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

We know that `f=(1)/(2 1)sqrt((T)/(mu))`
In air,`T=mg=(rho Vgrho_w)`
`thereforef=(1)/(2 1) sqrt((rhoVg)/(m))rho_w`
In water ` T=m g-`up thrust =`rho_wVrhog-v/2rho_w g=(vg)/2(2p-1)rho_w`
Therefore
`therefore f'=(1)/(2 1)sqrt(((vg)/2(2rho-rho_w))/(mu))=(1)/(2 1)sqrt(Vgrho)/(mu)rho_w sqrt(((2rho-
1_w))/(2rho))=300[(2rho-1)/2rho]^(1/2)` therefore rho_w=1 g//infty` and from Eq (i)