JEE Main 2023PhysicsRotational MotionHardMCQ

JEE Main 2023 — Rotational Motion Question with Solution

JEE Main 2023 (25 Jan Shift 1)

Question

An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is:

(Take g=10 m s-2)

 

Choose an option

Show full solutionCorrect option: C
Correct answer
C300 N

Step-by-step explanation

Let T 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 C

T sin30°× 60=mg×50+8g×100

T2×60=20×50+80×100

⇒3T=100+800

⇒300 N.

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About this question

This is a previous-year question from JEE Main 2023, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.