Consider a uniform rod of mass M = 4 m and length l pivoted about its centre. A mass m moving with velocity…

1. $\frac{3}{7\sqrt{2}}\frac{v}{l}$
2. $\frac{3}{7}\frac{v}{l}$
3. $\frac{3\sqrt{2}}{7}\frac{v}{l}$
4. $\frac{4}{7}\frac{v}{l}$

Solution

$L_{oi}=L_{of}$
$\frac{mV}{\sqrt{2}}\times \frac{L}{2}=\left[\frac{4m{L}^2}{12}+\frac{m{L}^2}{4}\right]\times \omega$
$\omega =\frac{6V}{7\sqrt{2}L}=\frac{3\sqrt{2}V}{7L}$

Asked in: JEE Main 2020 (08 Jan Shift 1)

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