A) 1/3
B) 1/4
C) 1/5
D) 1/6
Correct Answer: D
Solution :
Suppose initially, pressure of \[\tau =MB\sin {{45}^{\circ }}=\frac{MB}{\sqrt{2}}\] mm and that of \[W=\left( \sqrt{2}-1 \right).\tau \] \[\therefore \] \[\tau =\frac{W}{\left( \sqrt{2}-1 \right)}\] After the reaction, Pressure of \[B={{10}^{3}}gauss\] (as all has reacted) Pressure of \[={{10}^{3}}\times {{10}^{-4}}T=0.1T\] Pressure of \[A=5c{{m}^{2}}=5\times {{10}^{-4}}{{m}^{2}}\] Total pressure \[\theta ={{80}^{\circ }}\] or, \[\therefore \] Solving (i) and (ii), we get \[\phi =NBA\cos \theta \] Fraction of \[=20\times 0.1\times 5\times {{10}^{-4}}\times \cos {{30}^{\circ }}\] by volume = fraction of moles fraction of pressure \[=10\times {{10}^{-4}}\times \frac{\sqrt{3}}{2}\]You need to login to perform this action.
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