A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their

Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60 °$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $\frac{2 - \sqrt{3}}{\sqrt{10}}$ s, then the height of the top of the inclined plane, in metres, is _______. Take $g = 10 {ms}^{- 2}$ . [2018]

Smart Achievers · Accepted Answer

(0.75)

The time taken to reach the ground or descend

$\frac{2 - \sqrt{3}}{\sqrt{10}}$

$For ring$ $t_{1} = \frac{1}{\sin 60 °} \sqrt{\frac{2 h}{g} (1 + 1)} = \frac{4}{\sqrt{3}} \sqrt{\frac{h}{g}}$ $∵$ $\frac{K^{2}}{R^{2}} = 1$

$For disc$ $t_{2} = \frac{1}{\sin 60 °} \sqrt{\frac{2 h}{g} (1 + \frac{1}{2})} = \sqrt{\frac{4 h}{g}}$ $∵$ $\frac{K^{2}}{R^{2}} = \frac{1}{2}$

$Given t_{1} - t_{2} = \frac{2 - \sqrt{3}}{\sqrt{10}}$

$∴$ $\frac{4}{\sqrt{3}} \sqrt{\frac{h}{g}} - \sqrt{\frac{4 h}{g}} = \frac{2 - \sqrt{3}}{\sqrt{10}}$ $\Rightarrow 2 \sqrt{h} - \sqrt{3} \sqrt{h} = \sqrt{3} - \frac{3}{2}$

$\Rightarrow \sqrt{h} (2 - 1.732) = 1.732 - 1.5 \Rightarrow \sqrt{h} = \frac{0.232}{0.268}$

$∴$ $h \approx 0.75 m (Height of the top of the inclined plane)$

Solution

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