91Ó°ÊÓ

The frequency of a sonometer wire is related to the tension (T) in the wire by the following formula: \[f = \sqrt{\frac{T}{mL}}\] where f is the frequency, m is the mass per unit length, and L is the length of the wire. When the weights are in water, the frequency is 80 Hz, and when the weights are in the unknown liquid, the frequency is 60 Hz. We can find the ratio of tensions in the wire in these two situations using the given frequencies: \[\frac{T_{water}}{T_{liquid}} = \frac{f_{water}^2}{f_{liquid}^2} = \frac{80^2}{60^2} = \frac{4}{3}\]

Since the tensions in the wire depend on atmospheric pressure, the ratio of specific gravities can be obtained using Archimedes' principle: \[\frac{SG_{liquid}}{SG_{water}} = \frac{T_{water} - T_0}{T_{liquid} - T_0}\] where SG is the specific gravity and T0 is the tension in the wire when there is no immersion. Rearranging the equation, we get: \[\frac{SG_{liquid} - 1}{SG_{water} - 1} = \frac{T_{water} - T_0}{T_{liquid} - T_0}\] Since SG of water is 1, the equation simplifies to: \[\frac{SG_{liquid} - 1}{1- 1} = \frac{T_{water}}{T_{liquid}}\] Thus, the equation becomes: \[SG_{liquid} = 1 + \frac{T_{water}}{T_{liquid}}\]

Now, we can find the specific gravity of the unknown liquid using the relationship we derived in step 2: \[SG_{liquid} = 1 + \frac{T_{water}}{T_{liquid}} = 1 + \frac{4}{3} = \frac{7}{3} \approx 2.33\] This value doesn't match any of the given answer choices, so we may need to rethink the problem. The potential cause of the discrepancy may be the way the frequencies are calculated. In the sonometer wire problem, a standing wave is formed, whose frequency depends on the string's tension, length, and linear mass density. For a given length and linear mass density, we are interested in determining the relationship between the frequency and tension when the wire is immersed in different fluids. If we consider that the end of the wire is free, the frequency is given by the following equation: \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\) where T is the tension in the wire, L is the length of the wire, and μ is the linear mass density. Based on this equation, we can now rewrite step 1: \[\frac{T_{water}}{T_{liquid}} = \frac{f_{water}^2}{f_{liquid}^2} = \frac{80^2}{60^2} = \frac{16}{9}\] Following the same calculations in step 2 and step 3, we find the specific gravity of the liquid: \[SG_{liquid} = 1 + \frac{T_{water}}{T_{liquid}} = 1 + \frac{16}{9} = \frac{25}{9} = 2.77\] This value does not match any of the given answer choices either, so there might be a mistake in the problem statement or the answer choices offered.