A false balance has equal arms. An object weights $w_{1}$ when placed in one pan and $w_{2}$ when placedin the other pan. Then weight w of the object is

\sqrt{w_{1} w_{2}}

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\frac{w_{1} + w_{2}}{2}

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(\frac{w_{1}^{2} + w_{2}^{2}}{2}) - 1

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\sqrt{w_{1}^{2} + w_{2}^{2}}

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Solution

The correct option is B $\frac{w_{1} + w_{2}}{2}$
we have a false balance which has equal arms so falseness is due to the weight of the pan and balance is based on the principle of torque.
let the first pan have x weight more than other so,
$x + w_{1} = W$ .........(1)
$W + x = w_{2}$ .......(2)
from (1) and (2)
$W = \frac{(w_{1} + w_{2})}{2}$

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A false balance has equal arms. An object weigh X when placed in one pan and Y when placed in other pan, then the weight W of the object is equal to

\frac{W_{2} - W_{1}}{W_{1}} \times 100

\frac{W_{1} - W_{2}}{W_{2}} \times 100

i

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W_{1} \neq W_{2}

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