Properties of Matter 2 Question 23
23. A column of mercury of length $10 cm$ is contained in the middle of a horizontal tube of length $1 m$ which is closed at both the ends. The two equal lengths contain air at standard atmospheric pressure of $0.76 m$ of mercury. The tube is now turned to vertical position. By what distance will the column of mercury be displaced? Assume temperature to be constant.
(1978)
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Answer:
Correct Answer: 23. 2.95 cm
Solution:
- Let area of cross-section of the tube be $A$.
(b)
Applying $p _1 V _1=p _2 V _2$ in $A$ and $B$ we have,
$$ \begin{aligned} & p _0(45)(A) & =p _1(45-x) A \\ & \text { or } \quad 76 \times 45 & =p _1(45-x) \cdots(i)\\ & p _0(45)(A) & =p _2(45+x) A \\ & 76 \times 45 & =p _2(45+x) \cdots(ii) \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ \left(p _1-p _2\right)=76 \times 45\left(\frac{1}{45-x}-\frac{1}{45+x}\right) $$
From figure (b),
$\left(p _1-p _2\right) A=$ Weight of $10 cm$ of Hg column
or $\quad p _1-p _2=$ Pressure equivalent to $10 cm$ of $Hg$ column
$$ 76 \times 45\left(\frac{1}{45-x}-\frac{1}{45+x}\right)=10 $$
Solving this equation, we get
$$ x=2.95 cm $$
