A uniform rod of mass 250 g having length 10 | vedclass

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(B) For the rod to be in rotational equilibrium,the net torque about the pivot point (at $40 \ cm$ mark) must be zero.Let $m$ be the mass to be suspen

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$190$

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(B) For the rod to be in rotational equilibrium,the net torque about the pivot point (at $40 \ cm$ mark) must be zero.Let $m$ be the mass to be suspended at the $90 \ cm$ mark.The weight of the rod acts at its center of mass,which is at the $50 \ cm$ mark.The distance of the $400 \ g$ mass from the pivot is $40 \ cm - 10 \ cm = 30 \ cm$.The distance of the rod's center of mass from the pivot is $50 \ cm - 40 \ cm = 10 \ cm$.The distance of the unknown mass $m$ from the pivot is $90 \ cm - 40 \ cm = 50 \ cm$.Taking the pivot as the reference point,the torque balance equation is:$(400 \ g 	imes 30 \ cm) = (250 \ g 	imes 10 \ cm) + (m 	imes 50 \ cm)$$12000 = 2500 + 50m$$50m = 12000 - 2500$$50m = 9500$$m = \frac{9500}{50} = 190 \ g$

$A$ uniform rod of mass $250 \ g$ having length $100 \ cm$ is balanced on a sharp edge at $40 \ cm$ mark. $A$ mass of $400 \ g$ is suspended at $10 \ cm$ mark. To maintain the balance of the rod,the mass to be suspended at $90 \ cm$ mark is: (in $g$)

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