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(C) For a uniform meter scale,the weight of the scale acts at its center of gravity,which is at the $50 \ cm$ mark. Let the weight of the scale be $W

Vedclass · Accepted Answer

$3.5$

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(C) For a uniform meter scale,the weight of the scale acts at its center of gravity,which is at the $50 \ cm$ mark. Let the weight of the scale be $W = mg \ N$.Taking moments about the pivot point at the $30 \ cm$ mark:The clockwise moment is due to the weight of the scale $(W)$ at $50 \ cm$ and the $10 \ N$ weight at $80 \ cm$.Clockwise moment = $W 	imes (50 - 30) + 10 	imes (80 - 30) = W 	imes 20 + 10 	imes 50 = 20W + 500$.The counter-clockwise moment is due to the $60 \ N$ weight at $10 \ cm$.Counter-clockwise moment = $60 	imes (30 - 10) = 60 	imes 20 = 1200$.For rotational equilibrium,clockwise moment = counter-clockwise moment:$20W + 500 = 1200$$20W = 700$$W = 35 \ N$.Since $W = mg$,taking $g = 10 \ m/s^2$,we have $m = 35 / 10 = 3.5 \ kg$.

$A$ uniform meter scale is balanced at the $30 \ cm$ mark on a knife edge,when weights of $60 \ N$ and $10 \ N$ are suspended at the $10 \ cm$ and $80 \ cm$ marks,respectively. The mass of the meter scale is: (in $kg$)

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