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An object of density $2000 \mathrm{~kg}-\mathrm{m}^{-3}$ is hung from a thin light wire. The fundamental frequency of the transverse waves in the wire is $200 \mathrm{~Hz}$. If the object is immersed in water such that half of its volume is submerged, then the fundamental frequency of the transverse waves in the wire is
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Verified Answer
The correct answer is:
173.2 Hz
Fundamental frequency, initially
$$
n_1=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{2 l} \sqrt{\frac{V g(2000)}{\mu}}
$$
Fundamental frequency, finally
$$
n_2=\frac{1}{2 l} \sqrt{\frac{V g\left(2000-\frac{1000}{2}\right)}{\mu}}
$$
(here we applied loss of weight due to upthrust)
$$
\begin{aligned}
& =\frac{1}{2 l} \sqrt{\frac{V g \times 1500}{\mu}} \\
\Rightarrow \quad \frac{n_1}{n_2} & =\sqrt{\frac{2000}{1500}}=\sqrt{\frac{4}{3}} \\
\Rightarrow \quad n_2 & =\frac{n_1 \times \sqrt{3}}{2}=100 \sqrt{3} \mathrm{~Hz}=173.2 \mathrm{~Hz}
\end{aligned}
$$
$$
n_1=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{2 l} \sqrt{\frac{V g(2000)}{\mu}}
$$
Fundamental frequency, finally
$$
n_2=\frac{1}{2 l} \sqrt{\frac{V g\left(2000-\frac{1000}{2}\right)}{\mu}}
$$
(here we applied loss of weight due to upthrust)
$$
\begin{aligned}
& =\frac{1}{2 l} \sqrt{\frac{V g \times 1500}{\mu}} \\
\Rightarrow \quad \frac{n_1}{n_2} & =\sqrt{\frac{2000}{1500}}=\sqrt{\frac{4}{3}} \\
\Rightarrow \quad n_2 & =\frac{n_1 \times \sqrt{3}}{2}=100 \sqrt{3} \mathrm{~Hz}=173.2 \mathrm{~Hz}
\end{aligned}
$$
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