Question

A liquid $X$ of density $3.36\text{g/cc}$ is poured in a $U-$ tube (containing mercury) in the right arm upto height $10\text{cm}$ above the mercury level. Another liquid $Y$ is poured in the left arm upto height $8\text{cm}$ as shown in figure. Upper levels of $X$ and $Y$ are the same. What is the density of liquid $Y$? [Take ${\rho }_{Hg}=13.6\text{g/cc}$]

• A. $0.8\text{g/cc}$
• B. $1.2\text{g/cc}$
• C. $1.4\text{g/cc}$
• D. $1.6\text{g/cc}$

Answer 1

The correct option is A $0.8\text{g/cc}$

Pressure at same level in the same liquid will be equal.
Hence, in the two arms of the $U-$tube, pressure will remain same on the surface $P{P}^{\prime }$,as shown in figure.


$\Rightarrow P_{atm}+{\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2=P_{atm}+{\rho }_{3}g{h}_3...\left(i\right)$
Here,
$P_{atm}=$ atmospheric pressure acting on the open ends of $U-$tube
${\rho }_1=$density of liquid Y
$h_1=8\text{cm}$
${\rho }_2={\rho }_{Hg}=13.6\text{g/cc}$
$h_2=10-8=2\text{cm}$
${\rho }_3={\rho }_X=3.36\text{g/cc}$
$h_3=10\text{cm}$
Substituing in Eq.$\left(i\right)$,
$P_{atm}+({\rho }_1\times g\times 8)+(13.6\times g\times 2)=P_{atm}+(3.36\times g\times 10)$
$\Rightarrow 8{\rho }_1+2\times 13.6=10\times 3.36$
$\Rightarrow {\rho }_1=\frac{33.6-27.2}{8}=0.8\text{g/cc}$