Physics - Practical physics Question with Solution | TestHub

The meter scale is supported at its centre of gravity, which is at the $50\text{ cm}$ mark.

The distance of the weight $w$ from the pivot is $50-20=30\text{ cm}$.

The distance of the $25\text{ g}$ weight from the pivot is $74-50=24\text{ cm}$.

For the meter scale to remain perfectly horizontal, the net torque about the pivot must be zero. Equating the anticlockwise torque to the clockwise torque:

$w\times 30=25\times 24$

$w=\frac{25\times 24}{30}$

$w=20\text{ gram wt.}$

Answer: 20 gram wt.