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

• A

  $\frac{1}{2} \mathrm{kg}$
• B

  $\frac{1}{3} \mathrm{kg}$
• Correct answer
  C

  $\frac{1}{12} \mathrm{kg}$
• D

  $\frac{1}{6} \mathrm{kg}$

Hence, the correct answer is option: 'C'.