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A uniform rod of length $L$ (in between the supports) and mass $m$ is placed on two supports $A$ and $B$ . The rod breaks suddenly at length $L / 10$ from the support $B$ . Find the reaction at support A immediately after the rod breaks.

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$\frac{9}{40} m g$

$\frac{19}{40} m g$

$\frac{m g}{2}$

$\frac{9}{20} m g$

answer is A.

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Detailed Solution

Torque $= τ = \frac{9}{10} m g = \frac{9}{20} L$

$= I α = \frac{m}{3} = \frac{9}{10} L \cdot \ {not}^{2} α; a = \frac{3 g}{2 L}$

Acceleration

$a_{C M} = α (A C) = \frac{3 g}{2 L} = \frac{9 L}{20} = \frac{27 g}{40} N_{A} = \frac{9}{40} m g$

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