A uniform rod of length 2 m , specific gravity 0.5 and mass 2 kg is hinged at one end to the bottom of a tank of water (specific gravity =1.0 ) filled upto a height of 1 m as shown in the figure. Taking the case 0^ , the force exerted by the hinge on the rod is

Step-by-step solution

Length of rod inside the water =1 = Upthrust, F= ( 2 2 )( ) ( 1 500 )(1000)(10) aligned & F=20 & Weight of rod, w=20 10=20 ~N aligned For rotational equilibrium of rod, net torque about O should be zero. F ( 2 )( )=w=(1 ) 20 2 ^2 =20 array ll & =45^ & F=20 45^ array F=20 2 ~N For vertical equilibrium of rod, force exerted by the hinge on the rod will be (20 2 -20) N downwards i.e. 8.3 N downwards