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 mass-less 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 equilibrium is:

Mg \over k( 1 - LA\sigma \over 2M )

Mg \over k( 1 - LA\sigma \over M )

Mg \over k( 1 + LA\sigma \over M )

Properties of Matter PYQ — JEE Main 2013 · Shift 0 · Q70 with Solution

View written solutionFree

Correct answer: C

The cylinder is suspended vertically by a spring and is half submerged in a liquid.

At equilibrium, the upward spring force and upward buoyant force together balance the weight:

$k x_0 + F_b = Mg$

where:

Buoyant force equals the weight of displaced liquid:

$F_b = \sigma g V_{\text{displaced}}$

Since the cylinder has cross-sectional area $A$ and length $L$ , its total volume is:

$V = AL$

It is half submerged, so displaced volume is:

$V_{\text{displaced}} = \frac{AL}{2}$

Hence,

$F_b = \sigma g \left(\frac{AL}{2}\right) = \frac{\sigma A L g}{2}$

Using

$k x_0 + \frac{\sigma A L g}{2} = Mg$

we get

$k x_0 = Mg - \frac{\sigma A L g}{2}$

So,

$x_0 = \frac{Mg - \frac{\sigma A L g}{2}}{k}$

Factor out $Mg$ :

$x_0 = \frac{Mg}{k}\left(1 - \frac{\sigma A L}{2M}\right)$

This is exactly:

$\boxed{\frac{Mg}{k}\left(1 - \frac{LA\sigma}{2M}\right)}$

So the correct option is C.

Stored correct answer: C

Derived answer: C

They agree.