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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.
(g=10 ms2)


A
12 Kg
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B
13 Kg
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C
16 Kg
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D
112 Kg
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Solution

The correct option is D 112 Kg
All the forces acting on the rod are as shown in the figure below:


For the rod to remain in equilibrium,

τnet=0

Calculating the net torque about the wedge, we have:

2g×20=0.5g×60+mg×120

m=0.56 Kg=112 Kg

Hence, option (D) is the correct answer.


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