A solid sphere of radius R and density ρ is attached to one end...

$\text{kx} + {\text{f}}_{\text{B}} = \text{mg}$
From equilibrium of sphere. \left(3 ρ\right)$<br >$\text{kx} + 2 ρ \text{vg} - 3 ρ \text{vg}$<br >$\text{kx} = ρ \text{vg}$<br >$\text{x} = \dfrac{ρ \text{vg}}{\text{k}} = ρ \text{g} \dfrac{4}{3} \pi {\text{R}}^{3} \dfrac{ρ \text{vg}}{\text{k}} = \dfrac{ρ \text{g} \dfrac{4}{3} \pi {\text{R}}^{3}}{\text{K}}$<br >Let${\text{v}_{1} =$volume submerged of lighter sphere.<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/EM0012545_AnsExp4_2.jpg"/><br >$\text{kx} + \text{mg} = {\text{F}}_{\text{B}}$<br >$ρ \text{vg} + ρ \text{vg} = \text{2} ρ {\text{v}_{1} \text{g}$<br >$= 2 ρ \text{vg} = 2 ρ {\text{v}_{1} \text{g}$<br >$⇒ {\text{v}_{1} = \text{v}