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Two masses m and $m\over2$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is $\tau=k\theta$ for angular displacement 0. If the rod is rotated by θ0 and released, the tension in it when it passes through its mean position will be: 

(1) $3k\theta_0^2\over l$

(2) $k\theta_0^2\over2l$

(3) $2k\theta_0^2\over l$

(4) $k\theta_0^2\over l$

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Ans: (4) $k\theta_0^2\over l$

Sol: $\omega=\sqrt{k\over I}$

$\omega=\sqrt{3k\over ml^2}$

$\Omega = \omega\theta_0 =$ average velocity

$T = m\Omega^2r_1$

$T = m\Omega^2{l\over3}$

$=m\omega^2\theta_0^2{l\over3}$

$=m{3k\over ml^2}\theta_0^2{l\over3}$

$={k\theta_0^2\over l}$

$I=\mu l^2={{m^2\over2}\over{3m\over2}}l^2$

$={ml^2\over3}$

${r_1\over r_2}={1\over2}\implies r_1={l\over3}$

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