A uniform rod of length 200 ,mathrm{~cm} and mass 500 ,mathrm{~g} is balanced on a wedge placed at 4

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6.System of Particles and Rotational Motion

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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

$\frac{1}{12}\,kg$

(NEET-2021)

Solution

By balancing torque

$2 \,g \times 20=0.5\, g \times 60+m g \times 120$

$m=\frac{0.5}{6}\, \mathrm{~kg}=\frac{1}{12}\, \mathrm{~kg}$

Standard 11

Physics

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