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  • A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it takes time . When

A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it takes time { }^{\prime} 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 _______.
 

Answers (1)

best_answer

Case (I) when disc slips down

s=u t+\frac{1}{2} a t^{2}

L=0+\frac{1}{2} (g \sin \theta) t_{1}^{2}

t_{1}=\sqrt{\frac{2 L}{g \sin \theta}} \rightarrow(1)

Case (II): When the disc rolls down

s=u t+\frac{1}{2} a t^{2}

a=\frac{g \sin \theta}{1+\frac{k^{2}}{R^{2}}} \quad \text { for } \operatorname{disc} ; k^{2}=\frac{R}{2}

\begin{aligned} &a=\frac{2}{3} g \sin \theta \\ &L=0+\frac{1}{2} \times \frac{2}{3} g \sin \theta \times t_{2}^{2} \\ &t_{2}=\sqrt{\frac{3 L}{g \sin \theta}} \rightarrow (2)\end{aligned}

 

\therefore x=2

Posted by

vishal kumar

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