A uniform rod of mass 15 kg and of length 5m is held stationary with the help of a light string as shown in the figure. The tension in the string is

The free-body diagram of the rod is as shown in the figure. (Force exerted by the hinge on the rod are not shown.) Given : OBsinθ=3 ⇒sinθ=35 OA=OB2=52Now, take torque about the hinge,T×OBsinθ−15g×OAcosθ=0⇒T×3−15g×52×45=0⇒T=100N