Rolling On Inclined Plane Questions in English

(C) The total kinetic energy of a rolling sphere is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} mu^2 + \frac{1}{2} I \omega^2$.
For a solid sphere, the moment of inertia $I = \frac{2}{5} mr^2$ and the rolling condition is $\omega = u/r$.
Substituting these, $K = \frac{1}{2} mu^2 + \frac{1}{2} (\frac{2}{5} mr^2) (u/r)^2 = \frac{1}{2} mu^2 (1 + \frac{2}{5}) = \frac{1}{2} mu^2 (\frac{7}{5}) = \frac{7}{10} mu^2$.
By the law of conservation of energy, the total kinetic energy is converted into potential energy at the maximum height $H$: $K = mgH$.
Therefore, $\frac{7}{10} mu^2 = mgH$.
Solving for $H$, we get $H = \frac{7 u^2}{10 g}$.

(A) According to the law of conservation of mechanical energy,the loss in potential energy $(PE)$ is equal to the gain in total kinetic energy $(KE_{total})$,which includes both translational and rotational kinetic energy.
Initial height $h_1 = 100 \ m$,final height $h_2 = 20 \ m$.
Change in height $\Delta h = h_1 - h_2 = 100 \ m - 20 \ m = 80 \ m$.
Loss in $PE = mg \Delta h = m \times 10 \times 80 = 800m \ J$.
Gain in $KE_{total} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$.
For a solid spherical ball,the moment of inertia $I = \frac{2}{5} mr^2$ and $\omega = \frac{v}{r}$.
Substituting these values: $KE_{total} = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mr^2) (\frac{v}{r})^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$.
Equating loss in $PE$ to gain in $KE_{total}$:
$800m = \frac{7}{10} mv^2$.
$v^2 = \frac{8000}{7}$.
$v = \sqrt{\frac{8000}{7}} = \sqrt{\frac{1600 \times 5}{7}} = 40 \sqrt{\frac{5}{7}} \ m/s$.

(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
Since the moment of inertia $I$ for a ring is $MR^2$,for a disc is $\frac{1}{2}MR^2$,and for a solid sphere is $\frac{2}{5}MR^2$,we have $I_{\text{ring}} > I_{\text{disc}} > I_{\text{sphere}}$.
Consequently,the acceleration follows the order $a_{\text{sphere}} > a_{\text{disc}} > a_{\text{ring}}$.
The speed at the bottom is $v = \sqrt{2ah} / \sqrt{\sin \theta} = \sqrt{\frac{2gh}{1 + \frac{I}{MR^2}}}$,so $v_{\text{sphere}} > v_{\text{disc}} > v_{\text{ring}}$.
The time of descent is $t = \sqrt{\frac{2s}{a}}$,which means $t \propto \frac{1}{\sqrt{a}}$. Therefore,$t_{\text{sphere}} < t_{\text{disc}} < t_{\text{ring}}$.
Comparing these with the given options,option $D$ is the wrong statement.

(A) The moment of inertia of a solid sphere is $I = \frac{2}{5} MR^2$.
Since the sphere rolls without slipping,the condition $v = R\omega$ holds.
The rotational kinetic energy $E_{rot}$ is given by $E_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{5} MR^2) \omega^2 = \frac{1}{5} MR^2 \omega^2 = \frac{1}{5} Mv^2$.
The total kinetic energy $E_{total}$ is the sum of translational and rotational kinetic energy: $E_{total} = \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = (\frac{5+2}{10}) Mv^2 = \frac{7}{10} Mv^2$.
The ratio of total kinetic energy to rotational kinetic energy is $\frac{E_{total}}{E_{rot}} = \frac{\frac{7}{10} Mv^2}{\frac{1}{5} Mv^2} = \frac{7}{10} \times 5 = \frac{7}{2}$.

(D) For a solid sphere,the moment of inertia is $I_{S} = \frac{2}{5} MR^2$.
For a solid cylinder,the moment of inertia is $I_{C} = \frac{1}{2} MR^2$.
The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
Therefore,the ratio of the acceleration of the cylinder '$a_{c}$' to that of the sphere '$a_{s}$' is:
$\frac{a_{c}}{a_{s}} = \frac{1 + \frac{I_{S}}{MR^2}}{1 + \frac{I_{C}}{MR^2}} = \frac{1 + \frac{2}{5}}{1 + \frac{1}{2}} = \frac{\frac{7}{5}}{\frac{3}{2}} = \frac{7}{5} \times \frac{2}{3} = \frac{14}{15}$.

(A) For a body sliding down a smooth inclined plane of height $h$,the potential energy is converted into kinetic energy: $mgh = \frac{1}{2}mV^2$,which gives $V = \sqrt{2gh}$.
For a ring rolling down the same inclined plane,the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a ring,the moment of inertia $I = mr^2$ and the rolling condition is $v = r\omega$,so $\omega = v/r$.
Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Thus,$v^2 = gh$,which means $v = \sqrt{gh}$.
Comparing the two velocities: $v = \sqrt{gh} = \frac{\sqrt{2gh}}{\sqrt{2}} = \frac{V}{\sqrt{2}}$.

(B) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a solid sphere,the moment of inertia $I$ about its center of mass is $\frac{2}{5}MR^2$.
Substituting this into the formula: $a = \frac{g \sin \theta}{1 + \frac{2/5 MR^2}{MR^2}} = \frac{g \sin \theta}{1 + 2/5} = \frac{g \sin \theta}{7/5} = \frac{5}{7} g \sin \theta$.
Given $\theta = 30^{\circ}$ and $\sin 30^{\circ} = 0.5$,we have:
$a = \frac{5}{7} g (0.5) = \frac{5}{7} g \left(\frac{1}{2}\right) = \frac{5g}{14}$.
Therefore,the correct option is $B$.

(A) For a solid sphere rolling without slipping, the total kinetic energy $K$ is the sum of translational and rotational kinetic energy: $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $I = \frac{2}{5}mr^2$ and $v = r\omega$, we have $K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
When the sphere moves up the ramp, this kinetic energy is converted into potential energy $U = mgh$, where $h = d \sin \theta$ and $d$ is the distance along the ramp.
Equating $K = U$: $\frac{7}{10}mv^2 = mgd \sin \theta$.
Solving for $d$: $d = \frac{7v^2}{10g \sin \theta}$.
Substituting the given values: $v = 2 \,m/s$, $g = 10 \,m/s^2$, $\theta = 30^{\circ}$:
$d = \frac{7 \times (2)^2}{10 \times 10 \times (1/2)} = \frac{7 \times 4}{100 \times 0.5} = \frac{28}{50} = 0.56 \,m$.

(B) For a body rolling down an inclined plane without slipping,mechanical energy is conserved.
Initial potential energy at the top is $PE = mgh$.
Initial kinetic energy at the top is $KE = 0$ (since it starts from rest).
At the bottom,the total kinetic energy is the sum of translational and rotational kinetic energy: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$.
Since it rolls without slipping,$v = r\omega$,so $\omega = v/r$.
Substituting these into the kinetic energy equation: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
By conservation of energy,$mgh = \frac{7}{10}mv^2$.
Solving for $v$,we get $v^2 = \frac{10gh}{7}$,which implies $v = \left(\frac{10gh}{7}\right)^{1/2}$.

(B) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula: $a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$.
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 $\theta = 30^{\circ}$ and $\sin 30^{\circ} = 0.5$ into the formula:
$a = \frac{g \sin 30^{\circ}}{1 + \frac{2}{5}} = \frac{g \times 0.5}{\frac{7}{5}} = \frac{0.5g \times 5}{7} = \frac{2.5g}{7} = \frac{5g}{14}$.

(A) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula:
$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$
For a solid sphere,the radius of gyration $K$ is related to the radius $R$ by $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.
Given $\theta = 30^{\circ}$ and $\sin 30^{\circ} = 0.5 = \frac{1}{2}$.
Substituting these values into the formula:
$a = \frac{g \sin 30^{\circ}}{1 + \frac{2}{5}}$
$a = \frac{g \times (1/2)}{7/5}$
$a = \frac{g}{2} \times \frac{5}{7} = \frac{5g}{14}$.

(D) Case $1$: For a sliding body,the potential energy is converted into translational kinetic energy: $\frac{1}{2} m V^2 = mgh$ ... $(i)$
Case $2$: For a rolling disc,the potential energy is converted into both translational and rotational kinetic energy: $\frac{1}{2} m (v')^2 + \frac{1}{2} I \omega^2 = mgh$
For a disc,the moment of inertia is $I = \frac{1}{2} mR^2$ and the rolling condition is $\omega = \frac{v'}{R}$.
Substituting these into the energy equation: $\frac{1}{2} m (v')^2 + \frac{1}{2} (\frac{1}{2} mR^2) (\frac{v'}{R})^2 = mgh$
$\frac{1}{2} m (v')^2 + \frac{1}{4} m (v')^2 = mgh$
$\frac{3}{4} m (v')^2 = mgh$ ... $(ii)$
Equating $(i)$ and $(ii)$: $\frac{1}{2} m V^2 = \frac{3}{4} m (v')^2$
$V^2 = \frac{3}{2} (v')^2$
$(v')^2 = \frac{2}{3} V^2$
$v' = V \sqrt{\frac{2}{3}}$

(A) For a solid cylinder rolling down an inclined plane,the total energy at the top is potential energy $PE = Mgh$.
At the bottom,the total energy is the sum of translational kinetic energy and rotational kinetic energy: $KE = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the rolling condition is $v = R\omega$,so $\omega = \frac{v}{R}$.
Substituting these into the energy equation: $Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$.
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2$.
$Mgh = \frac{3}{4} Mv^2$.
Solving for height $h$: $h = \frac{3 v^2}{4 g}$.

(A) For a solid cylinder rolling down an inclined plane,the total potential energy $Mgh$ is converted into translational kinetic energy and rotational kinetic energy.
Total energy $E = Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the rolling condition is $v = R\omega$.
Substituting these into the energy equation:
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$.
Thus,$Mv^2 = \frac{4}{3} Mgh$.
The rotational kinetic energy $K_{rot} = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{4} Mv^2$.
Substituting $Mv^2 = \frac{4}{3} Mgh$ into the expression for $K_{rot}$:
$K_{rot} = \frac{1}{4} (\frac{4}{3} Mgh) = \frac{Mgh}{3}$.

(A) For a disc rolling down an inclined plane,the total potential energy at the top is converted into translational and rotational kinetic energy at the bottom.
By the law of conservation of energy: $mgh = K_{trans} + K_{rot}$.
For a disc,the moment of inertia $I = \frac{1}{2}mr^2$.
Since the disc rolls without slipping,$v = r\omega$,so $\omega = \frac{v}{r}$.
The translational kinetic energy is $K_{trans} = \frac{1}{2}mv^2$.
The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{1}{4}mv^2$.
Total energy $mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Thus,$\frac{1}{2}mv^2 = \frac{2}{3}mgh$.
Substituting this into the expression for rotational kinetic energy: $K_{rot} = \frac{1}{4}mv^2 = \frac{1}{2}(\frac{1}{2}mv^2) = \frac{1}{2}(\frac{2}{3}mgh) = \frac{mgh}{3}$.

(A) By the law of conservation of energy,the total potential energy at the top is equal to the sum of translational and rotational kinetic energy at the bottom:
$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$
For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the condition for pure rolling is $v = R\omega$.
Substituting these into the energy equation:
$Mgh = \frac{1}{2} M(R\omega)^2 + \frac{1}{2} (\frac{1}{2} MR^2)\omega^2$
$Mgh = \frac{1}{2} MR^2\omega^2 + \frac{1}{4} MR^2\omega^2 = \frac{3}{4} MR^2\omega^2$
Thus,the rotational kinetic energy $K_{rot} = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} MR^2)\omega^2 = \frac{1}{4} MR^2\omega^2$.
From the energy equation,$MR^2\omega^2 = \frac{4}{3} Mgh$.
Substituting this into the expression for $K_{rot}$:
$K_{rot} = \frac{1}{4} (\frac{4}{3} Mgh) = \frac{Mgh}{3}$.

(D) Given,height of the inclined plane,$h = 7 \ m$.
Acceleration due to gravity,$g = 10 \ m/s^2$.
By the law of conservation of energy,the potential energy lost by the solid sphere equals the total kinetic energy gained (translational + rotational).
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5}mR^2$ and the condition for rolling without slipping is $\omega = \frac{v}{R}$.
Substituting these values:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{2}{5}mR^2\right) \left(\frac{v}{R}\right)^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2$
$mgh = \frac{7}{10}mv^2$
$v^2 = \frac{10}{7}gh$
$v = \sqrt{\frac{10}{7} \times 10 \times 7} = \sqrt{100} = 10 \ m/s$.

(C) For a body sliding down a smooth inclined plane,the potential energy is converted entirely into translational kinetic energy. Thus,$v = \sqrt{2gh}$.
For a sphere rolling down a rough inclined plane,the potential energy is converted into both translational and rotational kinetic energy. The velocity at the bottom is given by $v_{CM} = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$.
Substituting $v = \sqrt{2gh}$,we get $v_{CM} = \frac{v}{\sqrt{1 + \frac{K^2}{R^2}}}$.
For a solid sphere,the radius of gyration $K$ satisfies $\frac{K^2}{R^2} = \frac{2}{5}$.
Substituting this value,$v_{CM} = \frac{v}{\sqrt{1 + \frac{2}{5}}} = \frac{v}{\sqrt{\frac{7}{5}}} = \sqrt{\frac{5}{7}} v$.

(C) The velocity $v$ of a body rolling without slipping down an inclined plane of height $h$ is given by the formula:
$v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$
where $K$ is the radius of gyration and $R$ is the radius of the body.
For a ring,$K^2 = R^2$,so $v_{R} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.
For a solid cylinder,$K^2 = \frac{R^2}{2}$,so $v_{C} = \sqrt{\frac{2gh}{1+1/2}} = \sqrt{\frac{4gh}{3}} \approx 1.15 \sqrt{gh}$.
For a solid sphere,$K^2 = \frac{2}{5}R^2$,so $v_{S} = \sqrt{\frac{2gh}{1+2/5}} = \sqrt{\frac{10gh}{7}} \approx 1.19 \sqrt{gh}$.
Comparing the values,we find that $v_{R} < v_{C} < v_{S}$.

(D) The total kinetic energy $(KE)$ of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.
$KE = KE_{\text{trans}} + KE_{\text{rot}}$
$KE = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2}$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^{2}$ and the condition for rolling without slipping is $v = R \omega$,which implies $\omega = \frac{v}{R}$.
Substituting these into the equation:
$KE = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m R^{2}) (\frac{v}{R})^{2}$
$KE = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m v^{2})$
$KE = \frac{1}{2} m v^{2} (1 + \frac{2}{5})$
$KE = \frac{1}{2} m v^{2} (\frac{7}{5})$
$KE = \frac{7}{10} m v^{2}$

(B) For a solid cylinder rolling without slipping,the total kinetic energy $(K.E.)$ is the sum of translational and rotational kinetic energy:
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2$ and the condition for rolling without slipping is $v = R\omega$ (or $\omega = v/R$):
$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$K.E. = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
By the law of conservation of energy,the loss in potential energy $(P.E.)$ equals the gain in kinetic energy:
$mgh = \frac{3}{4}mv^2$
$v^2 = \frac{4gh}{3}$
$v = \sqrt{\frac{4gh}{3}}$

(D) For a solid sphere rolling down an inclined plane without sliding,the acceleration is given by $a_{roll} = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$.
For a solid sphere,the moment of inertia $I = \frac{2}{5} MR^2$,so $a_{roll} = \frac{g \sin \theta}{1 + 2/5} = \frac{5}{7} g \sin \theta$.
Given $a_{roll} = 3.5 \ m \ s^{-2}$,we have $3.5 = \frac{5}{7} g \sin \theta$,which implies $g \sin \theta = 3.5 \times \frac{7}{5} = 4.9 \ m \ s^{-2}$.
When the sphere slides down without rolling,there is no rotational motion,so the acceleration is simply $a_{slide} = g \sin \theta$.
Therefore,$a_{slide} = 4.9 \ m \ s^{-2}$.

(D) When a disc rolls down an inclined plane without slipping,the loss in gravitational potential energy is converted into translational and rotational kinetic energy.
By the law of conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a disc,the moment of inertia is $I = \frac{1}{2}mr^2$ and the rolling condition is $\omega = \frac{v}{r}$.
Substituting these values: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$.
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Canceling $m$ from both sides: $gh = \frac{3}{4}v^2$.
Solving for $v$: $v = \sqrt{\frac{4gh}{3}}$.
Thus,the velocity $v$ depends only on the height of the incline $h$ and the acceleration due to gravity $g$.

(A) Let the radius of gyration of the body be $K$,mass be $m$,height be $h$,and radius be $R$.
According to the law of conservation of mechanical energy,the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = mK^2$ and $\omega = v/R$,we have:
$mgh = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$
$v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$
For a ring,$K^2 = R^2$,so $v_{\text{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.
For a solid cylinder,$K^2 = \frac{R^2}{2}$,so $v_{\text{cylinder}} = \sqrt{\frac{2gh}{1+0.5}} = \sqrt{\frac{4gh}{3}} \approx 1.15\sqrt{gh}$.
For a solid sphere,$K^2 = \frac{2}{5}R^2$,so $v_{\text{sphere}} = \sqrt{\frac{2gh}{1+0.4}} = \sqrt{\frac{10gh}{7}} \approx 1.19\sqrt{gh}$.
Comparing the velocities,the ring has the minimum velocity at the bottom of the inclined plane.

(C) For a rigid body rolling without slipping on an inclined plane,the force of friction $f$ is given by the formula: $f = \frac{Mg \sin \theta}{1 + \frac{MR^2}{I}}$.
$(A)$ For a ring,$I = MR^2$. Thus,$f = \frac{Mg \sin \theta}{1 + 1} = \frac{Mg \sin \theta}{2}$.
$(B)$ For a solid sphere,$I = \frac{2}{5}MR^2$. Thus,$f = \frac{Mg \sin \theta}{1 + 2.5} = \frac{Mg \sin \theta}{3.5}$.
$(C)$ For a solid cylinder,$I = \frac{1}{2}MR^2$. Thus,$f = \frac{Mg \sin \theta}{1 + 2} = \frac{Mg \sin \theta}{3}$.
$(D)$ For a hollow cylinder,$I = MR^2$. Thus,$f = \frac{Mg \sin \theta}{1 + 1} = \frac{Mg \sin \theta}{2}$.
Matching the results: $A-II, B-I, C-III, D-II$.

(C) Let $h$ be the height attained by the sphere before it stops. According to the law of conservation of energy, the total initial kinetic energy (translational + rotational) is converted into potential energy at the maximum height.
$K_{\text{rot}} + K_{\text{trans}} = mgh$
$\frac{1}{2} I \omega^2 + \frac{1}{2} m v_{CM}^2 = mgh$
Since the sphere is rolling, $v_{CM} = R\omega$ and for a solid sphere, the moment of inertia $I = \frac{2}{5} mR^2$.
Substituting these values:
$\frac{1}{2} \left( \frac{2}{5} mR^2 \right) \left( \frac{v_{CM}}{R} \right)^2 + \frac{1}{2} m v_{CM}^2 = mgh$
$\frac{1}{5} m v_{CM}^2 + \frac{1}{2} m v_{CM}^2 = mgh$
Dividing by $m$ and simplifying:
$v_{CM}^2 \left( \frac{1}{5} + \frac{1}{2} \right) = gh$
$v_{CM}^2 \left( \frac{2+5}{10} \right) = gh$
$\frac{7}{10} v_{CM}^2 = gh$
Given $v_{CM} = 10 \,m/s$ and $g = 10 \,m/s^2$:
$\frac{7}{10} \times (10)^2 = 10 \times h$
$\frac{7}{10} \times 100 = 10h$
$70 = 10h$
$h = 7 \,m$

(C) Given:
Initial velocity of the solid spherical ball is $v = 10 \ m \ s^{-1}$.
Mass of the ball is $m = 11 \ kg$.
Since frictional losses are negligible,the total mechanical energy is conserved.
The initial total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energy:
$K_i = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^2$ and $\omega = \frac{v}{R}$.
$K_i = \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2 = \frac{1}{2} m v^2 (1 + \frac{2}{5}) = \frac{1}{2} m v^2 (\frac{7}{5}) = \frac{7}{10} m v^2$.
At the maximum height $h$,the ball momentarily stops,so its final kinetic energy $K_f = 0$.
The potential energy gained by the ball is $U_f = m g h$.
By the law of conservation of energy,$K_i = U_f$:
$\frac{7}{10} m v^2 = m g h$
$h = \frac{7 v^2}{10 g} = \frac{7 \times (10)^2}{10 \times 10} = \frac{7 \times 100}{100} = 7 \ m$.
Thus,the value of $h$ is $7 \ m$.

(B) The acceleration of a body rolling down an inclined plane without slipping is given by $a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$, where $I$ is the moment of inertia about the center of mass.
For a solid sphere, $I_{sphere} = \frac{2}{5} MR^2$. Thus, $a_{sphere} = \frac{g \sin \theta}{1 + 2/5} = \frac{g \sin \theta}{7/5} = \frac{5}{7} g \sin \theta = 3 \,ms^{-2}$.
This implies $g \sin \theta = \frac{3 \times 7}{5} = 4.2 \,ms^{-2}$.
For a thin uniform circular disc, $I_{disc} = \frac{1}{2} MR^2$. Thus, $a_{disc} = \frac{g \sin \theta}{1 + 1/2} = \frac{g \sin \theta}{3/2} = \frac{2}{3} g \sin \theta$.
Substituting the value of $g \sin \theta$, we get $a_{disc} = \frac{2}{3} \times 4.2 = 2.8 \,ms^{-2}$.

(A) For a solid sphere rolling down an inclined plane without slipping, the velocity $v$ at the bottom is given by the formula:
$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$
Where $g = 10 \text{ ms}^{-2}$, $h = 28 \text{ m}$, and for a solid sphere, the radius of gyration $k$ satisfies $k^2 = \frac{2}{5}R^2$, so $\frac{k^2}{R^2} = \frac{2}{5}$.
Substituting the values:
$v = \sqrt{\frac{2 \times 10 \times 28}{1 + \frac{2}{5}}}$
$v = \sqrt{\frac{560}{\frac{7}{5}}}$
$v = \sqrt{\frac{560 \times 5}{7}}$
$v = \sqrt{80 \times 5} = \sqrt{400} = 20 \text{ ms}^{-1}$.

(D) For an object rolling down an inclined plane of length $l$ and angle $\theta$,the acceleration is given by $a = \frac{g \sin \theta}{1 + K^2/R^2}$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$ with $u=0$ and $s=l$,we get $l = \frac{1}{2} \left( \frac{g \sin \theta}{1 + K^2/R^2} \right) t^2$.
Thus,$t = \sqrt{\frac{2l(1 + K^2/R^2)}{g \sin \theta}}$,which implies $t \propto \sqrt{1 + K^2/R^2}$.
For a hollow cylinder,the radius of gyration $K^2 = R^2$,so $K^2/R^2 = 1$. Thus,$t_1 \propto \sqrt{1 + 1} = \sqrt{2}$.
For a solid cylinder,the radius of gyration $K^2 = R^2/2$,so $K^2/R^2 = 1/2$. Thus,$t_2 \propto \sqrt{1 + 1/2} = \sqrt{3/2}$.
Taking the ratio,$\frac{t_2}{t_1} = \frac{\sqrt{3/2}}{\sqrt{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Given $t_1 = 2 \ s$,we have $t_2 = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \approx 1.732 \ s$.

(B) Since the cylinder rolls without slipping,the work done by friction is zero. Therefore,we can apply the principle of conservation of mechanical energy.
At the top,the total energy is purely potential: $E_i = mgh$.
At the bottom,the total energy is the sum of translational and rotational kinetic energy: $E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a solid cylinder,the moment of inertia $I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
Equating $E_i = E_f$:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
$v^2 = \frac{4gh}{3}$
$v = \sqrt{\frac{4 \times 10 \times 30}{3}} = \sqrt{400} = 20 \ m \ s^{-1}$.

(B) Given: Inclination $\theta = 30^{\circ}$, length $l = 60 \,cm = 0.6 \,m$, acceleration due to gravity $g = 10 \,m/s^2$.
Using the law of conservation of mechanical energy, the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Here, $h = l \sin \theta = 0.6 \times \sin 30^{\circ} = 0.6 \times 0.5 = 0.3 \,m$.
For a solid cylinder, the moment of inertia $I = \frac{1}{2}mr^2$ and for rolling without slipping, $\omega = \frac{v}{r}$.
Substituting these values:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$
$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
$gh = \frac{3}{4}v^2 \Rightarrow v = \sqrt{\frac{4gh}{3}}$
$v = \sqrt{\frac{4 \times 10 \times 0.3}{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2 \,m/s$.

(D) Given: Radius of sphere $= R$,mass of sphere $= m$,and angle of inclination $\theta = 45^{\circ}$.
For a solid sphere rolling without slipping on an inclined plane,the linear acceleration is given by:
$a = \frac{g \sin \theta}{1 + \frac{I}{m R^2}} = \frac{g \sin \theta}{1 + \frac{2/5 m R^2}{m R^2}} = \frac{5}{7} g \sin \theta$
Applying Newton's second law for linear motion along the incline:
$m g \sin \theta - f = m a$
$f = m g \sin \theta - m \left( \frac{5}{7} g \sin \theta \right) = \frac{2}{7} m g \sin \theta$
The normal force is $N = m g \cos \theta$.
The coefficient of static friction $\mu_s$ must satisfy:
$\mu_s \geq \frac{f}{N} = \frac{\frac{2}{7} m g \sin \theta}{m g \cos \theta} = \frac{2}{7} \tan \theta$
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$,so $\mu_s \geq \frac{2}{7} \approx 0.2857$.
Comparing the options:
$(A)$ $\frac{3}{7} \approx 0.428 > 0.2857$
$(B)$ $\frac{1}{2} = 0.5 > 0.2857$
$(C)$ $\frac{5}{8} = 0.625 > 0.2857$
$(D)$ $\frac{1}{7} \approx 0.1428 < 0.2857$
Since $\frac{1}{7}$ is less than $\frac{2}{7}$,the sphere will slip. Thus,option $(D)$ is the incorrect value for rolling without slipping.

(A) For a body undergoing pure rolling on an inclined plane,the forces acting along the plane are the component of gravity $mg \sin \theta$ and the frictional force $f$. The equation of motion is $ma = mg \sin \theta - f$.
The torque equation about the center of mass is $\tau = I \alpha = fR$.
Since the body is in pure rolling,$\alpha = \frac{a}{R}$. Also,the moment of inertia $I = mk^2$.
Substituting these into the torque equation: $mk^2 \cdot \frac{a}{R} = fR$,which gives $f = \frac{ma k^2}{R^2}$.
Substituting the expression for $f$ into the force equation: $ma = mg \sin \theta - \frac{ma k^2}{R^2}$.
Rearranging the terms: $ma(1 + \frac{k^2}{R^2}) = mg \sin \theta$.
Therefore,the acceleration is $a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}}$.

(C) Since the sphere is rolling without slipping and there is no friction,the total mechanical energy is conserved.
Let $R$ be the radius of the track and $r$ be the radius of the sphere. Given $r \ll R$.
The total energy at the lowest point (bottom) is the sum of translational and rotational kinetic energy:
$E_{bottom} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,$I = \frac{2}{5}mr^2$. Since it rolls without slipping,$\omega = \frac{v}{r}$.
$E_{bottom} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
At the maximum height $h$ reached by the center of mass,the velocity becomes zero,so the total energy is purely potential energy:
$E_{top} = mgh$
Equating the energies: $\frac{7}{10}mv^2 = mgh \Rightarrow h = \frac{7v^2}{10g}$
From the geometry of the track,$h = R(1 - \cos\theta)$.
Substituting the values $v = 10 \ m/s$,$g = 10 \ m/s^2$,and $R = 10 \ m$:
$h = \frac{7 \times (10)^2}{10 \times 10} = 7 \ m$
$7 = 10(1 - \cos\theta) \Rightarrow 0.7 = 1 - \cos\theta \Rightarrow \cos\theta = 0.3 = \frac{3}{10}$
Therefore,$\theta = \cos^{-1}\left(\frac{3}{10}\right)$.

(D) When a body of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping,its velocity at the bottom is given by $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$,where $k$ is the radius of gyration.
For a ring,$k^2 = R^2$,so $v_R = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.
For a solid disc,$k^2 = \frac{R^2}{2}$,so $v_D = \sqrt{\frac{2gh}{1+0.5}} = \sqrt{\frac{4gh}{3}} \approx 1.15 \sqrt{gh}$.
For a solid sphere,$k^2 = \frac{2R^2}{5}$,so $v_S = \sqrt{\frac{2gh}{1+0.4}} = \sqrt{\frac{10gh}{7}} \approx 1.19 \sqrt{gh}$.
Comparing the values,we find $V_S > V_D > V_R$.

(C) By the law of conservation of energy,the total initial mechanical energy equals the total final mechanical energy at the highest point.
Initial energy: $E_i = K.E_{trans} + K.E_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
For a solid sphere,$I = \frac{2}{5}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.
$E_i = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
At the highest point,the final kinetic energy is zero,and the potential energy is $mgh$.
So,$\frac{7}{10}mv^2 = mgh \Rightarrow h = \frac{7v^2}{10g}$
Given $v = 4 \ m/s$ and $g = 10 \ m/s^2$,$h = \frac{7 \times 4^2}{10 \times 10} = \frac{7 \times 16}{100} = 1.12 \ m$.
The distance $\ell$ along the plane is given by $\ell = \frac{h}{\sin 30^{\circ}} = \frac{1.12}{0.5} = 2.24 \ m = 224 \ cm$.

(C) For a solid sphere rolling without slipping down an inclined plane,the total kinetic energy $K$ at the bottom is the sum of translational and rotational kinetic energy.
$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since $I = \frac{2}{5}mR^2$ and $v = R\omega$,we have $K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.
By conservation of energy,the potential energy at height $h$ equals the total kinetic energy at the bottom:
$mgh = \frac{7}{10}mv^2 \implies v^2 = \frac{10}{7}gh$.
When the ball is thrown vertically upwards with velocity $v$,it undergoes motion under gravity. At the maximum height $h'$,the final velocity is $0$. Using the kinematic equation $v_f^2 = v_i^2 - 2gh'$:
$0 = v^2 - 2gh' \implies h' = \frac{v^2}{2g}$.
Substituting $v^2 = \frac{10}{7}gh$:
$h' = \frac{10/7 gh}{2g} = \frac{5}{7}h$.

(D) By the law of conservation of energy,the initial kinetic energy (translational + rotational) is converted into gravitational potential energy at the maximum height $h$.
Initial kinetic energy $K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2$.
Since the object rolls without slipping,$\omega = \frac{v_0}{R}$.
Substituting the moment of inertia $I = kmR^2$,we get:
$K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}(kmR^2)(\frac{v_0}{R})^2 = \frac{1}{2}mv_0^2(1+k)$.
At maximum height $h$,the potential energy is $U_f = mgh$.
Equating $K_i = U_f$,we have $\frac{1}{2}mv_0^2(1+k) = mgh$.
Thus,$h = \frac{v_0^2(1+k)}{2g}$.
Given $h = \frac{7v_0^2}{10g}$,we equate the two expressions:
$\frac{v_0^2(1+k)}{2g} = \frac{7v_0^2}{10g} \Rightarrow 1+k = \frac{14}{10} = 1.4 \Rightarrow k = 0.4$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mR^2$,so $k = 0.4$.
Therefore,the object is a solid sphere.