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

Options

A. `300((2rho-1)/(2rho))^(1//2)`
B. `300((2rho)/(2rho-1))^(1//2)`
C. `300((2rho)/(2rho-1))`
D. `300((2rho-1)/(2rho))`

Answer

A

Explanation

(a) In air : `T = mg =rhoVg`…..(i)
In water : `T = mg - upthrust`
`= Vrhog - (V)/(2)rhoomegag = (Vg)/(2)(2rho - rhoomega)`
:
: `f' = (1)/(2l)sqrt((Vg)/(2)(2rho - rhoomega)/(m)) = (1)/(2l)sqrt((Vgrho)/(m)) sqrt((2rho - rhoomega))/(2rho)`
`(f')/(f) = sqrt((2rho - rhoomega)/(2rho)) f' = f ((2rho - rhoomega)/(2rho))^(1//2)`
`= 300[(2rho - 1)/(2rho)]^(1//2) Hz`