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 ' $t_1{}^{\prime }$. 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_______.

Answer 1

Answer: 2

If disk slips on inclined plane, then it's acceleration
$a_1=g\sin \theta$
$L=\frac{1}{2}{a}_1{t}_1{}^2$
$\Rightarrow t_1=\sqrt{\frac{2L}{a_1}}\dots$(i)
If disk rolls on inclined plane, its acceleration,
$a_2=\frac{g\sin \theta }{1+\frac{I}{m{R}^2}}$
$a_2=\frac{g\sin \theta }{1+\frac{m{R}^2}{2m{R}^2}}$
$a_2=\frac{2}{3}g\sin \theta$
Now $L=\frac{1}{2}{a}_2\cdot t_2{}^2$
$\Rightarrow t_2=\sqrt{\frac{2L}{a_2}}\dots$ (ii)
Now $\frac{t_2}{t_1}=\sqrt{\frac{a_1}{a_2}}=\sqrt{\frac{3}{2}}$
$\Rightarrow x=2$