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

Let ` AB = L, AC = L//2, AD = l, A ` = area of cross-section of the rod
Weight of the rod ` = A L rho g ` acting through C.
Buoyancy force ` = A l rho _ 0 g `, acting through the midpoint of AD. Taking torque about A,
` ( l A rho _ 0 g ) (l)/(2 ) cos theta = (LA rho g ) (L)/(2) cos theta or (l ^ 2 )/(L ^ 2 ) = (rho )/(rho _ 0 ) `
Also , ` sin theta = (h)/(l) = (L)/( 2 l ) or (l)/(L ) = (1 ) /( 2 sin theta ) = sqrt((rho )/(rho _0)) or sin theta = (1)/(2) sqrt((rho_0)/(rho)) `