A solid sphere of radius $R$ and density $\rho$ is attached to one end of a mass-less spring of force…

1. The net elongation of the spring is $\frac{4 \pi R^{3} \rho g}{3k}$
2. The net elongation of the spring is $\frac{8 \pi R^3 \rho g}{3k}$
3. The light sphere is partially submerged
4. The light sphere is completely submerged

Solution

$\text{kx}+{\text{f}}_{\text{B}}=\text{mg}$
From equilibrium of sphere. $\left(3\rho \right)$
$\text{kx}+2\rho \text{vg}-3\rho \text{vg}$
$\text{kx}=\rho \text{vg}$
$\text{x}=\frac{\rho \text{vg}}{\text{k}}=\rho \text{g}\frac{4}{3}\pi {\text{R}}^3\frac{\rho \text{vg}}{\text{k}}=\frac{\rho \text{g}\frac{4}{3}\pi {\text{R}}^3}{\text{K}}$
Let ${\text{v}}_1=$ volume submerged of lighter sphere.

$\text{kx}+\text{mg}={\text{F}}_{\text{B}}$
$\rho \text{vg}+\rho \text{vg}=\text{2}\rho {\text{v}}_1\text{g}$
$=2\rho \text{vg}=2\rho {\text{v}}_1\text{g}$
$\Rightarrow {\text{v}}_1=\text{v}$ `

Asked in: JEE Advanced 2013 (Paper 1)

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