A metre 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. What is the mass of the metre stick?
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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 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 point R toward 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. $10\times\text{g}(45-12)-\text{m}'\text{g}(50-45)=0$ $\therefore\ \text{m}'=\frac{10\times33}{5}=66\text{g}$ Hence, the mass of the metre stick is 66g.
The mass of the metre stick is concentrated 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 point R toward 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. $10\times\text{g}(45-12)-\text{m}'\text{g}(50-45)=0$ $\therefore\ \text{m}'=\frac{10\times33}{5}=66\text{g}$ Hence, the mass of the metre stick is 66g.
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