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

Answer

A

Explanation

`f=(1)/(2L)sqrt((T)/(mu))`
`f'=(2)/(2L)sqrt((T)/(mu)(1+(1)/(2rho)))rArrf'=fsqrt((2 rho-1)/(2 rho))`