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

We know that
`f = (1)/(2l) sqrt((T)/(m))`
In air, `T = mg = rho Vg`
`:. f =(1)/(2l) sqrt((rho Vg)/(m))`
In water, `T = mg - "upthrust"`
=` V rhog -(V)/(2) rho_(omega)g =(Vg)/(2)(2 rho-rho_(omega))`
Therefore,
`:. f' =(1)/(2l) sqrt(((Vg)/(2)(2 rho-rho_(omega)))/(m))`...(i)
=`(1)/(2l) sqrt((Vg rho)/(m)) sqrt(((2 rho-rho omega))/(2 rho)) = 300 [(2 rho-1)/(2 rho)]^((1)/(2))`
`because rho_(omega) = 1 g// c c` and from Eqs (i).