JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 5)
A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is $8 \mathrm{~m} / \mathrm{s}$. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be :
$4 \sqrt{2} \mathrm{~m} / \mathrm{s}$
$8 \mathrm{~m} / \mathrm{s}$
$4 \mathrm{~m} / \mathrm{s}$
$8 \sqrt{2} \mathrm{~m} / \mathrm{s}$
Explanation
_4th_April_Evening_Shift_en_5_1.png)
If $V_B=2 V$
Point A is instantaneous center of rotation
Given $V_B=8 \mathrm{~m} / \mathrm{s}$
$$\begin{aligned} & \mathrm{V}=4 \mathrm{~m} / \mathrm{s} \\ & \mathrm{~V}_{\mathrm{P}}=\sqrt{2} \mathrm{v} \Rightarrow V_{\mathrm{p}}=4 \sqrt{2} \mathrm{~m} / \mathrm{s} \\ & \operatorname{correct}(1) \end{aligned}$$
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