A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is $8…
- $4 \sqrt{2} \mathrm{~m} / \mathrm{s}$
- $8 \mathrm{~m} / \mathrm{s}$
- $4 \mathrm{~m} / \mathrm{s}$
- $8 \sqrt{2} \mathrm{~m} / \mathrm{s}$
Solution

If $V_B=2 V$
Point A is instantaneous center of rotation
Given $V_B=8 \mathrm{~m} / \mathrm{s}$
$\begin{aligned}
& \mathrm{V}=4 \mathrm{~m} / \mathrm{s} \\ & \mathrm{~V}_{\mathrm{p}}=\sqrt{2} \mathrm{v} \Rightarrow \mathrm{~V}_{\mathrm{p}}=4 \sqrt{2} \mathrm{~m} / \mathrm{s} \\ & \operatorname{correct}(1)
\end{aligned}$
Asked in: JEE Main 2025 (04 Apr Shift 2)