JEE Mains · Physics · STD 11 - 3.2 motion in plane
A circular table is rotating with an angular velocity of \(\omega\) \(\mathrm{rad} / \mathrm{s}\) about its axis (see figure). There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of \(1 \mathrm{~m}\) on the groove. All the surface are smooth. If the radius of the table is \(3 \mathrm{~m}\), the radial velocity of the ball w.r.t. the table at the time ball leaves the table is \(x \sqrt{2} \omega\) \(\mathrm{m} / \mathrm{s}\), where the value of \(x\) is _______.
- A \(1\)
- B \(2\)
- C \(5\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(a_c=\omega^2 x\) \(\frac{v d v}{d x}=\omega^2 x\) \(\int_0^v v d v=\int_1^3 \omega^2 x d x\) \(\frac{v^2}{2}=\omega^2\left[\frac{x^2}{2}\right]\) \(\frac{v^2}{2}=\frac{\omega^2}{2}\left[3^2-1^2\right]\) \(v=2 \sqrt{2} \omega\) \(x=2\)
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