Properties Of Matter Ques 23
- An open glass tube is immersed in mercury in such a way that a length of $8 \mathrm{~cm}$ extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional $46 \mathrm{~cm}$. What will be length of the air column above mercury in the tube now? (Atmospheric pressure $=76 \mathrm{~cm}$ of $\mathrm{Hg}$ )
(2014 Main)
(a) $16 \mathrm{~cm}$
(b) $22 \mathrm{~cm}$
(c) $38 \mathrm{~cm}$
(d) $6 \mathrm{~cm}$
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Answer:
Correct Answer: 23.( a )
Solution:
- Key Idea In this question, the system is accelerating horizontally i.e. no component of acceleration in vertical direction. Hence, the pressure in the vertical direction will remain unaffected.
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Again, we have to use the concept that the pressure in the same level will be same.
For air trapped in tube, $p_1 V_1=p_2 V_2$
$ \begin{gathered} p_1=p_{\text {atm }}=\rho g h\ \text{where}\ h=76\ \text{cmHg} V_1=A \cdot 8 \quad[A=\text { area of cross-section }] \\ p_2=p_{\text {atm }}-\rho g(54-x)=\rho g(22+x) \\ V_2 = A \cdot x \rho g 76 \times 8 A = \rho g(22 + x) A x \\ x^2+22 x-624=0 \quad \Rightarrow x=16 \mathrm{~cm} \end{gathered} $
BETA
