Question

An open glass tube is immersed in mercury in such a way that a length of $8\text{cm}$ extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by an additional $46\text{cm}$. What will be length of the air column above mercury in the tube now?

[Atmospheric pressure $=76\text{cm of Hg}$ and assume temperature is constant]

• A. $16\text{cm}$
• B. $20\text{cm}$
• C. $14\text{cm}$
• D. $18\text{cm}$

Answer 1

The correct option is A $16\text{cm}$

After tube is raised vertically, total length of tube above water $=46+8=54\text{cm}$
Let '$x$' be the length of air column in the tube when open end is closed.


From diagram $B$,
$P_0=P_2+(54-x)$ ...$\left(1\right)$

From diagram $A$ and $B$,
$P_1{V}_1=P_2{V}_2$ [assuming temperature is constant]
$\Rightarrow P_0\times 8A=P_2\times xA$
where $A$ is cross-sectional area of tube.

$\Rightarrow P_2=\frac{8P_0}{x}$ ...$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$P_0=\frac{8P_0}{x}+(54-x)$
$\Rightarrow P_{0}x=8P_0+54x-x^2$
$\Rightarrow x^2-54x+P_{0}x-8P_0=0$
$(P_0=76\text{cm of Hg})$
$\Rightarrow x^2-54x+76x-8\times 76=0$
$\Rightarrow x^2+22x-608=0$
$\Rightarrow x^2-16x+38x-608=0$
$\Rightarrow x(x-16)+38(x-16)=0$
$\Rightarrow (x-16)(x+38)=0$
$\Rightarrow x=16\text{cm}orx=-38\text{cm}$

Height cannot be negative. Thus, length of the air column above mercury in the tube will be $16\text{cm}$.

Why this question? Concept involved - Variation of pressure with depth. Caution - Always take the pressure in the same unit. Importance in $JEE$ - Asked in Mains $2019$.