A rod bent at right angle along its centre line, is placed on a rough horizontal fixed cylinder of radius R as shown in figure. Mass of rod is 2 m and rod is in equilibrium. Assume that friction force on rod at A and B are equal in magnitude.

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f A ​ = f B ​ N A ​ + f B ​ = 2 m g N B ​ = f A ​ ​ Torque equation ​ f A ​ R + f B ​ R = m g R ⇒ ​ f A ​ + f B ​ = m g ⇒ ​ f A ​ = f B ​ = 2 m g ​ ​ ​ ​ ​ ⇒ N B ​ = 2 m g ​ N A ​ + 2 m g ​ = 2 m g N A ​ = 2 3 m g ​

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JEE Advanced (2028)