NEET · Physics · STD 11 - 6. system of particles and rotational motion
A uniform rod of length \(200 \,\mathrm{~cm}\) and mass \(500 \,\mathrm{~g}\) is balanced on a wedge placed at \(40\,cm\) 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. \(\left(\mathrm{g}=10 \,\mathrm{~m} / \mathrm{s}^{2}\right)\)

- A \(\frac{1}{2}\,kg\)
- B \(\frac{1}{3} \,kg\)
- C \(\frac{1}{6} \,kg\)
- D \(\frac{1}{12}\,kg\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{12}\,kg\)
Step-by-step Solution
Detailed explanation
By balancing torque \(2 \,g \times 20=0.5\, g \times 60+m g \times 120\)
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