| A horizontal uniform glass tube of 100 cm, length sealed at both ends contain 10 cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively \[81{}^\circ C\] and 76 cm of mercury. If the air column at one end is kept at \[0{}^\circ C\] and the other end at \[273{}^\circ C\], the pressure of air which is at \[0{}^\circ C\] is (in cm of Hg) |
 |
A) 76
B) 68.2
C) 102.4
D) 122
Correct Answer: C
Solution :
| \[{{\ell }_{1}}={{\ell }_{2}}=45\operatorname{cm}\] |
| The pressure must be same in both sides. |
| Hence |
| \[\frac{{{\ell }_{1}}}{{{T}_{1}}}=\frac{{{\ell }_{2}}}{{{T}_{2}}}\Rightarrow \frac{{{\ell }_{1}}}{273}=\frac{{{\ell }_{2}}}{273+273}\Rightarrow {{\ell }_{1}}\frac{{{\ell }_{2}}}{2}\] |
| Also \[{{\ell }_{1}}+{{\ell }_{2}}=90\]; |
| \[\therefore \]\[{{\ell }_{1}}=30\operatorname{cm}\]and\[{{\ell }_{2}}=60\operatorname{cm}\] |
| Now \[\frac{{{P}_{1}}({{\ell }_{1}}A)}{{{T}_{1}}}=\frac{P(\ell A)}{T}\] |
| or \[\frac{{{P}_{1}}\times 30}{273}=\frac{76\times 45}{273\times 81}\Rightarrow {{P}_{1}}=102.4\operatorname{cm}\] |
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