Rotational Motion

JEE Main level, single correct, medium

111 of 280

Question 111 of 280

An object of mass 8kg8kg is hanging from one end of a uniform rod CDCD of mass 2kg2kg and length 1m1m pivoted at its end CC on a vertical wall as shown in figure. It is supported by a cable ABAB such that the system is in equilibrium. The tension in the cable is:
(Take
g=10ms−2g=10m{s}^{-2})

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Answer: (C) 300N300N


Let
TT be the tension in cable.
Since the system is at equilibrium, the net torque, about an axis perpendicular to the plane and passing through the point of contact of the rod and the wall, is zero. Therefore, torque about point
CC
Tsin⁡30∘×60=mg×50+8g×100T\sin 30^{\circ}\times 60=mg\times 50+8g\times 100
T2×60=20×50+80×100\dfrac{T}{2}\times 60=20\times 50+80\times 100
⇒3T=100+800\Rightarrow 3T=100+800
⇒300N\Rightarrow 300N.

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