Question

A tube U - shaped has a uniform cross - section with arm length $l_1$ and $l_2(l_1>l_2)$. Tube has a liquid of density ${\rho }_1$ filled to a height h. Another liquid of density ${\rho }_2=\frac{{\rho }_1}{2}$ is poured in arm . Both liquids are immiscible. The length of the second liquid that should be poured in A so that first overtone of is in unison with fundamental tone of B is

• A. $\frac{3}{2}(l_1-3l_2-2h)$
• B. $\frac{3}{2}(l_1-3l_2+2h)$
• C. $\frac{2}{3}(l_1+3l_2+2h)$
• D. $\frac{2}{3}(l_1-3l_2+2h)$

Answer 1

The correct option is D

$\frac{2}{3}(l_1-3l_2+2h)$

Let be the length of second liquid poured in A. Let the first liquid come down by a level in arm and rises by x in arm B.

$\therefore$ Pressure at C = Pressure at D.

$\therefore {\rho }_{2}gy={\rho }_{1}g\left(2x\right)$

$\Rightarrow x=\frac{{\rho }_2}{2{\rho }_1}y=\frac{y}{4}$ $(\because {\rho }_2=\frac{{\rho }_1}{2})$

Length of air column in arm A is

$\Rightarrow l_4=(l_i-h)-(y-\frac{y}{4})=l_1-h-\frac{3y}{4}$

Length of air column in arm B is

$l_s=l_2-h+\frac{y}{4}$

Since first overtone of arm A is in unison with fundamental tone of B.

$\therefore 3\frac{V}{4l_4}=\frac{V}{4l_B}$

$\Rightarrow l_A=3l_B\Rightarrow l_1-h-\frac{3y}{4}=3(l_2-h+\frac{y}{4})\Rightarrow y=\frac{2}{3}(l_1-3l_2+2h)$