Step-by-step Derivation

1. Fundamental Frequency Formula: The fundamental frequency ($f$) of a vibrating string is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu'}}$ where $L$ is the length of the string, $T$ is the tension, and $\mu'$ is the mass per unit length.

2. Reference Frequency ($f_0$): The problem states that for string-1 (with mass per unit length $\mu$), at its free length $L_0$ and tension $T_0$, the fundamental frequency is $f_0$. So, we can write: $f_0 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}$

3. Condition for Highest Frequency: The frequency $f$ is inversely proportional to the length $L$ ($f \propto 1/L$). To achieve the highest possible fundamental frequency ($f_{max}$), the string must be vibrated at its minimum possible length. The given range for the free length is $[L_0, 2L_0]$. Therefore, the minimum length is $L = L_0$. The tension is constant for all strings at $T_0$. So, the highest fundamental frequency for a string with mass per unit length $\mu'$ is: $f_{max} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu'}}$

4. Calculate Highest Frequency for Each String: We now calculate $f_{max}$ for each string listed in List-I and express it in terms of $f_0$.

   • (I) String-1: $\mu' = \mu$ $f_{max, 1} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} = f_0$ This matches with (P) from List-II ($1 f_0$). Thus, I $\to$ P.

   • (II) String-2: $\mu' = 2\mu$ $f_{max, 2} = \frac{1}{2L_0} \sqrt{\frac{T_0}{2\mu}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{1}{\sqrt{2}} f_0$ This matches with (Q) from List-II ($1/\sqrt{2} f_0$). Thus, II $\to$ Q.

   • (III) String-3: $\mu' = 3\mu$ $f_{max, 3} = \frac{1}{2L_0} \sqrt{\frac{T_0}{3\mu}} = \frac{1}{\sqrt{3}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{1}{\sqrt{3}} f_0$ This matches with (R) from List-II ($1/\sqrt{3} f_0$). Thus, III $\to$ R.

   • (IV) String-4: $\mu' = 4\mu$ $f_{max, 4} = \frac{1}{2L_0} \sqrt{\frac{T_0}{4\mu}} = \frac{1}{\sqrt{4}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{1}{2} f_0$ This matches with (S) from List-II ($1/2 f_0$). Thus, IV $\to$ S.

5. Conclusion and Comparison: Our derived correct matching is: I $\to$ P, II $\to$ Q, III $\to$ R, IV $\to$ S. Let's compare this with the given options: A: I $\to$ P, II $\to$ Q, III $\to$ T, IV $\to$ S B: I $\to$ P, II $\to$ R, III $\to$ S, IV $\to$ Q C: I $\to$ Q, II $\to$ S, III $\to$ R, IV $\to$ P D: I $\to$ Q, II $\to$ P, III $\to$ R, IV $\to$ T

   None of the provided options match our derived result. Therefore, the question or the options are flawed.

   The stored correct answer is B. Let's analyze why it is incorrect. Option B states the matching is I $\to$ P, II $\to$ R, III $\to$ S, IV $\to$ Q. This implies the highest frequencies are $f_0$, $f_0/\sqrt{3}$, $f_0/2$, and $f_0/\sqrt{2}$ for strings 1, 2, 3, and 4 respectively. This contradicts our calculations for strings 2, 3, and 4. The mappings for strings 2, 3, and 4 seem to have been cyclically permuted.

   Based on a rigorous application of physics principles to the problem as stated, none of the options is correct.