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A horizontal rod of mass $' M'$ and length $' L'$ is tied to two vertical string symmetrically as shown in the figure. One of the strings at end $Q$ is cut at $t = 0$ and the rod starts rotating about the other end $P .$ Then

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At $t = 0,$ angular acceleration of rod about $P$ is $3 g / 2 L .$

At $t = 0,$ angular acceleration of rod about $C . M .$ of rod is $3 g / 2 L .$

At $t = 0,$ acceleration of $C . M .$ of rod is $3 g / 4$ in downward direction.

At $t = 0,$ tension in the string at $P$ is $M g / 4 .$

answer is A, B, C, D.

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Detailed Solution

$M g (\frac{L}{2}) = (\frac{M L^{2}}{3}) α$

Therefore $α = \frac{3 g}{2 L}$

$a_{c m} = \frac{α L}{2} = \frac{3 g}{4}$

$- T + M g = M (\frac{3 g}{4})$

Therefore $T = \frac{M g}{4}$

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