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Asked in NEET 2021 · Static equilibrium
Given: a uniform rod of length 200 cm and mass 0.5 kg pivoted at the 40 cm mark, with 2 kg hung at 20 cm and m at 160 cm.
Idea: in equilibrium the torques about the wedge must cancel; a uniform rod acts as though all its weight sits at its midpoint.
The rod's centre of mass is at the 100 cm mark, which is 60 cm to the right of the pivot.
Anticlockwise, from the 2 kg at 20 cm (0.20 m to the left): 2×10×0.20=4 N m.
Clockwise, from the rod: 0.5×10×0.60=3 N m; from m at 160 cm (1.20 m to the right): m×10×1.20=12m.
4=3+12m, so 12m=1 and m=1/(12) kg.
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