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 true weight $W$, of the object is

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

Answer 1

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

The beam balance is giving two different readings of weight of the same object, as the object is suspended by different arms. So, the balance is certainly faulty and has different arm lengths.


When $W$ is in pan $\left(1\right)$
$W{l}_1=W_1{l}_2-\left(i\right)$

in pan $\left(2\right)$
$W{l}_2=W_2{l}_1-\left(ii\right)$

On multiplying $\left(i\right)$ & $\left(ii\right)$,

$W^2{l}_1{l}_2=W_1{W}_2{l}_1{l}_2$

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

Hence, $\left(A\right)$ is the correct answer.