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 frequency 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

If `T = mg = vrhog`
`:. f = (1)/(2l) sqrt((T)/(mu)) = 300 ….(1)`
Now `T' = mg - f_(b) = vrhog - (V)/(2)g`.
`T' = vg((2rho - 1)/(2))`
`:. f' = 1/(2l)sqrt((vg)/(mu)((2rho-1))/(2))"……"(2)`
`:. (f')/(f) ((2rho - 1)/(2rho))^(1/2) , f' = 300 ((2rho - 1)/(2rho))^(1/2)`