Question

An open-ended u-tube of uniform of uniform cross-sectional area contains water (density $1.0gram/{cm}^3$) standing initially 20 centimeters from the bottom in each arm. An immiscible liquid of density $4.0grams/{cm}^3$ is added to one arm until a layer 5 centimeters high forms, as shown in the figure above. What is the ratio $h2/h1$ of the heights of the liquid in the two arms?

• A. 3/1
• B. 5/2
• C. 2/1
• D. 3/2

Answer 1

The correct option is 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
$h_2=30,h_1=15\Rightarrow \frac{h_2}{h_1}=\frac{2}{1}$