Question

An open-ended U-tube of uniform cross sectional area contains water (density $1.0{\text{gm/ cm}}^3$) standing initially $20\text{cm}$ from the bottom in each arm. An immiscible liquid of density $4.0{\text{gm/ cm}}^3$ is added to one arm until a layer $5\text{cm}$ high forms, as shown in the figure below. What is the ratio $\frac{h_2}{h_1}$ of the heights of the liquid in the two arms?

• A. $\frac{3}{1}$
• B. $\frac{5}{2}$
• C. $\frac{2}{1}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{2}{1}$

Given that initially water level in each limb is $20cm$, so total length of water column is $40cm$.

As liquid is poured in one limb, let water level in one limb be $x$, then water level in the other limb is $40-x$

Using Pascal's law,
$P_A=P_B$
Using the expression $P=h\rho g$,
$\Rightarrow 5\times 4\times g+x\times 1\times g=(40-x)\times 1\times g\Rightarrow x=10\text{cm}$

Now from diagram,
$h_1=x+5=15\text{cm}$
$h_2=40-x=30\text{cm}$
$\frac{h_2}{h_1}=2$