Question

A column of mercury of length $h=10cm$ is contained in the middle of a narrow horizontal tube of length $1m$ closed at ends. The air in both halves of the tube is under a pressure of $P_0=76cm$ of mercury. The tube is now slowly made vertical. The distance moved by mercury will be

• A. 4.5 cm
• B. 3.0 cm
• C. 2.5 cm
• D. 1.2 cm

Answer 1

Answer: (B)

Figure (a) shows the horizontal position and figure (b) shows the vertical position of the tube.
When the tube is horizontal, the volume of air at the two sides of mercury column
$=0.45x$ $\alpha$,
where $\alpha$ is the area of cross-section of the tube.
The pressure of air at each side $=76cm$ of $Hg$
$=0.76m$ of $Hg$
Now, for the vertical position of the tube, let the mercury be displaced by $x$ metre.
Then, the volume of the air at the upper part $=(0.45+x)\alpha$
If the new pressure of air at the upper part be $p_1$, then from Boyle's law, we get
$0.76\times 0.45\times \alpha =p_1\times (0.45+x)\times \alpha$
or, $p_1=\frac{0.76\times 0.5}{0.45-x}\dots$.(i)
Volume of air at the lower part of the tube $=(0.45-x)\alpha$
If the new pressure of air at this part be $p_2$,
then applying Boyle's law, we get
$0.76\times 0.45\times \alpha =p_2\times (0.45-x)\times \alpha$
or, $p_2=\frac{0.76\times 0.45}{0.45-x}\dots$ .(ii)
Now, obviously, $p_2>p_1$ and the difference in pressure between the lower and upper parts of the tube,
i.e. $\left(p_2-p_1\right)$ will be due to the mercury column of $0.1m$ in its vertical position.
$\therefore p_2-p_1=0.1\dots$ (iii)
From (i) and (ii), we get
$p_2-p_1=\frac{0.76\times 0.45}{0.45-x}-\frac{0.76\times 0.45}{0.45+x}\dots$ (iv)
$=\frac{(0.76\times 0.45)(0.45+x)-(0.76\times 0.45)(0.45-x)}{(0.45{)}^2-x^2}$
$=\frac{0.76\times 0.45\times 2x}{(0.45{)}^2-x^2}$
Now, from (iii) and (iv), we get
$\frac{0.76\times 0.45\times 2x}{(0.45{)}^2-x^2}=0.1$
or, $x^2+6.48x-0.2025=0$
or, $x=\frac{-6.84\pm \sqrt{(6.84{)}^2-4\times 1\times (0-0.2025)}}{2\times 1}$
or, $x=0.029,-6.87$.
Negative value of $x$ is discarded as it is absurd.
$\therefore x=0.029=2.9cm$.
So, mercury will be displaced by $2.9cm$ (nearly).