A false balance has equal arms. An object weights $w_{1}$ when placed in one pan and $w_{2}$ when placed in 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 A $\frac{w_{1} + w_{2}}{2}$
Given, weight of the object is recorded as $w_{1}$ and $w_{2}$ respectively and its true weight is $w$
Let the weight of the pans be x and y respectively
Balancing the object, pans and standard weights, we have
$(x + w) l = (y + w_{1}) l$
$∴ x + w = y + w_{1}$ ...........(1)
and $x + w_{2} = y + w$ ..................(2)
Subtracting eqn (2) from (1) , we get
$∴ w - w_{2} = w_{1} - w$
$\Rightarrow w = \frac{1}{2} (w_{1} + w_{2})$
Thus, the weight w of the object is $(\frac{w_{1} + w_{2}}{2})$

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