MHTCET 11 Apr 2026 Shift 2 Paper

Concept:The frequency of harmonics in a stretched wire depends on length, tension, and mass per unit length.Explanation:For a stretched wire, the $n^{\text{th}}$ harmonic frequency is given by:$$f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$Here, $\mu$ is mass per unit length, and for the same material, $\mu \propto r^2$.Given $r_1 = 2r_2$, we have:$$\frac{\mu_1}{\mu_2} = \frac{r_1^2}{r_2^2} = 4$$So, $\mu_1 = 4\mu_2$.Since tension $T$ is the same, the wave speed $v = \sqrt{\frac{T}{\mu}}$ gives:$$v_1 = \frac{v_2}{2}$$The first overtone of the first wire is its second harmonic:$$f_1 = \frac{2v_1}{2L_1} = \frac{v_1}{L_1}$$The second overtone of the second wire is its third harmonic:$$f_2 = \frac{3v_2}{2L_2}$$Given $f_1 = f_2$, we write:$$\frac{v_1}{L_1} = \frac{3v_2}{2L_2}$$Substitute $v_1 = \frac{v_2}{2}$:$$\frac{v_2}{2L_1} = \frac{3v_2}{2L_2}$$Cancelling common terms gives:$$\frac{1}{L_1} = \frac{3}{L_2}$$Therefore:$$\frac{L_1}{L_2} = \frac{1}{3}$$Answer:The ratio of the length of the first wire to the second wire is $\frac{1}{3}$, i.e., Option B.