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A glass tube of uniform area of cross section and open at one end encloses some air at 27^@C by a 4 cm long mercury thread that acts like a piston. When the tube is held vertical with its open end up length of the air column in the tube is 9 cm.When the open end is held downwards by turning the tube,the length of the enclosed air column becomes 10 cm. Find () the value of the atmospheric pressure (ii) the temperature at which the length of the air column becomes 9cm again,while the tube is still held inverted.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Let the <a href="https://interviewquestions.tuteehub.com/tag/atmospheric-887165" style="font-weight:bold;" target="_blank" title="Click to know more about ATMOSPHERIC">ATMOSPHERIC</a> pressure =pcmHg and the area of cross section of the tube =`a cm^2`[Fig.6.10] when the open end is kept upwards the <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of the confined air `V_1=9 a cm^3` and pressure `p_1=(p+4) cmHg`. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P2_U07_C06_SLV_033_S01.png" width="80%"/> <br/> On <a href="https://interviewquestions.tuteehub.com/tag/inverting-7690921" style="font-weight:bold;" target="_blank" title="Click to know more about INVERTING">INVERTING</a> the tube volume of the confined air `V_2=10 a cm^3` and pressure `p_2(p-4)`cmHg. <br/> Assuming temperature to be constant at `27^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` from Boyle.s law<br/> `p_1V_1=p_2V_2` <br/> or,`9a(p+4)=10a(p-4)` <br/> or,`9p+36=10p-40 , or p=76 cmHg` <br/> Suppose at `T_2K` the length of the air column becomes 9 cm again when the tube is still held inverted.Then it pressure `p_2=76-4=72 cmHg` and volume `V_2=9a cm^3`. <br/> `therefore(p_1V_1)/T_1=(p_2V_2)/T_2` <br/> or,`((76+4)times9a)/(273+27)=(72times9a)/T_2or,80/300=72/T_2`<br/>`thereforeT_2=270K=(270-273)^@C=-3^@C`</body></html>


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