Question

A circular disc reaches from top to bottom of an inclined plane of length $L$. When it slips down the plane, it takes time $t_1$. When it rolls down the plane, it takes time $t_2$. The value of $\frac{t_2}{t_1}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ will be ______.

• A. 2.00
• B. 2.0
• C. 2

Answer 1

Answer: (A), (B), (C)

If the disc slips on the inclined plane, then its acceleration,

$a_1=g\sin \theta$

So,

$L=\frac{1}{2}{a}_1{t}_1^2$

$\Rightarrow t_1=\sqrt{\frac{2L}{a_1}}....\left(i\right)$

If the disc rolls on the inclined plane, its acceleration,

$a_2=\frac{g\sin \theta }{1+\frac{I}{m{R}^2}}$

$\Rightarrow a_2=\frac{g\sin \theta }{1+\frac{m{R}^2}{2m{R}^2}}$

$\Rightarrow a_2=\frac{2}{3}g\sin \theta$

Now,

$L=\frac{1}{2}{a}_2{t}_2^2$

$\Rightarrow t_2=\sqrt{\frac{2L}{a_2}}$

So,

$\frac{t_2}{t_1}=\sqrt{\frac{a_1}{a_2}}=\sqrt{\frac{g\sin \theta }{\frac{2}{3}g\sin \theta }}=\sqrt{\frac{3}{2}}=\sqrt{\frac{3}{x}}$

$\therefore x=2$