Question

A column of mercury of 10cm length is contained in the middle of a narrow horizontal 1m long tube which is closed at both the ends. Both the halves of the tube contain air at a pressure of 76 cm of mercury. By what distance will the column of mercury be displaced if the tube is held vertically?

• A. x = 5.7 cm.
• B. x=2.95
• C. x=3.15
• D. x=4.1

Answer 1

Answer: (B)

x=2.95

M is the mid-point of tube AB.

At equilibrium

$p_1\times A+mg=p_2\times A$

$p_1\times A+10\times A\times d_{Hg}\times g=p_2\times A$

$=p_1+10d_{Hg}\times g=p_2$ …(i)

For air present in column AP

$p\times 45\times A=p_1\times (45+x)\times A$

$p1=\frac{45}{45+x}\times 76d_{Hg}\times g$ ...(ii)

For air present in column QB

$p\times 45\times A=p_2\times (45-x)\times A$

$p_2=\frac{45}{45-x}\times 76d_{Hg}\times g$ ...(iii)

From (i), (ii) and (iii)

$\frac{45\times 76\times d_{Hg}\times g}{45+x}+10d_{Hg}\times g=\frac{45}{45-x}\times 76\times d_{Hg}\times g$

$=\frac{45\times 76}{45+x}+10=\frac{45\times 76}{45-x}$

$x=2.95$cm.

Hence, te correct option is $\text{B}$