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An open pipe made up of glass is immersed in mercury in such a way that a length of 8 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 cm. What will be length of the air column above mercury in the tube now?(atm pressure =76 cm of HG)

Answer» <html><body><p>16cm<br/>22cm<br/>38cm<br/>6cm</p>Solution :Let a=cross sectional <a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> of the pipe (`i n cm^2`) <br/> So, initially volume of the above <a href="https://interviewquestions.tuteehub.com/tag/mercury-558950" style="font-weight:bold;" target="_blank" title="Click to know more about MERCURY">MERCURY</a> <br/> `V_1=<a href="https://interviewquestions.tuteehub.com/tag/8a-340116" style="font-weight:bold;" target="_blank" title="Click to know more about 8A">8A</a> cm^3`, pressure `=p_1 76 cmHg` <br/> When the top of the pipe is (46+8) or 54 cm higher than the outer mercury level [Fig.1.9] let x=rise of mercury level <a href="https://interviewquestions.tuteehub.com/tag/inside-1045864" style="font-weight:bold;" target="_blank" title="Click to know more about INSIDE">INSIDE</a> the pipe <br/> Then volume of air inside the pipe, `v_2=(54-x)a cm^3` <br/> and its pressure `p_2(76-x)cmHg` <br/> From Boyle.s law `p_1V_1=p_2V_2` <br/> or, `76 times 8a=(76-x)(54-x)a` <br/> or, `608=1404-130x+x^2` <br/> or, `x^2-130x+3496=0` <br/> or, `(x-38)(x-92)=0`, <br/> So x=38 cm or, x=92 cm <br/> The only physically meaningful solution is x=38 cm. Therefore the <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> of the air column in the pipe above mercury =`54-38=16 cm` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P2_U09_C01_E04_026_S01.png" width="80%"/></body></html>


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