A uniform rod AB having mass 2m and length $2 l$ is kept on a smooth horizontal plane. This rod is orthogonally connected to a horizontal taut string whose other end has a connected particle of mass m. If particle is given initial horizontal velocity $v_{0}$ , perpendicular to string, then just after giving this velocity, the tension in the string is $\frac{m v_{0}^{2}}{n l}$ . Then the value of $n$ is

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answer is C.

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Detailed Solution

$a = \frac{T}{2 m}; T l = \frac{2 m {(2 l)}^{2}}{12} α$

Acceleration of point
$A = a + l α = \frac{T}{2 m} + \frac{3 T}{2 m} = \frac{2 T}{m}$
w.r.t A particle M moves in circle
$∴$ from frame of A $T + m \times \frac{2 T}{m} = \frac{m v_{0}^{2}}{l}$
$∴ T = \frac{m v_{0}^{2}}{3 l}; α = \frac{v_{0}^{2}}{6 l}; α = \frac{v_{0}^{2}}{2 l^{2}}$

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