A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density `sigma` at equilibrium position. The extension `x_0` of the spring when it is in equlibrium is:

Options

A. ` (Mg )/( k ) (1 - (LA sigma )/(M)) `
B. ` (Mg ) /(k) (1 - (LA sigma )/( 2 M )) `
C. ` (Mg )/(k ) (1 + (LA sigma )/(M))`
D. ` (Mg )/(k) `

Answer

B

Explanation

Let `x_(0)` be the extension of the spring in equilibrium position. From FBD of cylinder in equilibrium,
`Kx_(0)+F_(B)=Mg`
`kx_(0)+sigma(L)/(2)Ag=Mg x_(0)=(Mg-(sigmaLAg)/(2))/(k)=(Mg)/(k)(1-(sigmaLA)/(2M))`