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 m/s²)

\(\frac{1}{12}\) kg

Torque due to mass 2 kg : \(\tau_2\) = r x F = 4 N m

Torque due to mass of rod : \(\tau_2\) = r x F = 3 N m

Torque due to mass m kg : \(\tau_2\) = r x F = 12m N m

For equilibrium, net torque = 0

4 N m - 3 N m - 12m N m = 0

m = \(\frac{1}{12}\) kg