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**Hint:**The question can be solved using the knowledge of law of conservation of momentum and work-energy theorem. According to the Law of conservation of momentum, the total momentum of a system is conserved.

**Complete step by step answer:**

Momentum is the product of mass and velocity, $p=mv$

According to the law of conservation of ${{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}$

Let ${{m}_{1}}=m$ and ${{m}_{2}}=3m$

Putting the values in the formula we get

$m{{v}_{1}}=3m{{v}_{2}}$

$\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{3}{1}$ - (i)

When the masses are compressed, they will possess some potential energy which will be converted to kinetic energy when the masses are released.

According to work-energy principle:

$\dfrac{1}{2}m{{v}^{2}}=F\times s$

Putting the values of both the masses in the formula

$\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}=f.{{s}_{1}}$

$\Rightarrow \dfrac{1}{2}{{m}_{2}}v_{2}^{2}=f.{{s}_{2}}$

We know that $f=\mu mg$

Therefore $\dfrac{1}{2}{{m}_{1}}v_{1}^{2}=\mu {{m}_{1}}gS\text{ -(ii)}$

and $\dfrac{1}{2}{{m}_{2}}v_{2}^{2}=\mu {{m}_{2}}gs\text{ - (iii)}$

Dividing equation (ii) by equation (iii) we get,

$\dfrac{v_{1}^{2}}{v_{2}^{2}}=\dfrac{S}{s}$

From equation (i),

${{\left( \dfrac{3}{1} \right)}^{2}}=\dfrac{S}{s}$

$\therefore \dfrac{S}{s}=\dfrac{9}{1}$

**Hence, the correct answer is option D.**

**Note:**Work energy theorem states that all the work done by the system is equal to the change in kinetic energy of the system. When the spring is compressed, the work is done on the system and is stored in the form of potential energy. When the spring is released, the potential energy gets converted to kinetic energy and the masses move due to that kinetic energy. The overall energy of the system is conserved. The energy can neither be created nor be destroyed; it can only be converted from one form to another.

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