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Step-by-Step Solution
Step 1: Express frequency in terms of tension
For a sonometer wire, the frequency (f) is proportional to the square root of the tension (T) in the wire and inversely proportional to the square root of its linear mass density ($\mu$). Symbolically:
$ f \propto \sqrt{\frac{T}{\mu}}.$
When the tension is provided by a weight of density $\rho$ and volume $V$, the tension is
$T = V\,\rho\,g,$
so
$ f \propto \sqrt{\frac{V\,\rho\,g}{\mu}}.$
Step 2: Frequency when weights are immersed in water
When the weight is completely immersed in water (density = 1), a buoyant force $V \times 1 \times g$ acts upward, reducing the effective tension to
$T_{\text{water}} = V(\rho - 1)\,g.$
The new frequency is given as $f_{\text{water}} = \frac{f}{2}.$
Using proportionality of frequency to the square root of tension:
$ \left(\frac{f_{\text{water}}}{f}\right)^2
= \frac{T_{\text{water}}}{T}
= \frac{V(\rho - 1)\,g}{V\,\rho\,g}
= \frac{\rho - 1}{\rho}.$
Given $f_{\text{water}} = \frac{f}{2},$ we have
$\left(\frac{1}{2}\right)^2 = \frac{\rho - 1}{\rho}.$
So:
$ \frac{1}{4} = \frac{\rho - 1}{\rho}
\implies \rho - 1 = \frac{\rho}{4}
\implies \rho - \frac{\rho}{4} = 1
\implies \frac{3\rho}{4} = 1
\implies \rho = \frac{4}{3}.$
Step 3: Frequency when weights are immersed in the unknown liquid
Let the unknown liquid have density $\sigma.$ The weight will now have effective tension
$T_{\text{liquid}} = V(\rho - \sigma)\,g.$
The observed frequency is $f_{\text{liquid}} = \frac{f}{3}.$
Again, using the same proportionality:
$ \left(\frac{f_{\text{liquid}}}{f}\right)^2
= \frac{T_{\text{liquid}}}{T}
= \frac{V(\rho - \sigma)\,g}{V\,\rho\,g}
= \frac{\rho - \sigma}{\rho}.$
Given $f_{\text{liquid}} = \frac{f}{3},$ we have
$\left(\frac{1}{3}\right)^2 = \frac{\rho - \sigma}{\rho} \implies \frac{1}{9} = \frac{\rho - \sigma}{\rho}.$
Substitute $\rho = \frac{4}{3}:$
Step 4: Solve for $\sigma$
$ \frac{4}{3} - \sigma = \frac{4}{3} \times \frac{1}{9} = \frac{4}{27}. $
Thus:
$ \sigma = \frac{4}{3} - \frac{4}{27}
= \frac{36}{27} - \frac{4}{27}
= \frac{32}{27}
\approx 1.185.$
The specific gravity of the liquid is therefore 1.185.
