Question

An open-ended U-tube of a uniform cross-secti0onal area contains water (density $1.0gram/{centimeter}^3$) standing initially $20$ centimeters from the bottom in each arm. An immiscible liquid of density $4.0gram/{centimeter}^3$ is added to one arm until a layer of $5$ centimeters high forms, as shown in the figure above. What is the ratio $h_2/h_1$ of the heights of the liquid in the two arms?

• A. $3/1$
• B. $5/2$
• C. $2/1$
• D. $3/2$

Answer 1

Answer: (C)

$2/1$

$h_1+h_2=20+20+5=45$ ...............$\left(I\right)$, since initially water was 20 cm in each arm.
Since pressure at bottom of both the limbs is equal after adding the immiscible liquid, we have
$5\times 4\times g+(h_1-5)\times 1\times g=h_2\times 1\times g$
$\Rightarrow h_2-h_1=15$ ...............$\left(II\right)$
from $\left(I\right)$ and $\left(II\right)$, we have
$h_2=30,h_1=15\Rightarrow \frac{h_2}{h_1}=\frac{2}{1}$