Question
A uniform rod of density $$\rho $$ is placed in a wide tank containing a liquid of density $${\rho _0}\left( {{\rho _0} > \rho } \right).$$ The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle $$\theta $$ with the horizontal, then:
A.
$$\sin \theta = \frac{1}{2}\sqrt {\frac{{{\rho _0}}}{\rho }} $$
B.
$$\sin \theta = \frac{1}{2}.\frac{{{\rho _0}}}{\rho }$$
C.
$$\sin \theta = \sqrt {\frac{\rho }{{{\rho _0}}}} $$
D.
$$\sin \theta = \frac{{{\rho _0}}}{\rho }$$
Answer :
$$\sin \theta = \frac{1}{2}\sqrt {\frac{{{\rho _0}}}{\rho }} $$
Solution :
Let $$L = PQ =$$ length of rod
$$\therefore SP = SQ = \frac{L}{2}$$
Weight of rod,
$$W = Al\rho g.$$ acting
At point $$S$$
And force of buoyancy, $${F_B} = Al{\rho _0}g.\left[ {l = PR} \right]$$
Which acts at mid-point of $$PR.$$
For rotational equilibrium.
$$\eqalign{ & Al{\rho _0}g \times \frac{\ell }{2}\cos \theta = AL\rho g \times \frac{L}{2}\cos \theta \cr & \Rightarrow \frac{{{l^2}}}{{{L^2}}} = \frac{\rho }{{{\rho _0}}} \Rightarrow \frac{l}{L} = \sqrt {\frac{\rho }{{{\rho _0}}}} \cr} $$$
From figure, $$\sin \theta = \frac{h}{l} = \frac{L}{{2l}} = \frac{1}{2}\sqrt {\frac{{{\rho _0}}}{\rho }} $$
Let $$L = PQ =$$ length of rod
$$\therefore SP = SQ = \frac{L}{2}$$
Weight of rod,
$$W = Al\rho g.$$ acting
At point $$S$$
And force of buoyancy, $${F_B} = Al{\rho _0}g.\left[ {l = PR} \right]$$
Which acts at mid-point of $$PR.$$
For rotational equilibrium.
$$\eqalign{ & Al{\rho _0}g \times \frac{\ell }{2}\cos \theta = AL\rho g \times \frac{L}{2}\cos \theta \cr & \Rightarrow \frac{{{l^2}}}{{{L^2}}} = \frac{\rho }{{{\rho _0}}} \Rightarrow \frac{l}{L} = \sqrt {\frac{\rho }{{{\rho _0}}}} \cr} $$$
From figure, $$\sin \theta = \frac{h}{l} = \frac{L}{{2l}} = \frac{1}{2}\sqrt {\frac{{{\rho _0}}}{\rho }} $$
