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 $m$ 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.

$(g=10{\text{m/s}}^2)$

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

Answer 1

The correct option is A $\frac{1}{12}\text{kg}$


On balancing the clockwise and anticlockwise torques, we get,

$mg(160-40)+W(100-40)=2g\left(20\right)$

$\Rightarrow mg\left(120\right)+0.5g\left(60\right)=40g$

$\Rightarrow m=\frac{1}{12}\text{kg}$

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