Question

A uniform rod of length $200\text{cm}$ and mass $500\text{g}$ is balanced on a wedge placed at $40\text{cm}$ mark. A mass of $2\text{Kg}$ is suspended from the rod at $20\text{cm}$ and another unknown mass ${}^{\prime }{m}^{\prime }$ is suspended from the rod at $160\text{cm}$ mark as shown in the figure. Find the value of $m$ such that the rod is in equilibrium.
$(g=10{\text{ms}}^{-2})$

• A. $\frac{1}{2}\text{Kg}$
• B. $\frac{1}{3}\text{Kg}$
• C. $\frac{1}{6}\text{Kg}$
• D. $\frac{1}{12}\text{Kg}$

Answer 1

The correct option is D $\frac{1}{12}\text{Kg}$

All the forces acting on the rod are as shown in the figure below:


For the rod to remain in equilibrium,

${\tau }_{net}=0$

Calculating the net torque about the wedge, we have:

$2g\times 20=0.5g\times 60+mg\times 120$

$m=\frac{0.5}{6}\text{Kg}=\frac{1}{12}\text{Kg}$

Hence, option $\left(D\right)$ is the correct answer.