Let W and W′ be the respective weights of the metre stick and the coin.
The mass of the metre stick is concentrated 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 point R toward 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
∴
m' = 66
g
Hence, the mass of the metre stick is 66 g.
