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JEE MATHEMATICS MCQ: DCS AND DRS

1 Easy
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Application +4 | -1
A uniform rod, 4 m4 {~m} in length and weight 2 kg2 {~kg} turns freely about a point distant one metre from one end and from that end a weight of 10 kg10 {~kg} is suspended. The weight that must be placed at the other end to produce equilibrium is
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Correct option: A
223 kg2 \frac{2}{3} {~kg}
Option A (223 kg2 \frac{2}{3} {~kg}) is correct because it is the weight required to balance the moments about the pivot point, considering the weight of the rod and the suspended weight.

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