Plz solve this.

Solution,
Frequency of the wave through the string, $$\begin{gathered} f_s\propto \sqrt{T} \\ {f}_s=k\sqrt{T} \\ where, k is a cons\tan t \\ Beats are formed by the superposition of two waves. \\ Beat frequency is the difference of the frequency between the tuning fork and the vibrating string \\ Now, \\ When T_1=100N \\ {f}_{s1}=k\sqrt{T_1}=10k \\ {f}_{s1}-f_T=5 \\ 10k-f_T=5 .............\left(1\right) \\ When T_1=64N \\ {f}_{s2}=k\sqrt{T_2}=5 \\ {f}_T-f_{s2}=5 \\ {f}_T-8k=5 .............\left(2\right) \\ Adding \left(1\right) and \left(2\right), we get \\ 2k=10 \\ k=5 \\ Also, \\ From \left(1\right) and \left(2\right) \\ 10k-f_T=f_T-8k \\ {f}_T=9k=9\times 5=45 Hz \end{gathered}$$
Ans (2)