$\begin{array}{1 1}(A)\;1.4\times 10^4sec\\(B)\;2.8\times 10^4sec\\(C)\;0.7\times 10^4sec\\(D)\;1.4\times 10^5sec\end{array} $

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Particles passing through hole outside per unit time is

$\large\frac{1}{2}$$n_1\sqrt{\large\frac{KT}{m}}\Delta ta$

$n_1$-number of particles per unit area

$a$-area of hole

Net out flow $\Rightarrow \large\frac{1}{2}$$(n_1-n_2)\sqrt{\large\frac{KT}{m}}\Delta a$

$n_1=\large\frac{P_1NA}{RT}$

After time $n_1'=\large\frac{P_1'NA}{RT}$

$n_1-n_1'$=Particles gone out

$\Rightarrow \large\frac{1}{2}$$(n_1-n_2)\sqrt{\large\frac{KT}{m}}$$zA$

$\large\frac{P_1N_AV}{RT}-\frac{P_1'N_AV}{RT}=\large\frac{1}{2}$$(P_1-P_2)\large\frac{N_A}{RT}\sqrt{\frac{KT}{m}}$$zA$

$z=2\bigg[\large\frac{P_1-P_1'}{P_1-P_2}\bigg]\frac{V}{a}\sqrt\frac{m}{KT}$

$z=1.38\times 10^5$s

$z\approx 1.4\times 10^5sec$

Hence (D) is the correct answer.

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