A uniform rod of density `rho` is placed in a wide tank containing a liquid of density `rho_0 (rho_0 gt rho)`. 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.

Options

A. `sin theta=1/2 sqrt(rho_0//rho)`
B. `sin theta=1/2 (rho_0)/(rho)`
C. `sin theta=sqrt(rho//rho_0)`
D. `sin theta= rho_0//rho`

Answer

A

Explanation

Force of buoyancy-
`F_B=Al rho_0 g` (l=PR)

For rotational equilibrium-
`Al rho_0 g xx l/2 cos theta = AL rho g xx L/2 cos theta`
`implies (l^2)/(L^2)=(rho)/(rho_0) implies l/L=sqrt((rho)/(rho_0))`
`implies sin theta=h/l (L)/(2l)=1/2 sqrt((rho_0)/(rho))`.