Q: A uniform cylindrical 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 σ at equilibrium position. The extension x_{0} of the spring when it is in equilibrium is

(a) $\large \frac{M g}{k}$

(b) $\large \frac{M g}{k} (1 – \frac{L A \sigma}{M})$

(c) $\large \frac{M g}{k} (1 – \frac{L A \sigma}{2 M})$

(b) $\large \frac{M g}{k} (1 + \frac{L A \sigma}{M})$

Ans: (c)

Sol: At equilibrium ,

Upward force = Downward force

$\large k x_0 + F_B = M g$

$\large kx_0 + A (L/2)\sigma g = M g$

$\large kx_0 = M g – A (L/2)\sigma g$

$\large x_0 = \frac{M g}{k} (1-\frac{AL\sigma}{2M})$