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

The Correct Option is D

To solve this problem, we need to apply the principles of torque and equilibrium. The rod is in equilibrium, meaning the net torque around the pivot point (at the 40 cm mark) must be zero. Let's calculate the torques due to each mass and set them equal to solve for the unknown mass \(m\).

1. Torque due to the 2 kg mass:
   • This mass is suspended at the 20 cm mark. The distance from the pivot (at 40 cm) is 20 cm = 0.2 m.
   • The force due to this mass is \(F = mg = 2 \times 10 = 20\, N\).
   • The torque is \(\tau_1 = F \times \text{distance} = 20 \times 0.2 = 4\, Nm\).
2. Torque due to the rod's own weight:
   • The rod is uniform, so its weight acts at its center of mass, which is at the 100 cm mark (midpoint of the rod).
   • The rod's weight is \(500\, g = 0.5\, kg\), and hence the force \(F = 0.5 \times 10 = 5\, N\).
   • The distance from the pivot is 60 cm = 0.6 m.
   • The torque is \(\tau_2 = 5 \times 0.6 = 3\, Nm\).
3. Torque due to the unknown mass \(m\):
   • This mass is suspended at the 160 cm mark. The distance from the pivot (at 40 cm) is 120 cm = 1.2 m.
   • Let the torque due to this mass be \(\tau_3 = m \times 10 \times 1.2\).
4. Setting the net torque to zero for equilibrium:
   • The clockwise torques (due to the 2 kg mass and the rod's weight) must equal the counterclockwise torque (due to the unknown mass).
   • Therefore, \(\tau_3 = \tau_1 + \tau_2\).
   • Substituting the known values: \(m \times 10 \times 1.2 = 4 + 3 = 7\).

Solving for \(m\): \(m \times 10 \times 1.2 = 7 \\ m = \frac{7}{12} \\ m = \frac{1}{12}\, kg\)

Thus, the value of the unknown mass \(m\) that keeps the rod in equilibrium is \(\frac{1}{12}\, kg\).