A solid sphere and a disc of same mass and radius starts rolling down a rough inclined plane, from the same height the ratio of the time taken in the two cases is

The correct option is √14:√15.

To find the ratio of the time taken by a solid sphere and a disc to roll down a rough inclined plane, we can analyze their acceleration during pure rolling.

The acceleration of a body rolling down an inclined plane of inclination angle $\theta$ without slipping is given by the formula:

$a=\frac{gsin\theta }{1+\frac{I}{M{R}^2}}$

where:
- $g$ is the acceleration due to gravity,
- $I$ is the moment of inertia of the rolling body,
- $M$ is the mass of the body, and
- $R$ is the radius of the body.

Step 1: Find the acceleration of the solid sphere ($a_1$)

The moment of inertia of a solid sphere is $I_1=\frac{2}{5}M{R}^2$.

Substituting this value into the acceleration formula:

$a_1=\frac{gsin\theta }{1+\frac{2}{5}}=\frac{gsin\theta }{\frac{7}{5}}=\frac{5}{7}gsin\theta$

Step 2: Find the acceleration of the disc ($a_2$)

The moment of inertia of a disc is $I_2=\frac{1}{2}M{R}^2$.

Substituting this value into the acceleration formula:

$a_2=\frac{gsin\theta }{1+\frac{1}{2}}=\frac{gsin\theta }{\frac{3}{2}}=\frac{2}{3}gsin\theta$

Step 3: Relate the time taken to the acceleration

Since both bodies start from rest and travel the same distance $s$ down the inclined plane, we use the equation of motion:

$s=ut+\frac{1}{2}a{t}^2$

With initial velocity $u=0$, this simplifies to:

$s=\frac{1}{2}a{t}^2\Rightarrow t=\sqrt{\frac{2s}{a}}$

Since the distance $s$ is the same for both cases, the time taken is inversely proportional to the square root of the acceleration:

$t\propto \frac{1}{\sqrt{a}}$

Step 4: Calculate the ratio of the times taken ($t_1:t_2$)

Let $t_1$ be the time taken by the solid sphere and $t_2$ be the time taken by the disc:

$\frac{t_1}{t_2}=\sqrt{\frac{a_2}{a_1}}$

Substituting the expressions for $a_1$ and $a_2$ calculated above:

$\frac{t_1}{t_2}=\sqrt{\frac{\frac{2}{3}gsin\theta }{\frac{5}{7}gsin\theta }}=\sqrt{\frac{2}{3}\times \frac{7}{5}}=\sqrt{\frac{14}{15}}=\frac{\sqrt{14}}{\sqrt{15}}$

Thus, the ratio of the time taken is $\sqrt{14}:\sqrt{15}$.