JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 27)
The centre of a wheel rolling on a plane surface moves with a speed v0. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $$\sqrt x {v_0}$$. Then the value of x is _____________.
Answer
02
Explanation
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$$\left| \omega \right| = {{{v_0}} \over R}$$
$${\overrightarrow v _p} = {v_0}\widehat i + \omega R( - \widehat j) = {v_0}\widehat i - {v_0}\widehat j$$
$$\left| {{{\overrightarrow v }_p}} \right| = \sqrt 2 {v_0}$$
$$x = 02$$
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