An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency in Hz is

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$300 (\frac{2 ρ - 1}{2 ρ})$

$300 {(\frac{2 ρ - 1}{2 ρ})}^{1 / 2}$

$300 {(\frac{2 ρ}{2 ρ - 1})}^{1 / 2}$

$300 (\frac{2 ρ}{2 ρ - 1})$

answer is A.

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Detailed Solution

In air : T = mg = ρVg

$∴ f = \frac{1}{2 l} \sqrt{\frac{ρ V g}{m}} . . . . . (i)$

In water : T = mg – upthrust

$= V ρ g - \frac{V}{2} ρ_{ω} g = \frac{V g}{2} (2 ρ - ρ_{ω})$

$∴ f' = \frac{1}{2 l} \sqrt{\frac{\frac{V g}{2} (2 ρ - ρ_{ω})}{m}} = \frac{1}{2 l} \sqrt{\frac{V g ρ}{m}} \sqrt{\frac{(2 ρ - ρ_{ω})}{2 ρ}} \frac{f'}{f} = \sqrt{\frac{2 ρ - ρ_{ω}}{2 ρ}} \Rightarrow f' = f {(\frac{2 ρ - ρ_{ω}}{2 ρ})}^{\frac{1}{2}} = 300 {[\frac{2 ρ - 1}{2 ρ}]}^{\frac{1}{2}} H z$

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