A body weighing $8 g$ when placed in one pan and $18 g$ when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

13 g

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9 g

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22 g

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12 g

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Solution

The correct option is D $12 g$
Let, The left pan has weight $m_{1}$ and distance $l_{1}$ from the pivot.
Let, The left pan has weight $m_{2}$ and distance $l_{2}$ from pivot.
Let $X$ be the unknown weight of the body,
Balancing the torques:
$m_{1} \times l_{1} = m_{2} \times l_{2}$ (given that beam is horizontal when pan is empty)
So $\frac{l_{1}}{l_{2}} = \frac{m_{2}}{m_{1}}$
$(X + m_{1}) l_{1} = (8 + m_{2}) l_{2} . . . . . . . . . . . . . . . . . . . . . (1)$
$(18 + m_{1}) l_{1} = (X + m_{2}) l_{2} . . . . . . . . . . . . . . . . . . . . . (2)$
Now, By subsitution of $m_{1} l_{1} = m_{2} . l_{2}$ in both equation,
$X l_{1} = 8 l_{2} . . . . . . . . . . . . . . . . . . (3)$
$18 l_{1} = X l_{2} . . . . . . . . . . . . . . . . . . . (4)$
From equation (3) and (4),
$\frac{X}{18} = \frac{8}{X}$
$X^{2} = 18 \times 8$
$X = 144$
$X = 12 g$
So, The true weight of the body is $12 g$ .

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