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 x0 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
$\large k x_0 + F_B = M g$
$\large kx_0 + A (L/2)\sigma g = M g$
$\large kx_0 = M g – A (L/2)\sigma g$
$\large x_0 = \frac{M g}{k} (1-\frac{AL\sigma}{2M})$