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  • A uniform rod of lenght 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark.A mass of 2 kg is suspended form the rod at 20 cm and another unkonwn mass 'm' is suspended from th

A uniform rod of lenght 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark.A mass of 2 kg is suspended form the rod at 20 cm and another unkonwn 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=10m/s^{2})

Option: 1

\frac{1}{2}kg


Option: 2

\frac{1}{3}kg


Option: 3

\frac{1}{6}kg


Option: 4

\frac{1}{12}kg


Answers (1)

best_answer

for rotational equilibrium

T_{1}=T_{2}

\left ( 2\times 10 \right )r_{1}=\left ( m\times 10 \right )r_{2}+\left ( 0.5\times 10 \right )\left ( 0.6 \right )

r_{1}=0.2m \: \: \: \: r_{2}=1.2m

2\times 0.2-0.3=m\left ( 1.2 \right )

\left ( m=\frac{1}{12} kg\right )

Posted by

Ritika Jonwal

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