Question

A meter stick is balanced on a knife edge at its centre. When two coins, each of mass $5g$ are put one on top of the other at the $12.0cm$ mark, the stick is found to be balanced at $45.0cm$ . The mass of the meter stick is

• A. $13g$
• B. $33g$
• C. $66g$
• D. $77g$

Answer 1

Answer: (C)

Let $w$ and $w^{\prime }$ be the respective weight on the meter stick and the coin

The mass of meter stick is connected at its mid-point, i.e. at the $50cm$ mark.
Mass of the meter stick $=m$
Mass of each coin, $m=5g$
When the coins are placed $12cm$ away from the end $P$, the centre of mass gets shifted by $5cm$ from $R$ towards the end $P$. The centre of mass is located at a distance of $45cm$ from point $P$.
The net torque will be conserved for rotational equilibrium about point $R$
$10+g(45-12)-m^{\prime }g(50-45)=0$
$\therefore m^{\prime }=\frac{10\times 33}{5}=66g$
Hence, the mass of meter stick is $66g$.