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Topic: Equilibrium of rigid bodies, rigid body rotation, equations of rotational motion· Rotational Motion

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…

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. (g = 10 m/s\(^2\))

rod on wedge at 40 cm with 2 kg at 20 cm and m at 160 cm

1. \(\frac{1}{2}\) kg

2. \(\frac{1}{3}\) kg

3. \(\frac{1}{6}\) kg

4. \(\frac{1}{12}\) kg

1 answer

Best answer

Answer: 4. \(\frac{1}{12}\) kg

Key idea: take torques about the wedge (40 cm mark). The rod's weight acts at its centre (100 cm).

  • Anticlockwise (left): \(2\times10\times20=400\) N cm.
  • Clockwise (right): rod \(0.5\times10\times60=300\) N cm, and mass m: \(m\times10\times120=1200\,m\) N cm.

\(400=300+1200m\Rightarrow m=\dfrac{100}{1200}=\dfrac1{12}\) kg. Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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