A tube U-shaped has a uniform cross-sectionn with arm length `l_(1) and l_(2) (l_(1)gtl_(2))`. Tube has a liquid of density `rho_(1)` filled to a height h. another liquid of density `rho_(2)=(rho_1)(1)/(4)` is poured in arm A. both liquids are immiscible. the length of second liquid that should be poured in A so that first overtone of A is in unisom with fundamental tone of B is:

Options

A. `2(J_(1)-3J_(2)-2h)`
B. `2(J_(1)-3J_(2)+2h)`
C. `2(J_(1)+3J_(2)+2h)`
D. `2(J_(1)-3J_(2)+2h)`

Answer

B

Explanation

First overtone of A=fundamental tone of B
`implies (3V)/(4l_(A))=(V)/(4l_(B))implies (l_(A))/(l_(B))=3`
Let l length of liquid be poured.
Using `P_(A)=P_(B)`
`implies P_(0)+(rho_(1))/(4)gl=P_(0)+rho_(1)g(2x) implies x=(l)/(8)`
Now, `l_(A)+l+(h-x)=l_(1)`
And `l_(B)+x+h=l_(2)` Solving we can get: `l=2(l_(1)-3l_(2)+2h)`.