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**Hint:**Use formula for electrostatic force to relate it with the tension force produced in the ring. Use the relation between the tensile strength and tension force to relate the tension force with tensile strength of the ring.

**Complete step by step answer:**

The charges placed on the ring will develop repulsive forces on each other and the net effect of which is the rupturing of the ring.

The force acting on the charges is the electrostatic force. The electrostatic force is proportional to the square of the charge and inversely proportional to the square of the distance. Therefore, we can write,

\[ \Rightarrow{F_e} \propto \dfrac{{{q^2}}}{{{r^2}}}\]

Here, r is the radius of the ring.

This force is balanced by the tension force in the ring. Therefore, the tension force is proportional to the electrostatic force. Thus, we can write,

\[ \Rightarrow T \propto \dfrac{{{q^2}}}{{{r^2}}}\]

The tensile stress \[\sigma \] is the ratio of tension force and the area of cross-section of the ring.

Therefore,

\[ \Rightarrow\sigma = \dfrac{T}{A}\]

\[ \Rightarrow \sigma \propto T\]

Therefore, we can write,

\[ \Rightarrow\sigma \propto \dfrac{{{q^2}}}{{{r^2}}}\]

\[ \Rightarrow q = \sqrt \sigma r\]

We have given that the tensile strength of the second ring is k times of the first and the radius is n times. Therefore, the charge on the second ring will be,

\[ \Rightarrow q' = qn\sqrt k \]

Therefore, the maximum charge on the second ring should be less than the above charge.

**So, the correct answer is option (B).**

**Note:**Do not equate Hooke's law with the electrostatic force on the charges. It will give different results. While solving this question, you don’t need to take the values of the constants as it concerns only dependence on tensile strength, charge and radius of the ring.

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