Question

In the sonometer experiment, a tuning fork of frequency $256 Hz$ is in resonance with $0.4 m$ length of the wire when the iron load attached to free end of wire is $2 kg$. If the load is immersed in water, the length of the wire in resonance would be: (specific gravity of iron $=8$ )

• A. 0.37 m
• B. 0.43 m
• C. 0.31 m
• D. 0.2 m

Answer 1

Answer: (A)

For a sonometer
$v=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$
$256=\frac{1}{2(0.4 m )} \sqrt{\frac{2 g}{\mu}}$
Weight of iron $=V \rho g=V 8 g=2 g$
$\Rightarrow V=\left(\frac{1}{4}\right)$
Apparent weight of iron in water $=V(\rho-\sigma) g$
$=V(8-1) g=V 7 g=\frac{7}{4} g$
$\Rightarrow \frac{1}{2 l^{\prime}} \sqrt{\frac{\frac{7}{4} g}{\mu}}=\frac{1}{2(0.4)} \sqrt{\frac{2 g}{\mu}}$
or $l^{\prime}=(0.4) \sqrt{\frac{7}{8}}=0.37 m$