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(C) The meter scale is supported at its center of gravity,which is the $50 	ext{ cm}$ mark.For the scale to be in equilibrium,the clockwise torque mu

Vedclass · Accepted Answer

$20$

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(C) The meter scale is supported at its center of gravity,which is the $50 	ext{ cm}$ mark.For the scale to be in equilibrium,the clockwise torque must equal the counter-clockwise torque about the pivot point ($50 	ext{ cm}$ mark).The distance of the first weight '$w$' from the pivot is $d_1 = |50 	ext{ cm} - 20 	ext{ cm}| = 30 	ext{ cm}$.The distance of the second weight $(25 	ext{ g})$ from the pivot is $d_2 = |74 	ext{ cm} - 50 	ext{ cm}| = 24 	ext{ cm}$.Applying the principle of moments: $w 	imes d_1 = 25 	ext{ g} 	imes d_2$.$w 	imes 30 	ext{ cm} = 25 	ext{ g} 	imes 24 	ext{ cm}$.$w = (25 	imes 24) / 30 = 600 / 30 = 20 	ext{ g}$.Thus,the weight of the body is $20 	ext{ g}$.

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