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Q.

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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a

940mg

b

1940mg

c

mg2

d

920mg

answer is A.

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Detailed Solution

Torque =τ=910mg920L=Iα=m3910L2αα=3g2L Acceleration, aCM=α(AC) aCM=3g2L9L20=27g40 Now,  910mg−NA=maCM=m⋅27g40 or  NA=940mg
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