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(B) In the first case,the pressure of the gas $P_1$ is given by $P_1 = P_0 + \frac{Mg}{A}$.In the second case,when the tube is inverted,the pressure o

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$\frac{101}{99}$

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(B) In the first case,the pressure of the gas $P_1$ is given by $P_1 = P_0 + \frac{Mg}{A}$.In the second case,when the tube is inverted,the pressure of the gas $P_2$ is given by $P_2 = P_0 - \frac{Mg}{A}$.Since the process is isothermal,$P_1 V_1 = P_2 V_2$. Since $V = A 	imes L$,we have $P_1 L_1 = P_2 L_2$.Therefore,$\frac{L_2}{L_1} = \frac{P_1}{P_2} = \frac{P_0 + \frac{Mg}{A}}{P_0 - \frac{Mg}{A}} = \frac{P_0 A + Mg}{P_0 A - Mg}$.Substituting the values: $P_0 A = 10^5 	imes 20 	imes 10^{-6} = 2 \, N$ and $Mg = 0.002 	imes 10 = 0.02 \, N$.$\frac{L_2}{L_1} = \frac{2 + 0.02}{2 - 0.02} = \frac{2.02}{1.98} = \frac{202}{198} = \frac{101}{99}$.

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