SATHEE: Properties of Matter 2 Question 23

Properties of Matter 2 Question 23

23. A column of mercury of length $10 c m$ 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,

$\begin{aligned} p_{0} (45) (A) & = p_{1} (45 - x) A \\ or 76 \times 45 & = p_{1} (45 - x) \dots (i) \\ p_{0} (45) (A) & = p_{2} (45 + x) A \\ 76 \times 45 & = p_{2} (45 + x) \dots (i i) \end{aligned}$

From Eqs. (i) and (ii), we get

$(p_{1} - p_{2}) = 76 \times 45 (\frac{1}{45 - x} - \frac{1}{45 + x})$

From figure (b),

$(p_{1} - p_{2}) A =$ Weight of $10 c m$ of Hg column

or $p_{1} - p_{2} =$ Pressure equivalent to $10 c m$ of $H g$ column

$76 \times 45 (\frac{1}{45 - x} - \frac{1}{45 + x}) = 10$

Solving this equation, we get

$x = 2.95 c m$

Properties of Matter 2 Question 23

Answer:

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Properties of Matter 2 Question 23

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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