A solid sphere of radius R and density `rho` is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density `3rho`. The complete arrangement is placed in a liquid of density `2rho` and is allowed to reach equilibrium. The correct statements(s) is (are)

Options

A. the net elongation of the spring is `(4pi R^(3)pg)/(3k)`
B. the net elongation of the spring is `(8pi R^(3)pg)/(3k)`
C. the light sphere is partially submerged
D. the light sphere is completely submerged.

Answer

A, D

Explanation

Sphere of density 3p will be below the sphere of density p as shown in figure.
Forces acting on lower sphere are weight downwards, spring force upward and buoyant force upward. For equilibrium we can write the following
`kx +4/3 pi R(2) (2p)g=4/3 pi R^(3) (3p)g`
Where x is a elongation of spring.
Solving above equation we get: `x=(4pi R^(2)pg)/(3k)`.
Net downward forces (Spring fore and weight) acting on lighter body can be written as follows:
`F=kx +4/3 pi R^(3)pg`
Substituting value of kx calculate above we get
`F=4/3 pi R^(2) (2p)g`
Above calculated force is equal to byoyant force on lighter body when it is completely immersed, hence we can understant that for equilibrium lighter body will be completely immersed.