A circular disc reaches from top to bottom of an inclined plane of length L. When it slips down the plane, it takes time t₁. When it rolls down the plane, it takes time t₂. The value of $\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}\text{ is }\sqrt{\frac{3}{\mathrm{x}}}$. The value of x will be …… .

According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as

When the disc slips down the inclined plane, it takes time t₁. Therefore, in this case its acceleration, ${\mathrm{a}}_1=\mathrm{gsin}\mathrm{\theta }$

$\because \mathrm{s}={\mathrm{ut}}_1+\frac{1}{2}{\mathrm{a}}_1{\mathrm{t}}_1^2$

$\Rightarrow \mathrm{s}=\frac{1}{2}\mathrm{gsin}{\mathrm{\theta t}}_1^2$              ...(i)

And when the disc rolls down the inclined plane, it takes time t₂.

Therefore in this case, its acceleration,

${\mathrm{a}}_2=\frac{\mathrm{gsin}\mathrm{\theta }}{1+\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}=\frac{\mathrm{gsin}\mathrm{\theta }}{1+\frac{1}{2}}=\frac{2}{3}\mathrm{gsin}\mathrm{\theta }\left[\because \text{ for disc, }\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}=\frac{1}{2}\right]$

$\because \mathrm{s}={\mathrm{ut}}_2+\frac{1}{2}{\mathrm{a}}_2{\mathrm{t}}_2^2=\frac{1}{2}\cdot \frac{2}{3}\mathrm{gsin}{\mathrm{\theta t}}_2^2$         ...(ii)

On dividing Eq. (i) by Eq. (ii), we get

$\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}=\sqrt{\frac{3}{2}}$         ...(iii)

According to question, value of $\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}\text{ is }\sqrt{\frac{3}{\mathrm{x}}}$.

Comparing it with Eq. (iii), we get

x = 2