Ncert Exercises & Solutions

Question 7. 17. A meter stick is balanced on a knife-edge at its center. When two coins, each of mass 5g 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. What is the mass of the meter stick?

Let W and W′ be the respective weights of the meter stick and the coin.

The mass of the meter 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 = 5g

When the coins are placed 12 cm away from the end P, the center of mass gets shifted by 5 cm from point R toward the end P. The center of mass is located at a distance of 45 cm from point P.

The net torque about point R,

\begin{array}{l} 10 \times g (45 - 12) - m^{'} g (50 - 45) = 0 \\ \Rightarrow m^{'} = \frac{10 \times 33}{5} = 66 g \end{array}

Hence, the mass of the meter stick is 66g.

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