A uniform rod of length $200$ cm and mass $500$ g is balanced on a wedge placed at $40$ cm mark. A mass of $2$ kg is suspended from the rod at $20$ cm and another unknown mass 'm' is suspended from the rod at $160$ cm mark as shown in the figure. Find the value of 'm' such that the rod is in equilibrium. ($g = 10\ \text{m/s}^2$)
Answer: (A) $\dfrac{1}{12}$ kg
Take torques about the wedge at the $40$ cm mark.
Anticlockwise: the $2$ kg mass at $20$ cm, lever arm $20$ cm: $2g \times 20$.
Clockwise: the rod's weight at its centre ($100$ cm), lever arm $60$ cm: $0.5g \times 60$. And $m$ at $160$ cm, lever arm $120$ cm: $mg \times 120$.
$$2 \times 20 = 0.5 \times 60 + 120m \;\Rightarrow\; 40 = 30 + 120m \;\Rightarrow\; m = \frac{1}{12}\ \text{kg}$$
Solution by Sreeraj P, M.Sc Physics