The fundamental frequency of a sonometer wire is 50 Hz for some length and tension. If the length is increased by $25 \%$ by keeping tension same then frequency change of second harmonic is

The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

With an initial fundamental frequency of 50 Hz, increasing the length by 25% gives $L_2 = \frac{5}{4}L_1$. Since tension and linear mass density remain constant, the new fundamental frequency becomes $f_1' = \frac{1}{2L_2}\sqrt{\frac{T}{\mu}} = \frac{1}{2(5L_1/4)}\sqrt{\frac{T}{\mu}} = \frac{4}{5} \times 50 = 40$ Hz.

The second harmonic frequency is always twice the fundamental frequency. Initially $f_{2,\text{initial}} = 2 \times 50 = 100$ Hz, and after the change $f_{2,\text{final}} = 2 \times 40 = 80$ Hz.

The percentage change in the second harmonic frequency is $\frac{80 - 100}{100} \times 100\% = -20\%$, indicating a 20% decrease.