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Q.

Two blocks each of the mass m are attached to the ends of a massless rod which pivots as shown in the figure. Initially, the rod is held in the horizontal position and then released. Calculate the net torque on this system above pivot.

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a

ml2g−ml1gk^

b

ml1g−ml2gk^

c

ml1g+ml2gk^

d

−ml1g+ml2gk^

answer is B.

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Detailed Solution

Torque due to the weight of Left m:τ→1=mgl1k^(∴ this is anti-clock wise )Torque due to weight of right m:τ→2=−mgl2k^(∴ this is clockwise)  So net torque: τ→=τ→1+τ→2=mgl1−mgl2k^
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