Correct answer option is (D)

Only the translational kinetic energy of disc changes into gravitational potential energy. And rotational $\mathrm{KE}$ remains unchanged as there is no friction.

$\begin{array}{rl}& \frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}\\ \\ & \Rightarrow \mathrm{h}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}\end{array}$

Correct answer is 3

To find the value of $\alpha$ from the given problem, we need to analyze the motion of a circular disc moving down an inclined plane in two different modes: slipping and rolling.

Slipping:

When the disc slips without rolling, it is primarily subjected to kinetic friction and gravity, without any rolling friction or torque affecting rotational motion. The motion can be considered as purely translational.

1. Equation for Time in Slipping Mode:

The acceleration $a$ of the disc while slipping is given by:

$a=g\sin \theta$

where $g$ is the acceleration due to gravity and $\theta$ is the angle of the inclined plane.

The time $t$ to travel down the incline of length $l$ with this acceleration from rest is:

$l=\frac{1}{2}a{t}^2\Rightarrow t=\sqrt{\frac{2l}{a}}=\sqrt{\frac{2l}{g\sin \theta }}$

Rolling:

When the disc rolls, both translational and rotational motions are involved, and the rolling motion means that there is a rotational inertia factor that affects the acceleration.

1. Equation for Time in Rolling Mode:

For a solid disc, the moment of inertia $I$ is $\frac{1}{2}M{R}^2$, where $M$ is the mass and $R$ is the radius of the disc. The acceleration $a$ when rolling down without slipping is reduced due to the rotational inertia:

$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}=\frac{g\sin \theta }{1+\frac{1}{2}}=\frac{2g\sin \theta }{3}$

The time to travel the same distance $l$ is:

$t_{\text{roll}}=\sqrt{\frac{2l}{a_{\text{roll}}}}=\sqrt{\frac{2l}{\frac{2g\sin \theta }{3}}}=\sqrt{\frac{3l}{g\sin \theta }}$

Compare Times:

Given in the problem is the relation:

$t_{\text{roll}}={\left(\frac{\alpha }{2}\right)}^{1/2}t$

From the derived formulas:

$\sqrt{\frac{3l}{g\sin \theta }}={\left(\frac{\alpha }{2}\right)}^{1/2}\sqrt{\frac{2l}{g\sin \theta }}$

Solving for $\alpha$:

$\sqrt{3}={\left(\frac{\alpha }{2}\right)}^{1/2}\sqrt{2}$

$3=\frac{\alpha }{2}\times 2$

$3=\alpha$

Conclusion:

Thus, $\alpha$ is 3.

Correct answer is 7

To solve this problem, we first note that for both the solid sphere and the hollow cylinder, the total mechanical energy is conserved as they roll up the inclined plane without slipping. The initial kinetic energy (comprised of both translational and rotational kinetic energy) is converted into potential energy at the maximum height.

Kinetic Energy for Each Body at the Start:

For the solid sphere, the moment of inertia $I$ is given by $I=\frac{2}{5}m{r}^2$, where $m$ is mass and $r$ is the radius of the sphere. The kinetic energy is the sum of translational kinetic energy $\left(\frac{1}{2}m{v}^2\right)$ and rotational kinetic energy $\left(\frac{1}{2}I{\omega }^2\right)$, where $\omega$ is the angular velocity. Since the sphere rolls without slipping, $v=r\omega$.

The total initial kinetic energy for the solid sphere is:

$K{E}_{\text{sphere}}=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{v}{r}\right)}^2=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2$

For the hollow cylinder, the moment of inertia $I$ is $m{r}^2$. Thus, its total kinetic energy is:

$K{E}_{\text{cylinder}}=\frac{1}{2}m{v}^2+\frac{1}{2}m{r}^2{\left(\frac{v}{r}\right)}^2=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=m{v}^2$

Potential Energy at Maximum Height:

For both bodies, the potential energy at the maximum height is given by $PE=mgh$, where $h$ is the height reached.

Applying Conservation of Energy:

For the solid sphere, the energy conservation equation is:

$\frac{7}{10}m{v}^2=mg{h}_1$

Solving for $h_1$ gives:

$h_1=\frac{7v^2}{10g}$

For the hollow cylinder, the conservation of energy gives:

$m{v}^2=mg{h}_2$

Thus, $h_2=\frac{v^2}{g}$.

Finding the Ratio $h_1:h_2$:

The ratio of $h_1$ to $h_2$ is:

$\frac{h_1}{h_2}=\frac{\frac{7v^2}{10g}}{\frac{v^2}{g}}=\frac{7}{10}$

Therefore, the value of $n$, which represents the numerator in the ratio $\frac{n}{10}$, is $7$. Thus, the ratio of maximum heights $h_1:h_2$ reached by the solid sphere and the hollow cylinder, respectively, is $\frac{7}{10}$.

Correct answer is 10

To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $\theta$, the component of gravitational acceleration along the plane is $g\sin \theta$. However, because the cylinder is rolling and not sliding, not all of this component accelerates the center of mass; some of it goes into causing rotational acceleration about the center of mass.

For a rolling cylinder, the moment of inertia $I$ is $I=\frac{1}{2}m{r}^2$, where $m$ is the mass of the cylinder and $r$ is the radius. The condition for rolling without slipping is that the linear acceleration $a$ of the center of mass is equal to the radius $r$ times the angular acceleration $\alpha$, i.e., $a=r\alpha$.

To find the linear acceleration $a$, we use the torque $\tau$ about the center of mass caused by the gravitational force down the incline. The torque due to gravity is $\tau =mg\sin \theta \cdot r$, and from Newton's second law for rotation, the angular acceleration is given by

$\alpha =\frac{\tau }{I}=\frac{mg\sin \theta \cdot r}{\frac{1}{2}m{r}^2}=\frac{2g\sin \theta }{r}$

Now using $a=r\alpha$:

$a=r\left(\frac{2g\sin \theta }{r}\right)=2g\sin \theta$

But we must account for the fact that only a portion of the gravitational acceleration goes into translating the cylinder down the plane due to the rolling condition. This is where we apply the concept of the ${\displaystyle }$rolling factor'' for a cylinder, which is $\frac{2}{3}$ for a solid cylinder, meaning $\frac{2}{3}$ of the gravitational component is used for translation.

The acceleration of the center of mass for the cylinder is therefore:

$a=\frac{2}{3}g\sin \theta$

Now we plug in the values of $\theta ={60}^{\circ }$ (which has $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$) and $g=10m/s^2$:

$a=\frac{2}{3}\cdot 10\cdot \frac{\sqrt{3}}{2}=\frac{10\sqrt{3}}{3}$

To match the given expression $\frac{x}{\sqrt{3}}m/s^2$, let's manipulate our expression for $a$:

$a=\frac{10\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{10\cdot 3}{3\cdot \sqrt{3}}=\frac{10}{\sqrt{3}}m/s^2$

Hence, the value of $x$ is $10$.

Correct answer is 7

In pure rolling work done by friction is zero. Hence potential energy is converted into kinetic energy. Since initially the ring and the sphere have same potential energy, finally they will have same kinetic energy too.

$\therefore$ Ratio of kinetic energies $=1$

$\Rightarrow \frac{7}{x}=1\Rightarrow x=7$