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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$300 (\frac{2 ρ - 1}{2 ρ})$

$300 {(\frac{2 ρ - 1}{2 ρ})}^{1 / 2}$

$300 {(\frac{2 ρ}{2 ρ - 1})}^{1 / 2}$

$300 (\frac{2 ρ}{2 ρ - 1})$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In air : T = mg = ρVg

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

In water : T = mg – upthrust

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

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