A liquid of density `rho_0` is filled in a wide tank to a height h. A solid rod of length L, cross-section A and density `rho` is suspended freely in the tank. The lower end of the rod touches the base of the tank and h=L/n (where n gt 1). Then the angle of inclination `theta` of the rod with the horizontal in equilibrium position is

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)) `