Question

A solid sphere of radius $R$ and density $\rho $ is attached to one end of a massless spring of force constant $k$. The other end of the spring is connected to another solid sphere of radius $R$and density $3\rho $. The complete arrangement is placed in a liquid of density $2\rho $ and is allowed to reach equilibrium. The correct statement(s) is(are)
(This question has multiple correct options)
A. The net elongation of the spring is $\dfrac{{4\pi {R^3}\rho g}}{{3k}}$
B. The net elongation of the spring is $\dfrac{{8\pi {R^3}\rho g}}{{3k}}$
C. The light sphere is partially submerged.
D. The light sphere is completely submerged.

Answer 1

Hint: First we need to find the weight of the system. The buoyant force is the consequence of liquid displaced by the weight of the system. Equating these forces, we may obtain the required value of elongation in the spring.

Complete step-by-step answer:
The given arrangement consists of a solid sphere of radius $R$ and density $\rho $ which is attached to one end of a massless spring of force constant $k$. The other end of the spring is connected to another solid sphere of radius $R$ and density $3\rho $. If the mass of each of the spheres is taken to be m, then the weight of the system can be given as ${\text{mg}}$.
So, if ${\text{W}}$ is the weight of the system, then,
$W = (\rho \times \dfrac{4}{3}\pi {R^3} + 3\rho \times \dfrac{4}{3}\pi {R^3})g$
Here we calculate the mass of each sphere by multiplying the density of the sphere with the volume of the sphere.
Therefore, the weight of the system will be
$W = \dfrac{{16\rho }}{3}\pi {R^3}g$
Now, buoyant force=${F_g}$
${\rho _1} = 2\rho $

where, ${\rho _1}$ is the density of the liquid in which the system is placed.

Therefore,

${F_g} = V \times 2\rho g$

Now,

Buoyant force= weight of system

Therefore,
$
  V \times 2\rho g = \dfrac{{16\rho }}{3}\pi {R^3}g \\
   \Rightarrow V = \dfrac{{8\pi {R^3}}}{3} \\
 $
That is,
Submerged volume=$\dfrac{{8\pi {R^3}}}{3}$
Total volume of system=$\dfrac{4}{3}\pi {R^3} + \dfrac{4}{3}\pi {R^3} = \dfrac{8}{3}\pi {R^3}$
That is, total volume = submerged volume

It means both sphere are submerged completely

Now, if $x$ is the net elongation in the spring, then
$
  kx + \dfrac{4}{3}\pi {R^3} \times \rho \times g = \dfrac{4}{3}\pi {R^3} \times 2\rho \times g \\
   \Rightarrow x = \dfrac{4}{{3k}}\pi {R^3}\rho g \\
 $
Therefore, from the above calculations we can conclude that the answer to this question is option A. The net elongation of the spring is $\dfrac{{4\pi {R^3}\rho g}}{{3k}}$
and option D. The light sphere is completely submerged.

So, the correct answer is “Option A and D”.

Note: When the spring and mass arrangement is placed in the liquid then the elongation produced in the spring is directly proportional to the density of the liquid. Larger than density of the liquid, greater is the elongation produced.