A sonometer wire vibrating in two loops is in resonance with a tuning fork. If tension applied to the string is increased to 16 times the original tension and length of the wire is doubled. It can still be in resonance with the tuning fork provided it vibrates in

The frequency of a vibrating string is given by $f = \dfrac{p}{2L} \sqrt{\dfrac{T}{\mu}}$, where $p$ is the number of loops, $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

Let the initial frequency be $f_1$. Given $p_1 = 2$, we have:

$f_1 = \dfrac{2}{2L_1} \sqrt{\dfrac{T_1}{\mu}}$

For the final state, the tension is increased to $16$ times the original tension ($T_2 = 16T_1$) and the length is doubled ($L_2 = 2L_1$). The frequency $f_2$ must be equal to $f_1$ for the wire to remain in resonance with the same tuning fork.

$f_2 = \dfrac{p_2}{2L_2} \sqrt{\dfrac{T_2}{\mu}}$

Equating $f_1$ and $f_2$:

$\dfrac{2}{2L_1} \sqrt{\dfrac{T_1}{\mu}} = \dfrac{p_2}{2(2L_1)} \sqrt{\dfrac{16T_1}{\mu}}$

$\dfrac{1}{L_1} \sqrt{\dfrac{T_1}{\mu}} = \dfrac{p_2 \times 4}{4L_1} \sqrt{\dfrac{T_1}{\mu}}$

$\dfrac{1}{L_1} = \dfrac{p_2}{L_1}$

$p_2 = 1$

Therefore, the wire will vibrate in one loop.

Answer: one loop