Question

In the sonometer experiment, a tuning fork of frequency $$256Hz$$ is in resonance with $$0.4m$$ length of wire when the iron load attached to the free end of the wire is $2$kg. If the load is immersed in the water, the length of the wire in resonance would be (specific gravity of iron $8$)
$$\begin{array}{rl}& A.0.37m\\ & B.0.43m\\ & C.0.31m\\ & D.0.2m\end{array}$$

Answer 1

Hint: A stretched string, when plucked produces a transverse wave which travels along the stretched string. These transversal waves reach the end of the strings and are often reflected back. The superposition of the transverse wave with the reflected wave produces a stationary wave. This is the principle behind the sonometer.

Complete step by step answer:
We know that the frequency of the wave is given as $f\propto {\displaystyle \frac{1}{L}}\sqrt{{\displaystyle \frac{T}{\mu }}}$where, the length of the string is $L$and mass $m$is attached to it, and $\mu$ is the mass per unit length of the string . Then the string experiences a tension$T$.
Given that, $f=256Hz$, $L=0.4m$ and $m=2kg$
Clearly, the frequency doesn’t vary, if the load is immerse in the water, then we can say that the frequency is a constant here, then we get ${\displaystyle \frac{\sqrt{T}}{L}}=k$
We know that the tension in the string, when the setup is in air is given as $T_a=mg=2g$
The tension in the string, when the setup is in water is given as $T_w=mg-V\rho g$ where $V$ is the volume of the water and $\rho$ is its density. To simplify let us rewrite $m=Vd$, where $d$ is the specific gravity of the material, here iron.
Then we get, $T_w=Vdg-V\rho g=Vdg(1-{\displaystyle \frac{\rho }{d}})$
We know that $\rho =1$ and given that $d=8$
Then, $T_w=mg(1-{\displaystyle \frac{1}{8}})=mg\left({\displaystyle \frac{7}{8}}\right)$
Now, let us take ${\displaystyle \frac{{\sqrt{T}}_a}{L_a}}={\displaystyle \frac{{\sqrt{T}}_w}{L_w}}$
Then substituting the values we get,$${\displaystyle \frac{\sqrt{mg}}{0.4}}={\displaystyle \frac{\sqrt{{\displaystyle \frac{7mg}{8}}}}{L_w}}$$
Or, $L_w={\displaystyle \frac{\sqrt{7}}{2\sqrt{2}}}\times 0.4$
Or, $L_w=0.37m$
Hence the answer is $$A.0.37m$$

Note:
The frequency of the wave is independent of the medium. Since the same setup is immersed in water, the mass per unit length is also a constant. Also, note that the frequency of the waves is given by the laws of stretched strings.