$A$ sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same,the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?

(B) 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.
Initially,the wire vibrates in its fundamental mode (first harmonic) with frequency $f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = f$,where $f$ is the frequency of the tuning fork.
When the length is doubled $(L' = 2L)$ and tension $T$ remains constant,the new fundamental frequency becomes $f_1' = \frac{1}{2(2L)} \sqrt{\frac{T}{\mu}} = \frac{1}{2} f$.
For the tuning fork to still be in resonance,the wire must vibrate at a higher harmonic such that its frequency matches $f$. The $n$-th harmonic frequency is $f_n = n \times f_1'$.
Setting $f_n = f$,we get $n \times (\frac{1}{2} f) = f$,which implies $n = 2$.
Therefore,the tuning fork will be in resonance with the wire when the wire vibrates in its second harmonic (first overtone).