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 .... .

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Answer & Solution

Correct Answer

(D) \(2\)

Detailed explanation

If disk slips on inclined plane, then it's aceleration

\(a_{1}=g \sin \theta\)

\(L=\frac{1}{2} a_{1} t_{1}^{2}\)

\(\Rightarrow t_{1}=\sqrt{\frac{2 L}{a_{1}}}\ldots \ldots({i} )\)

If disk rols on inclined plane, its acceleration,

\(a_{2} =\frac{g \sin \theta}{1+\frac{I}{m R^{2}}}\)

\(a_{2} =\frac{g \sin \theta}{1+\frac{m R^{2}}{2 m R^{2}}}\)

\(a_{2} =\frac{2}{3} g \sin \theta\)

Now \({L}=\frac{1}{2} {a}_{2} \cdot {t}_{2}^{2}\)

\(\Rightarrow {t}_{2}=\sqrt{\frac{2 {L}}{{a}_{2}}} \quad \ldots \ldots({ii} )\)

Now \(\frac{t_{2}}{t_{1}}=\sqrt{\frac{a_{1}}{a_{2}}}=\sqrt{\frac{3}{2}}\)

\(\Rightarrow {x}=2\)

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