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.
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
