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An iron load of $2 \mathrm{~kg}$ is suspended in air from the free end of a sonometer wire of length $1 \mathrm{~m}$. A tuning fork of frequency $256 \mathrm{~Hz}$, is in resonance with $\frac{1}{\sqrt{7}}$ times the length of the sonometer wire. If the load is immensed in water, the length of the wire in metre that will be in resonance with the same tuning fork is
$($ Specific gravity of iron $=8)$
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$($ Specific gravity of iron $=8)$
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The correct answer is:
$\frac{1}{\sqrt{8}}$
We know $I \propto \sqrt{T}$
$$
\therefore \quad \frac{I_{\text {air }}}{I_{\text {water }}}=\sqrt{\frac{T_{\text {air }}}{T_{\text {water }}}}
$$ $\begin{array}{llrl}\text { But specific gravity } 8 & =\frac{T_{\text {air }}}{T_{\text {air }}-T_{\text {water }}} \\ \Rightarrow & T_{\text {water }} =\frac{7}{8} T_{\text {air }} \\ \therefore & \frac{l_{\text {air }}}{l_{\text {water }}} =\sqrt{\frac{8}{7}} \\ \text { But } & l_{\text {air }} =\frac{1}{\sqrt{7}} l \text { (given) } \\ \therefore & l_{\text {water }} =\frac{1}{\sqrt{8}} l\end{array}$
$$
\therefore \quad \frac{I_{\text {air }}}{I_{\text {water }}}=\sqrt{\frac{T_{\text {air }}}{T_{\text {water }}}}
$$ $\begin{array}{llrl}\text { But specific gravity } 8 & =\frac{T_{\text {air }}}{T_{\text {air }}-T_{\text {water }}} \\ \Rightarrow & T_{\text {water }} =\frac{7}{8} T_{\text {air }} \\ \therefore & \frac{l_{\text {air }}}{l_{\text {water }}} =\sqrt{\frac{8}{7}} \\ \text { But } & l_{\text {air }} =\frac{1}{\sqrt{7}} l \text { (given) } \\ \therefore & l_{\text {water }} =\frac{1}{\sqrt{8}} l\end{array}$
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