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

Answer

A, B, C

Explanation

The diagrammatic representation of the given problem is shown in figure. The expression of fundamental frequency is

`v=1/(2l) sqrt(T/(mu))`
In air, `T=mg= (Vrho)g`
` :. v = 1/(2l) sqrt(Vrhog)/(mu)`
When the object is half immersed in water, then
`T' = mg-upthrust =Vrhog - (V/2)(rho)_(w)g=((v/2))g(2rho-rho_(w))`
the new fundemental frequency is
`v=1/(2l) xx sqrt((T')/(mu))=1/(2l)(sqrt(((Vg)/2)(2rho-rho_w))/(mu))`
` :. (v'/v)= (sqrt((2rho-rho_w)/(2rho)))` or `v' =v((2rho-rho_w)/(2rho))^1//2`
`=300((2rho-1)/(2rho))^1//2 Hz`
`:. The correct option is (a).`