$f=\frac{v}{2 \ell}=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}=\frac{1}{2 \ell} \sqrt{\frac{T}{A \rho}}\left[\because v=\sqrt{\frac{T}{\mu}} \text { and } \mu=\frac{m}{\ell}\right]$
Also, $Y=\frac{T \ell}{A \Delta \ell} \Rightarrow \frac{T}{A}=\frac{Y \Delta \ell}{\ell} \Rightarrow f=\frac{1}{2 \ell} \sqrt{\frac{\gamma \Delta \ell}{\ell \rho}}$ $...(i)$
Putting the value of $\ell, \frac{\Delta \ell}{\ell}, \rho$ and $\gamma$ in eq $^{n} .$ $(i)$ we get,
$f=\sqrt{\frac{2}{7}} \times \frac{10^{3}}{3}$ or, $f \approx 178.2 \mathrm{Hz}$
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$[1]$ The graph of kinetic energy $E_k$ of the ball against height $h$ is shown in figure $1$
$[2]$ The graph of height $h$ against time $t$ is shown in figure $2$
$[3]$ The graph of gravitational energy $E_g$ of the ball against height $h$ is shown in figure $3$
Which of $A, B, C, D, E$ shows the correct answers?
