Question

A graduated uniform metre rod is balance horizontally on a knife edge at a distance of 10cm from its centre of gravity when mass of 50g and 30g are suspended at 5cm and 90cm divisions of the rod find the weight of the rod

Answer 1

Torque on the rod due to the weight 30 gm= (w×10 + 30×50)

{in clockwise direction}

Torque on the rod due to 50 gm weight= (35×50) {in anti-clockwise direction}

∴ W×10 + 30×50×g= 50×35×g

=> 10W= 50(35-30)×g

=> W= (50×5×g)/10 = (25×g) Newton

=> W= (25/1000)× 10

        =0.25 N