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A ring and a disc are initially at rest,...

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 `(2 - sqrt(3)) //sqrt(10s)` 2– 3 / 10s, then the height of the top of the inclined plane, in metres, is _____________ . Take `g = 10ms^(-2)`

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To solve the problem step by step, we will analyze the motion of both the ring and the disc rolling down the inclined plane. ### Step 1: Identify the parameters and equations We have: - Angle of inclination, \( \theta = 60^\circ \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) - Time difference between the ring and disc reaching the ground, \( \Delta t = \frac{2 - \sqrt{3}}{\sqrt{10}} \, \text{s} \) ...
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