A physical balance has its arms of unequal le | vedclass

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(B) Let the true weight of the body be $W$ and the lengths of the two arms be $L_1$ and $L_2$.For the first case,the body is in one pan,so the torque

Vedclass · Accepted Answer

$12$

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(B) Let the true weight of the body be $W$ and the lengths of the two arms be $L_1$ and $L_2$.For the first case,the body is in one pan,so the torque balance equation is $W 	imes L_1 = 18 	imes L_2$,which gives $\frac{L_1}{L_2} = \frac{18}{W}$.For the second case,the body is in the other pan,so the torque balance equation is $W 	imes L_2 = 8 	imes L_1$,which gives $\frac{L_1}{L_2} = \frac{W}{8}$.Equating the two expressions for $\frac{L_1}{L_2}$,we get $\frac{18}{W} = \frac{W}{8}$.This simplifies to $W^2 = 18 	imes 8 = 144$.Therefore,the true weight $W = \sqrt{144} = 12 \,kg$.

$A$ physical balance has its arms of unequal length. $A$ body weighs $18 \,kg$ if kept in one pan and weighs $8 \,kg$ if kept in the other pan. The true weight of the body is ............. $kg$

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