Question

The beam and pans of a balance have negligible mass. An object weighs $W_1$ when placed in one pan and $W_2$ when placed in the other pan. The weight $W$ of the object is :

• A. $\sqrt{W_1{W}_2}$
• B. $\sqrt{(W_1+W_2)}$
• C. $W_1^2+W_2^2$
• D. $(W_1^{-1}+W_2^{-1})/2$

Answer 1

The correct option is A $\sqrt{W_1{W}_2}$

As pan and beam are weightless, error must be in length of beams.

So, both pans are at different distance from support.

Let beam on left be $l_1$ and beam on right be $l_2$.

Suppose we put weight $W$ in pan 1, then weight reads $W_1$, that means, weight in pan 2 is $W_1$.

So, $W{l}_1=W_1{l}_2$-------(1)

Now when weight $W$ is kept in pan 2, it reads $W_2$, so weight in pan 1 now is $W_2$.

So, $W_2{l}_1=W{l}_2$--------(2)

Dividing,

We get,

$\frac{W}{W_2}=\frac{W_1}{W}$

$W^2=W_1{W}_2$

$W=\sqrt{W_1{W}_2}$

So, option A is correct.