A sonometer wire carrying a 4 kg load vibrates in its fundamental...

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

The tension $T$ is provided by the load $M$, so $T = Mg$.

Since the frequency $f$ and linear mass density $\mu$ remain constant, we have $\dfrac{\sqrt{M}}{L} = \text{constant}$.

Therefore, $\dfrac{\sqrt{M_1}}{L_1} = \dfrac{\sqrt{M_2}}{L_2}$.

Given $M_1 = 4 \text{ kg}$, $L_1 = L$, and $L_2 = 2L$, substituting these values gives:

$\dfrac{\sqrt{4}}{L} = \dfrac{\sqrt{M_2}}{2L}$

$\dfrac{2}{L} = \dfrac{\sqrt{M_2}}{2L}$

$\sqrt{M_2} = 4$

$M_2 = 16 \text{ kg}$