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The frequency of a sonometer wire is $100 \mathrm{~Hz}$. When the weights producing the tensions are completely immersed in water, the frequency becomes $80 \mathrm{~Hz}$ and on immersing the weights in a certain liquid, the frequency becomes $60 \mathrm{~Hz}$. The specific gravity of the liquid is
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The correct answer is:
1.77
As we know, frequency $f \propto \sqrt{m g}$ or $f \propto \sqrt{g}$
In water, $\mathrm{f}_{\mathrm{w}}=0.8 \mathrm{f}_{\text {air }}$
$$
\begin{array}{l}
\frac{\mathrm{g}^{\prime}}{\mathrm{g}}(0.8)^{2}=0.64 \\
\Rightarrow 1-\frac{\rho_{\mathrm{w}}}{\rho_{\mathrm{m}}}=0.64 \\
\Rightarrow \frac{\rho_{\mathrm{w}}}{\rho_{\mathrm{m}}}=0.36
\end{array}
$$
In liquid, $\frac{g^{\prime}}{g}=(0.6)^{2}=0.36$
$$
1-\frac{\rho_{1}}{\rho_{\mathrm{m}}}=0.36 \frac{\rho_{1}}{\rho_{\mathrm{m}}}=0.64
$$
From eq. (1) and (2) $\frac{\rho_{l}}{\rho_{n}}=\frac{0.64}{0.36} \therefore \rho_{l}=1.77$
In water, $\mathrm{f}_{\mathrm{w}}=0.8 \mathrm{f}_{\text {air }}$
$$
\begin{array}{l}
\frac{\mathrm{g}^{\prime}}{\mathrm{g}}(0.8)^{2}=0.64 \\
\Rightarrow 1-\frac{\rho_{\mathrm{w}}}{\rho_{\mathrm{m}}}=0.64 \\
\Rightarrow \frac{\rho_{\mathrm{w}}}{\rho_{\mathrm{m}}}=0.36
\end{array}
$$
In liquid, $\frac{g^{\prime}}{g}=(0.6)^{2}=0.36$
$$
1-\frac{\rho_{1}}{\rho_{\mathrm{m}}}=0.36 \frac{\rho_{1}}{\rho_{\mathrm{m}}}=0.64
$$
From eq. (1) and (2) $\frac{\rho_{l}}{\rho_{n}}=\frac{0.64}{0.36} \therefore \rho_{l}=1.77$
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