A thin uniform rod mass `M` and length `L` is hinged at its upper end. And released from rest from a horizontal position. The tension at a point located at a distance `L//3` from the hinge point, when the rod become vertical will

Options

A. `(W)/(W_1)`
B. `(Wl_1)/(Wl - W_1 l_2)`
C. `(l_1)/(l_1 - l_2)`
D. `l_1/l_2`

Answer

C

Explanation

In first case, Let O be midpoint of the rod. For rotational equilibrium of the rod,

Taking moments about O, we get
`W_1l_1 = Wl` ....(i)
In second case, When metal piece W is immersed in water, it experiences a buoyant force.
` therefore ` Apparent weight = Weight - Buoyant force
`W_(app) = W-F_B =Vrho_( metal) g` (where V is the volume of the metal piece) `W_(app) = V rho_(metal)g [ 1 - (rho_(water))/(rho_(metal))] = W [ 1- (rho_(water))/(rho_(metal))` ....(ii)
Again, for the rotational equilibrium of the rod, Taking moments about O, we get
`W_1 l_2 = W_(app) l = Wl [ 1- (rho_(water))/(rho_(metal))]` (using (ii) )
`W_1l_2 = W_1l_1[ - (rho_(water))/(rho_(metal)) ]` using (i)
`l_2/l_1 = [1-(rho_(water))/(rho_(metal))]`
`(rho_(water))/(rho_(metal)) = 1 - l_2/l_1 = (l_1 - l_2)/(l_1)`
`therefore `Specific gravity of the metal piece ` = (rho_(metal))/(rho_(water))= (l_1)/(l_1 -l_2)`