Loss in potential energy of the wheel = Gain in potential energy of the mass $m$ + Gain in kinetic energy of the system.
$Mgh = mgh + \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$
Assuming the wheel is a disc,$I = \frac{1}{2} Mr^2$ and $\omega = \frac{v}{r}$.
$(M - m)gh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} Mr^2) (\frac{v}{r})^2 + \frac{1}{2} mv^2$
$(12 - 3)gh = \frac{1}{2} (12)v^2 + \frac{1}{4} (12)v^2 + \frac{1}{2} (3)v^2$
$9gh = 6v^2 + 3v^2 + 1.5v^2 = 10.5v^2$
$v^2 = \frac{9gh}{10.5} = \frac{90gh}{105} = \frac{6}{7} gh$
$v = \sqrt{\frac{6}{7} gh} = \frac{1}{2} \sqrt{\frac{24}{7} gh}$.
Comparing with $\frac{1}{2} \sqrt{xgh}$,we get $x = \frac{24}{7} \approx 3.43$. Given the options,the closest integer value is $3$.

(D) The velocity $V$ of a body rolling down an inclined plane of height $H$ without slipping is given by the formula: $V = \sqrt{\frac{2gH}{1 + k^2/R^2}}$,where $k$ is the radius of gyration.
For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2/R^2 = 1/2$.
For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $k^2/R^2 = 2/5$.
The ratio of velocities is given by: $\frac{V_{\text{cylinder}}}{V_{\text{sphere}}} = \sqrt{\frac{1 + k_{\text{sphere}}^2/R^2}{1 + k_{\text{cylinder}}^2/R^2}}$.
Substituting the values: $\frac{V_{\text{cylinder}}}{V_{\text{sphere}}} = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$.

(C) By the principle of conservation of energy,the loss in potential energy equals the gain in kinetic energy (translational + rotational).
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid cylinder,the moment of inertia $I = \frac{1}{2}mR^2$. Since the string unwinds without slipping,the condition $v = R\omega$ holds,which implies $\omega = \frac{v}{R}$.
Substituting these into the energy equation:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$
$mgh = \frac{3}{4}mv^2$
$gh = \frac{3}{4}v^2$
Given $g = 10\,ms^{-2}$ and $v = 4\,ms^{-1}$:
$10h = \frac{3}{4}(4)^2$
$10h = \frac{3}{4} \times 16$
$10h = 12$
$h = 1.2\,m = 120\,cm$.

(B) At $P_{1}$,the ring is moving down and its angular velocity $\omega$ is increasing to satisfy the rolling condition $(v_{CM} = R\omega)$. The friction force $f$ acts upwards (opposite to $v_{CM}$) to provide the necessary torque to increase $\omega$.
At $P_{2}$,the velocity $v_{CM}$ is horizontal and the acceleration of the center of mass is purely centripetal. There is no tangential acceleration,so the frictional force is zero.
At $P_{3}$,the ring is moving up and its angular velocity $\omega$ is decreasing to satisfy the rolling condition. The friction force $f$ acts upwards (in the same direction as $v_{CM}$) to provide the torque required to decrease $\omega$.
Therefore,the frictional force is opposite to $v_{CM}$ at $P_{1}$ and in the same direction as $v_{CM}$ at $P_{3}$.

(B) Since the ring is rolling without any slippage,the total initial kinetic energy is equal to the total final potential energy at height $h$.
Total initial kinetic energy = Kinetic energy of translation + Kinetic energy of rotation
$\Rightarrow K.E. = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2}$
For a ring,the moment of inertia $I = m R^{2}$ and the condition for pure rolling is $v = R \omega$,which implies $\omega = v / R$.
Substituting these into the energy equation:
$K.E. = \frac{1}{2} m v^{2} + \frac{1}{2} (m R^{2}) (v / R)^{2}$
$K.E. = \frac{1}{2} m v^{2} + \frac{1}{2} m v^{2} = m v^{2}$
At the maximum height $h$,the final kinetic energy is zero,so the total initial kinetic energy is converted into potential energy:
$m v^{2} = m g h$
$h = \frac{v^{2}}{g}$

(B) For cylinder $P$ (rolling without slipping):
By the principle of conservation of energy,the initial potential energy is converted into translational and rotational kinetic energy at the bottom:
$mgh = \frac{1}{2}mv_p^2 + \frac{1}{2}I\omega^2$
Since $I = \frac{1}{2}mr^2$ and $v_p = r\omega$,we have:
$mgh = \frac{1}{2}mv_p^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v_p}{r})^2$
$mgh = \frac{1}{2}mv_p^2 + \frac{1}{4}mv_p^2 = \frac{3}{4}mv_p^2$
$v_p = \sqrt{\frac{4gh}{3}}$
For cylinder $Q$ (sliding without friction):
Since there is no friction,the cylinder does not rotate. All potential energy is converted into translational kinetic energy:
$mgh = \frac{1}{2}mv_q^2$
$v_q = \sqrt{2gh}$
The ratio of the speeds is:
$\frac{v_q}{v_p} = \frac{\sqrt{2gh}}{\sqrt{\frac{4gh}{3}}} = \sqrt{2 \cdot \frac{3}{4}} = \sqrt{\frac{3}{2}}$

(A) Let the initial horizontal speed of the sphere be $v$.
Since the sphere rolls without slipping,its total kinetic energy on the horizontal surface is the sum of its translational and rotational kinetic energies:
$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^2$ and the condition for rolling without slipping is $v = R \omega$,so $\omega = v/R$.
Substituting these values:
$KE_{total} = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2 = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 = \frac{7}{10} m v^2$
At the maximum height $h$,the sphere momentarily comes to rest before rolling back down. By the principle of conservation of mechanical energy,the initial kinetic energy equals the potential energy at height $h$:
$\frac{7}{10} m v^2 = m g h$
Solving for $v$:
$v^2 = \frac{10}{7} g h$
$v = \sqrt{\frac{10 g h}{7}}$

(D) For a sphere in pure rolling,the equations of motion are:
$N - Mg \cos \theta = 0 \quad \dots(i)$
$Mg \sin \theta - f = Ma_{CM} \quad \dots(ii)$
$\tau = f \times R = I \alpha = I \frac{a_{CM}}{R} \quad \dots(iii)$
Since $I = \frac{2}{5} MR^2$,substituting into $(iii)$ gives $f = \frac{2}{5} Ma_{CM}$.
Substituting $f$ into $(ii)$:
$Mg \sin \theta - \frac{2}{5} Ma_{CM} = Ma_{CM} \implies Mg \sin \theta = \frac{7}{5} Ma_{CM} \implies a_{CM} = \frac{5}{7} g \sin \theta$.
Now,substituting $a_{CM}$ back into the expression for $f$:
$f = \frac{2}{5} M \left( \frac{5}{7} g \sin \theta \right) = \frac{2}{7} Mg \sin \theta$.
For pure rolling without slipping,the static friction must satisfy $f \leq \mu_s N$.
Substituting $f = \frac{2}{7} Mg \sin \theta$ and $N = Mg \cos \theta$:
$\frac{2}{7} Mg \sin \theta \leq \mu_s Mg \cos \theta$
$\tan \theta \leq \frac{7}{2} \mu_s$
$\theta \leq \tan^{-1} \left( \frac{7}{2} \mu_s \right)$.

(D) The time taken for a body to roll down an inclined plane is given by $t = \sqrt{\frac{2s(1 + \beta)}{g \sin \theta}}$,where $\beta = \frac{I}{MR^2}$. For a thin spherical shell,$I = \frac{2}{3}MR^2$,so $\beta = 2/3$. However,for a shell with thickness $t$,the moment of inertia is $I = \frac{2}{3}M \frac{R_o^5 - R_i^5}{R_o^3 - R_i^3}$. Given $R_o = 20 \,cm$ and $t \ll R_o$,$I \approx \frac{2}{3}MR^2(1 + \frac{t}{R})$. Thus,$\beta \approx \frac{2}{3}(1 + \frac{t}{R})$. The time $t \propto \sqrt{1 + \beta}$. Since the time difference is $1 \%$,$\frac{\Delta t}{t} = \frac{1}{2} \frac{\Delta \beta}{1 + \beta} \approx 0.01$. Given $\beta \approx 2/3$,$1 + \beta \approx 5/3$. Then $\frac{1}{2} \frac{\Delta \beta}{5/3} = 0.01 \implies \Delta \beta = 0.033$. Since $\beta = \frac{2}{3}(1 + \frac{t}{R})$,$\Delta \beta = \frac{2}{3} \frac{\Delta t}{R}$. Substituting values: $0.033 = \frac{2}{3} \frac{\Delta t}{20} \implies \Delta t = 0.033 \times 30 = 0.99 \,cm$. The total thickness is $0.5 + 0.99 = 1.49 \,cm$,which is closest to $1.5 \,cm$.

(D) For an object rolling down an inclined plane without slipping,the conservation of energy states that the potential energy at the top is equal to the total kinetic energy at the bottom.
$mgh = K.E._{trans} + K.E._{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a disc,the moment of inertia $I = \frac{1}{2}mr^2$ and the rolling condition is $\omega = \frac{v}{r}$.
Substituting these into the energy equation:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
$v = \sqrt{\frac{4gh}{3}}$
Since the final velocity $v$ depends only on the height $h$ and the acceleration due to gravity $g$,and is independent of the mass $m$ and radius $r$ of the disc,both discs will have the same linear velocity upon reaching the ground.
Therefore,the ratio of their linear velocities is $1:1$.

(C) The total kinetic energy $(K_{total})$ during pure rolling is the sum of translational kinetic energy $(K_{trans})$ and rotational kinetic energy $(K_{rot})$.
$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $v = R\omega$,we have $K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{I}{mR^2})$.
The ratio of rotational kinetic energy to total kinetic energy is:
$\text{Ratio} = \frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2} = \frac{\frac{1}{2}I(\frac{v}{R})^2}{\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2} = \frac{I}{mR^2 + I} = \frac{1}{1 + \frac{mR^2}{I}}$.
For a Disc $(I = \frac{1}{2}mR^2)$: $\text{Ratio} = \frac{1}{1 + 2} = \frac{1}{3} \approx 0.33$.
For a Solid Sphere $(I = \frac{2}{5}mR^2)$: $\text{Ratio} = \frac{1}{1 + 2.5} = \frac{2}{7} \approx 0.28$.
For a Ring $(I = mR^2)$: $\text{Ratio} = \frac{1}{1 + 1} = \frac{1}{2} = 0.5$.
For a Hollow Sphere $(I = \frac{2}{3}mR^2)$: $\text{Ratio} = \frac{1}{1 + 1.5} = \frac{2}{5} = 0.4$.
Comparing the values,the Ring has the maximum percentage of rotational kinetic energy. Thus,option $C$ is correct.

(C) By the law of conservation of energy,the potential energy at the top is equal to the total kinetic energy at the bottom.
$PE = Mgh$
$KE = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the condition for pure rolling is $v = R\omega$.
Substituting these into the kinetic energy equation:
$KE = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$
Equating potential energy and kinetic energy:
$Mgh = \frac{3}{4} Mv^2$
$v^2 = \frac{4gh}{3} \implies v = 2 \sqrt{\frac{gh}{3}}$
Since $\omega = \frac{v}{R}$,the angular velocity is:
$\omega = \frac{2}{R} \sqrt{\frac{gh}{3}}$

(B) For an object of mass $M$,radius $R$,and moment of inertia $I = kMR^2$ rolling down an inclined plane of angle $\theta$,the force of friction $f$ is given by the formula:
$f = \frac{Mg \sin\theta}{1 + \frac{MR^2}{I}} = \frac{Mg \sin\theta}{1 + \frac{1}{k}}$
For a solid cylinder,$I = \frac{1}{2}MR^2$,so $k = \frac{1}{2}$. Thus,$f_{cylinder} = \frac{Mg \sin\theta}{1 + 2} = \frac{1}{3}Mg \sin\theta$.
For a solid sphere,$I = \frac{2}{5}MR^2$,so $k = \frac{2}{5}$. Thus,$f_{sphere} = \frac{Mg \sin\theta}{1 + \frac{5}{2}} = \frac{Mg \sin\theta}{3.5} = \frac{2}{7}Mg \sin\theta$.
Comparing the two,$\frac{1}{3} \approx 0.333$ and $\frac{2}{7} \approx 0.285$.
Since $\frac{1}{3} > \frac{2}{7}$,the force of friction is more for the cylinder.

(D) When the sphere is thrown up the incline,it has both translational velocity $v$ and angular velocity $\omega$. For pure rolling up the incline,the friction must act to oppose the tendency of slipping. As the sphere moves up,the friction acts in the upward direction to provide the necessary torque to maintain the rolling condition.
When the sphere reaches its highest point and starts moving down,the direction of its translational velocity reverses. To maintain pure rolling while moving down the incline,the friction again acts in the upward direction to oppose the tendency of the sphere to slip down the incline.
Therefore,the direction of rolling friction is upward during both the upward and downward motion.

(D) For a body rolling down an inclined plane,the acceleration of the center of mass is given by:
$a_{cm} = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$
For a solid sphere,the moment of inertia is $I = \frac{2}{5} MR^2$.
Substituting this into the acceleration formula:
$a_{cm} = \frac{g \sin \theta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 2/5} = \frac{5}{7} g \sin \theta$
The time taken to travel a distance $L$ along the incline is given by $L = \frac{1}{2} a_{cm} t^2$,which implies $t = \sqrt{\frac{2L}{a_{cm}}}$.
Substituting $a_{cm}$:
$t = \sqrt{\frac{2L}{\frac{5}{7} g \sin \theta}} = \sqrt{\frac{14L}{5g \sin \theta}}$
As seen from the expression,the time $t$ depends only on the length of the incline $L$,the angle of inclination $\theta$,and the acceleration due to gravity $g$. It is independent of the mass $M$,radius $R$,and density $\rho$ of the sphere. Thus,the time is independent of all these factors.

(C) Case $(i)$: Slipping down a smooth incline.
In this case,the ring undergoes pure translational motion. The acceleration is $a_1 = g \sin \theta$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$ with initial velocity $u = 0$,we get $s = \frac{1}{2} (g \sin \theta) t_1^2$,so $t_1 = \sqrt{\frac{2s}{g \sin \theta}}$.
Case $(ii)$: Rolling down a rough incline.
In this case,the ring undergoes rolling motion. The acceleration is $a_2 = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a thin circular ring,the moment of inertia $I = MR^2$. Substituting this,$a_2 = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
Using the same equation of motion,$s = \frac{1}{2} a_2 t_2^2$,so $t_2 = \sqrt{\frac{2s}{a_2}} = \sqrt{\frac{4s}{g \sin \theta}}$.
Ratio of times:
$\frac{t_1}{t_2} = \frac{\sqrt{\frac{2s}{g \sin \theta}}}{\sqrt{\frac{4s}{g \sin \theta}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$.

(A) For a body of mass $m$ and radius $r$ rolling down an inclined plane of angle $\theta$,the acceleration $a$ is given by $a = \frac{mg \sin \theta}{m + \frac{I}{r^2}}$.
The frictional force $f$ required for pure rolling is $f = ma_{cm} \cdot \frac{I}{mr^2} = \frac{mg \sin \theta}{1 + \frac{mr^2}{I}}$.
For a solid sphere,the moment of inertia $I = \frac{2}{5} mr^2$.
Substituting $I$ into the friction equation: $f = \frac{mg \sin \theta}{1 + \frac{mr^2}{(2/5)mr^2}} = \frac{mg \sin \theta}{1 + 2.5} = \frac{mg \sin \theta}{3.5} = \frac{2}{7} mg \sin \theta$.
Since $f \le \mu N$ and $N = mg \cos \theta$,we have $\mu \ge \frac{f}{N} = \frac{(2/7) mg \sin \theta}{mg \cos \theta} = \frac{2}{7} \tan \theta$.
Thus,the minimum coefficient of friction is $\mu_{min} = \frac{2}{7} \tan \theta$.

(B) For an object sliding down a smooth incline,the potential energy is converted entirely into translational kinetic energy:
$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}$.
For a ring rolling down an inclined plane,the potential energy is converted into both translational and rotational kinetic energy:
$mgh = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2$.
Since for a ring $I = mr^2$ and $\omega = \frac{v'}{r}$,we have:
$mgh = \frac{1}{2}mv'^2 + \frac{1}{2}(mr^2)(\frac{v'^2}{r^2}) = \frac{1}{2}mv'^2 + \frac{1}{2}mv'^2 = mv'^2$.
Equating the two expressions for $mgh$:
$mv'^2 = \frac{1}{2}mv^2 \implies v'^2 = \frac{v^2}{2} \implies v' = \frac{v}{\sqrt{2}}$.

(D) For a body rolling down an inclined plane,the force of friction $f$ acts in the upward direction along the incline.
$1$. Since the sphere is moving downward,the friction $f$ acts upward,which opposes the translational motion.
$2$. The torque produced by the friction $f$ about the center of the sphere is $\tau = f \times R$. This torque causes an angular acceleration $\alpha$ in the direction of rolling,meaning friction supports the rotational motion.
$3$. The equation of motion for translation is $mg \sin \theta - f = ma$,and for rotation is $fR = I\alpha$. For a hollow sphere,$I = \frac{2}{3}mR^2$. Solving these,we find $f = \frac{mg \sin \theta}{1 + \frac{mR^2}{I}} = \frac{mg \sin \theta}{1 + \frac{3}{2}} = \frac{2}{5} mg \sin \theta$.
$4$. Since $f = \frac{2}{5} mg \sin \theta$,it is clear that $f \propto \sin \theta$. Therefore,if $\theta$ decreases,$\sin \theta$ decreases,and consequently,the frictional force $f$ decreases.
Thus,all the given statements are correct.

(D) For a cylinder rolling down an inclined plane of height $h$,the mechanical energy is conserved. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since $I = \frac{1}{2}mr^2$ and $v = r\omega$,we have $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{3}{4}mv^2$. Thus,$v = \sqrt{\frac{4gh}{3}}$. Since $h$ is the same,the final speed $v$ is the same in both cases.
The acceleration of a body rolling down an incline is $a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}}$. Since the inclination $\theta$ is different,the acceleration $a$ will be different. The time of descent is given by $t = \sqrt{\frac{2s}{a}}$,where $s = \frac{h}{\sin \theta}$. Substituting $a$,we get $t = \sqrt{\frac{2h}{g \sin^2 \theta} (1 + \frac{I}{mr^2})}$. Since $\theta$ is different,the time of descent $t$ will be different. Therefore,the speed is the same,but the time of descent is different.

(B) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into translational and rotational kinetic energy at the bottom.
$mgh = K_{trans} + K_{rot}$
For a disc,the moment of inertia $I = \frac{1}{2} mR^2$ and the rolling condition is $v = R\omega$.
$K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{2} mR^2) (\frac{v}{R})^2 = \frac{1}{4} mv^2$.
Thus,$mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2$.
Given $m = 3 \, kg$,$h = 5 \, m$,and $g = 10 \, m/s^2$:
$3 \times 10 \times 5 = \frac{3}{4} mv^2$
$150 = \frac{3}{4} mv^2$
$mv^2 = 200$
Translational kinetic energy $K_{trans} = \frac{1}{2} mv^2 = \frac{1}{2} (200) = 100 \, J$.

(A) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + K^2 / R^2}$,where $K$ is the radius of gyration and $R$ is the radius.
$(i)$ For a solid sphere,$I = \frac{2}{5} MR^2 = MK^2$,so $K^2 = \frac{2}{5} R^2$. The acceleration is $a_s = \frac{g \sin \theta}{1 + 2/5} = \frac{5}{7} g \sin \theta$.
$(ii)$ For a disc,$I = \frac{1}{2} MR^2 = MK^2$,so $K^2 = \frac{1}{2} R^2$. The acceleration is $a_d = \frac{g \sin \theta}{1 + 1/2} = \frac{2}{3} g \sin \theta$.
$(iii)$ Since $a_s > a_d$,the solid sphere reaches the bottom earlier. The difference in acceleration is due to the different values of the radius of gyration $K$ for the two bodies.

(C) The acceleration $a$ of a body rolling without slipping on an inclined plane of angle $\theta$ is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$
where $K$ is the radius of gyration and $R$ is the radius of the sphere.
For a uniform solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$. Since $I = MK^2$,we have $K^2 = \frac{2}{5}R^2$,which implies $\frac{K^2}{R^2} = \frac{2}{5}$.
Substituting this into the acceleration formula:
$a = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5}{7} g \sin \theta$
Since the acceleration $a$ is independent of both the mass $M$ and the radius $R$,both spheres will have the same acceleration. Given they start from rest from the same height,they will reach the bottom at the same time. Thus,option $C$ is correct.

(B) The formula for the velocity $v$ of a body rolling down an inclined plane without slipping is given by $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$.
Here,$h$ is the vertical height of the inclined plane,$h = L \sin \theta = 60 \times 10^{-2} \times \sin 30^{\circ} = 0.6 \times 0.5 = 0.3\,m$.
For a solid cylinder,the radius of gyration $k$ satisfies $k^2 = \frac{R^2}{2}$,so $\frac{k^2}{R^2} = \frac{1}{2}$.
Substituting these values into the formula:
$v = \sqrt{\frac{2 \times 10 \times 0.3}{1 + 0.5}} = \sqrt{\frac{6}{1.5}} = \sqrt{4} = 2\,ms^{-1}$.

(A) At the highest point,the final kinetic energy $KE_f = 0$.
Initial kinetic energy $KE_i$ is the sum of translational and rotational kinetic energy:
$KE_i = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$
For a hollow spherical ball,the moment of inertia $I = \frac{2}{3} mR^2$. In the case of pure rolling,$v = R\omega$,so $\omega = \frac{v}{R}$.
Substituting these values:
$KE_i = \frac{1}{2} mv^2 + \frac{1}{2} \times (\frac{2}{3} mR^2) \times (\frac{v}{R})^2$
$KE_i = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$
Applying the law of conservation of energy,the initial kinetic energy is converted into potential energy at the maximum height $h$:
$KE_i = PE_f$
$\frac{5}{6} mv^2 = mgh$
$h = \frac{5v^2}{6g}$
Given $v = 3\, m/s$ and taking $g = 10\, m/s^2$:
$h = \frac{5 \times (3)^2}{6 \times 10} = \frac{5 \times 9}{60} = \frac{45}{60} = 0.75\, m$
Converting to centimeters: $0.75\, m = 75\, cm$.

(B) In pure rolling motion,the work done by static friction is zero because the point of contact is instantaneously at rest.
By the law of conservation of energy,the initial potential energy $(PE = mgh)$ is converted into the total kinetic energy $(KE)$ at the bottom of the incline.
Since both objects start from rest at the same height $h$,their total kinetic energies at the bottom must be equal,regardless of their moments of inertia.
Therefore,$KE_{\text{ring}} = KE_{\text{sphere}}$.
The ratio of their kinetic energies is $\frac{KE_{\text{ring}}}{KE_{\text{sphere}}} = 1$.
Given that the ratio is $\frac{7}{x}$,we have $\frac{7}{x} = 1$,which implies $x = 7$.

(D) For a body rolling down an inclined plane without slipping,the acceleration $a$ is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}}$
For a solid cylinder,the moment of inertia about its central axis is $I_{cm} = \frac{1}{2} MR^2$.
Substituting this into the formula:
$a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} MR^2}{MR^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2}{3} g \sin \theta$
Given $g = 10 \ m/s^2$ and $\theta = 60^{\circ}$:
$a = \frac{2}{3} \times 10 \times \sin(60^{\circ}) = \frac{20}{3} \times \frac{\sqrt{3}}{2} = \frac{10 \sqrt{3}}{3} = \frac{10}{\sqrt{3}} \ m/s^2$
Comparing this with the given expression $\frac{x}{\sqrt{3}} \ m/s^2$,we get $x = 10$.

(C) The disc is rolling on a horizontal surface without slipping,so it possesses both translational kinetic energy $(K_t = \frac{1}{2} Mv^2)$ and rotational kinetic energy $(K_r = \frac{1}{2} I\omega^2)$.
When the disc moves up a smooth inclined surface,there is no friction. Therefore,the torque due to friction is zero,and the rotational kinetic energy remains constant throughout the motion.
Only the translational kinetic energy is converted into gravitational potential energy as the disc climbs the incline.
Applying the law of conservation of energy for the translational part:
$\frac{1}{2} Mv^2 = Mgh$
Solving for $h$:
$h = \frac{v^2}{2g}$

(B) Using the principle of conservation of energy,the loss in kinetic energy equals the gain in potential energy.
$mgh = K.E._{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the object rolls without slipping,$\omega = \frac{v}{R}$ and $I = Mk^2$,where $k$ is the radius of gyration.
$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$
$h = \frac{v^2}{2g}(1 + \frac{k^2}{R^2})$
Thus,$h \propto (1 + \frac{k^2}{R^2})$.
For a solid sphere,$\frac{k^2}{R^2} = \frac{2}{5}$,so $h_1 \propto (1 + \frac{2}{5}) = \frac{7}{5}$.
For a hollow cylinder,$\frac{k^2}{R^2} = 1$,so $h_2 \propto (1 + 1) = 2$.
Therefore,$\frac{h_1}{h_2} = \frac{7/5}{2} = \frac{7}{10}$.
Given $\frac{h_1}{h_2} = \frac{n}{10}$,we find $n = 7$.

(A) For slipping down the plane without friction,the acceleration is $a = g \sin \theta$. The time taken to cover distance $l$ is $t = \sqrt{\frac{2l}{g \sin \theta}}$.
For rolling down the plane,the acceleration is $a' = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$. For a circular disc,the radius of gyration $k$ is $\frac{R}{\sqrt{2}}$,so $\frac{k^2}{R^2} = \frac{1}{2}$.
Thus,$a' = \frac{g \sin \theta}{1 + 1/2} = \frac{2}{3} g \sin \theta$.
The time taken for rolling is $t' = \sqrt{\frac{2l}{a'}} = \sqrt{\frac{2l}{\frac{2}{3} g \sin \theta}} = \sqrt{\frac{3}{2}} \sqrt{\frac{2l}{g \sin \theta}} = \sqrt{\frac{3}{2}} t$.
Comparing this with $t' = \left(\frac{\alpha}{2}\right)^{1/2} t$,we get $\sqrt{\frac{\alpha}{2}} = \sqrt{\frac{3}{2}}$,which implies $\alpha = 3$.

(D) For an object rolling without slipping,the total initial kinetic energy is the sum of translational and rotational kinetic energy: $K_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + \frac{k^2}{r^2})$.
At the maximum height $h$,the kinetic energy is converted into potential energy: $K_i = mgh$.
Equating the two: $\frac{1}{2}mv^2(1 + \frac{k^2}{r^2}) = mgh$.
Given $h = \frac{3v^2}{4g}$,we substitute this into the equation:
$\frac{1}{2}v^2(1 + \frac{k^2}{r^2}) = g(\frac{3v^2}{4g})$
$\frac{1}{2}(1 + \frac{k^2}{r^2}) = \frac{3}{4}$
$1 + \frac{k^2}{r^2} = \frac{3}{2}$
$\frac{k^2}{r^2} = \frac{1}{2}$.
Since the radius of gyration $k$ for a disc is given by $k^2 = \frac{1}{2}r^2$,the object is a disc.

(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$.
For a hollow cylinder,$I = mR^2$,so $a_{hollow} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
For a solid cylinder,$I = \frac{1}{2}mR^2$,so $a_{solid} = \frac{g \sin \theta}{1 + 0.5} = \frac{g \sin \theta}{1.5} = \frac{2g \sin \theta}{3}$.
Since $a_{solid} > a_{hollow}$,the solid cylinder reaches the bottom first. Thus,$STATEMENT-1$ is False.
By the principle of conservation of energy,the potential energy lost $(mgh)$ is converted into total kinetic energy $(K_{total} = K_{trans} + K_{rot})$. Since both have the same mass and height,their total kinetic energies at the bottom are identical. Thus,$STATEMENT-2$ is True.

(C) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$.
For a ring, $k^2 = R^2$, so $a_R = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$.
For a disc, $k^2 = \frac{R^2}{2}$, so $a_D = \frac{g \sin \theta}{1 + 0.5} = \frac{2g \sin \theta}{3}$.
The distance traveled along the incline is $d = \frac{h}{\sin \theta}$.
Using $d = \frac{1}{2} a t^2$, we have $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2h}{a \sin \theta}}$.
For the ring: $t_R = \sqrt{\frac{2h}{(g \sin \theta / 2) \sin \theta}} = \sqrt{\frac{4h}{g \sin^2 \theta}}$.
For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin \theta / 3) \sin \theta}} = \sqrt{\frac{3h}{g \sin^2 \theta}}$.
Given $\theta = 60^{\circ}$, $\sin \theta = \frac{\sqrt{3}}{2}$, so $\sin^2 \theta = \frac{3}{4}$.
$t_R = \sqrt{\frac{4h}{g(3/4)}} = \sqrt{\frac{16h}{3g}}$ and $t_D = \sqrt{\frac{3h}{g(3/4)}} = \sqrt{\frac{4h}{g}}$.
Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$, we have $\sqrt{\frac{h}{g}} \left( \frac{4}{\sqrt{3}} - 2 \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{\frac{h}{10}} \left( \frac{4-2\sqrt{3}}{\sqrt{3}} \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} \right) = 2-\sqrt{3} \Rightarrow \sqrt{h} = \frac{\sqrt{3}}{2} \Rightarrow h = \frac{3}{4} = 0.75 \,m$.

(B) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$.
For a ring,$k^2 = R^2$,so $a_R = \frac{g \sin \theta}{1+1} = \frac{g \sin \theta}{2}$.
For a disc,$k^2 = R^2/2$,so $a_D = \frac{g \sin \theta}{1+0.5} = \frac{2g \sin \theta}{3}$.
The distance to be covered along the incline is $d = \frac{h}{\sin \theta}$.
Using $d = \frac{1}{2} a t^2$,we have $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2h}{a \sin \theta}}$.
For the ring: $t_R = \sqrt{\frac{2h}{(g \sin \theta / 2) \sin \theta}} = \sqrt{\frac{4h}{g \sin^2 \theta}} = \sqrt{\frac{4h}{g (3/4)}} = \sqrt{\frac{16h}{3g}}$.
For the disc: $t_D = \sqrt{\frac{2h}{(2g \sin \theta / 3) \sin \theta}} = \sqrt{\frac{3h}{g \sin^2 \theta}} = \sqrt{\frac{3h}{g (3/4)}} = \sqrt{\frac{4h}{g}}$.
Given $t_R - t_D = \frac{2-\sqrt{3}}{\sqrt{10}}$,we have $\sqrt{\frac{h}{g}} \left( \sqrt{\frac{16}{3}} - 2 \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{\frac{h}{10}} \left( \frac{4 - 2\sqrt{3}}{\sqrt{3}} \right) = \frac{2-\sqrt{3}}{\sqrt{10}}$.
$\sqrt{h} \left( \frac{2(2-\sqrt{3})}{\sqrt{3}} \right) = 2-\sqrt{3}$.
$\sqrt{h} = \frac{\sqrt{3}}{2} \implies h = \frac{3}{4} = 0.75 \ m$.

(D) Let $a_c$ be the acceleration of the center of mass and $f$ be the frictional force.
From Newton's second law for translation: $F - f = ma_c$ $(1)$
From the torque equation about the center of mass: $fR = I_c \alpha$ $(2)$
For rolling without slipping: $a_c = \alpha R$,so $\alpha = \frac{a_c}{R}$.
Substituting $\alpha$ in $(2)$: $fR = I_c \frac{a_c}{R} \implies f = \frac{I_c a_c}{R^2}$.
Substituting $f$ in $(1)$: $F - \frac{I_c a_c}{R^2} = ma_c \implies a_c = \frac{F}{m + \frac{I_c}{R^2}}$.
$(A)$ Incorrect: $a_c$ depends on $I_c$,which differs for solid $(I_c = \frac{1}{2}mR^2)$ and hollow $(I_c = mR^2)$ cylinders.
$(D)$ Correct: For a thin-walled hollow cylinder,$I_c = mR^2$. Thus,$a_c = \frac{F}{m + \frac{mR^2}{R^2}} = \frac{F}{2m}$.
$(C)$ Incorrect: The frictional force $f = \frac{I_c a_c}{R^2} = \frac{I_c F}{R^2(m + I_c/R^2)}$,which is not necessarily $\mu mg$.
$(B)$ Correct: For rolling without slipping,$f \leq \mu mg$. Since $f = \frac{I_c a_c}{R^2}$,we have $\frac{I_c a_c}{R^2} \leq \mu mg \implies a_c \leq \frac{\mu mgR^2}{I_c}$. For a solid cylinder,$I_c = \frac{1}{2}mR^2$,so $a_c \leq \frac{\mu mgR^2}{0.5mR^2} = 2\mu g$.

(D) The moment of inertia $I$ of a body depends on the distribution of mass. Since cylinder $P$ has mass concentrated near its surface,it has a larger moment of inertia $(I_P > I_Q)$.
The linear acceleration $a$ of a rolling body on an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$.
Since $I_P > I_Q$,it follows that $a_P < a_Q$. This means cylinder $Q$ accelerates faster.
Using the kinematic equation $s = \frac{1}{2}at^2$,the time taken to reach the bottom is $t = \sqrt{\frac{2s}{a}}$. Since $a_P < a_Q$,we have $t_P > t_Q$,meaning $Q$ reaches the ground first.
The final velocity $v$ is given by $v^2 = 2as$. Since $a_Q > a_P$,$v_Q > v_P$. The translational kinetic energy is $\frac{1}{2}mv^2$,so $Q$ has higher translational kinetic energy.
Finally,since $v = \omega R$,the angular speed $\omega = \frac{v}{R}$. Since $v_Q > v_P$,it follows that $\omega_Q > \omega_P$. Thus,cylinder $Q$ reaches the ground with a larger angular speed.

(C) The total mechanical energy of a disc rolling without slipping is the sum of its translational and rotational kinetic energies,plus its potential energy.
Total energy $E = K_{trans} + K_{rot} + U = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + mgh$.
For a uniform disc,$I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
Thus,$E = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 + mgh = \frac{3}{4}mv^2 + mgh$.
Applying the law of conservation of energy between points $A$ and $B$,and $C$ and $D$:
For path $AB$: $E_A = E_B \implies \frac{3}{4}mv_1^2 + mgh_1 = \frac{3}{4}mv_f^2 + 0$,where $h_1 = 30 \ m$.
For path $CD$: $E_C = E_D \implies \frac{3}{4}mv_2^2 + mgh_2 = \frac{3}{4}mv_f^2 + 0$,where $h_2 = 27 \ m$.
Equating the two expressions for $\frac{3}{4}mv_f^2$:
$\frac{3}{4}mv_1^2 + mgh_1 = \frac{3}{4}mv_2^2 + mgh_2$.
Substituting the given values: $\frac{3}{4}(3)^2 + 10(30) = \frac{3}{4}v_2^2 + 10(27)$.
$\frac{27}{4} + 300 = \frac{3}{4}v_2^2 + 270$.
$6.75 + 30 = \frac{3}{4}v_2^2 \implies 36.75 = \frac{3}{4}v_2^2$.
$v_2^2 = \frac{36.75 \times 4}{3} = 12.25 \times 4 = 49$.
$v_2 = \sqrt{49} = 7 \ m/s$.

(D) Given: Mass $M = 1 kg$,Radius $R = 1 m$,$\theta = 30^{\circ}$,$g = 10 m s^{-2}$.
Component of weight down the plane: $Mg \sin \theta = 1 \times 10 \times \sin 30^{\circ} = 10 \times 0.5 = 5 N$.
Let $f$ be the friction force acting up the plane. The two external forces of $1 N$ each act in opposite directions relative to the center,creating a net torque.
Equation of motion for translation: $Mg \sin \theta - f - F_{ext} = Ma$. Here,the two $1 N$ forces act such that one aids motion and one opposes it,but they create a net torque. Looking at the diagram,the net force along the incline is $Mg \sin \theta = 5 N$. The two $1 N$ forces cancel each other out in terms of net translational force.
So,$5 - f = 1 \times a \Rightarrow f = 5 - a$.
Equation of motion for rotation about the center of mass $(COM)$: $\tau_{net} = I \alpha$.
The torque due to friction $f$ is $fR$ (counter-clockwise). The torque due to the two $1 N$ forces is $2 \times (1 N \times 0.5 m) = 1 N m$ (clockwise).
Thus,$fR - 1 = I \alpha$,where $I = \frac{2}{5} MR^2 = 0.4 kg m^2$ and $\alpha = \frac{a}{R} = a$.
$f(1) - 1 = 0.4 a \Rightarrow f = 1 + 0.4 a$.
Equating the two expressions for $f$: $5 - a = 1 + 0.4 a$.
$4 = 1.4 a \Rightarrow a = \frac{4}{1.4} = \frac{40}{14} = \frac{20}{7} \approx 2.86 m s^{-2}$.

(A) Using the Work-Energy Theorem $(WET)$:
$W_g = K_f - K_i$
Since the sphere starts from rest,$K_i = 0$. The work done by gravity is $W_g = mgL \sin \theta$.
The total kinetic energy $(K.E.)$ in pure rolling is given by:
$K_f = \frac{1}{2} mv^2 + \frac{1}{2} I_{cm} \omega^2$
For a solid sphere,$I_{cm} = \frac{2}{5} mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.
$K_f = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v^2}{r^2}) = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$.
Equating work done to change in kinetic energy:
$mgL \sin \theta = \frac{7}{10} mv^2$
$v^2 = \frac{10}{7} gL \sin \theta$
This implies $v^2 \propto \sin \theta$.
Therefore,the ratio is:
$\frac{v_1^2}{v_2^2} = \frac{\sin 30^{\circ}}{\sin 45^{\circ}} = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio $v_1^2 : v_2^2$ is $1 : \sqrt{2}$.

(C) The linear acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}}$.
For a uniform solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} mr^2$.
Substituting this into the acceleration formula: $a = \frac{g \sin \theta}{1 + \frac{1/2 mr^2}{mr^2}} = \frac{g \sin \theta}{1 + 1/2} = \frac{g \sin \theta}{3/2} = \frac{2}{3} g \sin \theta$.
Given the angle of inclination $\theta = 45^{\circ}$,we have $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Substituting the value of $\sin 45^{\circ}$ into the expression for $a$: $a = \frac{2}{3} g \left( \frac{1}{\sqrt{2}} \right) = \frac{\sqrt{2} g}{3}$.

(A) The time taken to reach the bottom of an inclined plane of length $\ell$ is given by $t = \sqrt{\frac{2\ell}{a_{cm}}}$.
The acceleration of a body rolling down an inclined plane is $a_{cm} = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}}$.
For a solid sphere,$I_{cm} = \frac{2}{5}MR^2$,so $a_1 = \frac{g \sin \theta}{1 + 2/5} = \frac{5g \sin \theta}{7}$.
For a hollow sphere,$I_{cm} = \frac{2}{3}MR^2$,so $a_2 = \frac{g \sin \theta}{1 + 2/3} = \frac{3g \sin \theta}{5}$.
Comparing the accelerations,$a_1 = \frac{5}{7}g \sin \theta \approx 0.714 g \sin \theta$ and $a_2 = \frac{3}{5}g \sin \theta = 0.6 g \sin \theta$. Thus,$a_1 > a_2$.
Since $t \propto \frac{1}{\sqrt{a_{cm}}}$,a higher acceleration results in a shorter time. Therefore,$t_1 < t_2$.

(A) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into translational and rotational kinetic energy at the bottom: $Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
Since the object rolls without slipping,$\omega = \frac{v}{R}$,so $Mgh = \frac{1}{2} Mv^2 (1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.
Thus,$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$.
For a ring,$I = MR^2$,so $k^2 = R^2$ and $v_{\text{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.
For a solid sphere,$I = \frac{2}{5}MR^2$,so $k^2 = \frac{2}{5}R^2$ and $v_{\text{sphere}} = \sqrt{\frac{2gh}{1 + 2/5}} = \sqrt{\frac{10gh}{7}}$.
The ratio of velocities is $\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \frac{\sqrt{gh}}{\sqrt{10gh/7}} = \sqrt{\frac{7}{10}}$.
To match the form $\sqrt{\frac{x}{5}}$,we write $\sqrt{\frac{7}{10}} = \sqrt{\frac{3.5}{5}}$.
Rounding to the nearest integer,$x = 4$.

(A) For a body rolling down an incline without slipping,the linear velocity $v$ at the bottom is given by the formula: $v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$.
For a solid sphere,the radius of gyration $K$ is given by $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.
Substituting the given values $g = 10 \ \text{m/s}^2$ and $h = 14 \ \text{m}$:
$v = \sqrt{\frac{2 \times 10 \times 14}{1 + \frac{2}{5}}} = \sqrt{\frac{280}{\frac{7}{5}}} = \sqrt{\frac{280 \times 5}{7}}$.
$v = \sqrt{40 \times 5} = \sqrt{200} = 10 \sqrt{2} \ \text{m/s}$.