Question

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

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

Answer 1

Answer: (C)

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