Question

$A$ 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. $\left( g =10\, m / s ^{2}\right)$

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

Answer 1

Answer: (D)

From principle of moments
$2 \times 20=0.5 \times 60+ m \times 120$
$2=1.5+6 \,m $
$0.5=6 \,m$
$m =\frac{1}{12} \,kg$