Question 257 of 280

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 $\dfrac{{t}_{2}}{{t}_{1}}$ is $\sqrt{\dfrac{3}{x}}$. The value of $x$ will be

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Answer: 2

If disk slips on inclined plane, then its acceleration, ${a}_{1}=g\sin \theta$
$L=\dfrac{1}{2}{a}_{1}{t}_{1}^{2}$
$\Rightarrow {t}_{1}=\sqrt{\dfrac{2L}{{a}_{1}}}\ldots \left(i\right)$
If disk rolls on inclined plane, its acceleration,
${a}_{2}=\dfrac{g\sin \theta }{1+\dfrac{I}{m{R}^{2}}}$
${a}_{2}=\dfrac{g\sin \theta }{1+\dfrac{m{R}^{2}}{2m{R}^{2}}}$
${a}_{2}=\dfrac{2}{3}g\sin \theta$
Now $L=\dfrac{1}{2}{a}_{2}\cdot {t}_{2}^{2}$
$\Rightarrow {t}_{2}=\sqrt{\dfrac{2L}{{a}_{2}}}\ldots \left(ii\right)$
Now, $\dfrac{{t}_{2}}{{t}_{1}}=\sqrt{\dfrac{{a}_{1}}{{a}_{2}}}=\sqrt{\dfrac{3}{2}}$
$\Rightarrow x=2$