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(B) The capillary rise $h_0$ under atmospheric pressure $P_0$ is given by:$h_0 = \frac{2 T \cos 	heta}{ho g r} = \frac{2 	imes 0.075 	imes 1}{10^

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(B) The capillary rise $h_0$ under atmospheric pressure $P_0$ is given by:$h_0 = \frac{2 T \cos 	heta}{ho g r} = \frac{2 	imes 0.075 	imes 1}{10^3 	imes 10 	imes 10^{-4}} = 0.15 \,m = 15 \,cm$When the air is isothermally compressed from $V_0$ to $\frac{100}{101} V_0$,the new pressure $P$ inside the container is:$P_0 V_0 = P \left( \frac{100}{101} V_0 ight) \Rightarrow P = \frac{101}{100} P_0 = 1.01 P_0$The pressure balance at the liquid surface inside the capillary is:$P_0 - \frac{2 T \cos 	heta}{r} + ho g h = P$Substituting $P = 1.01 P_0$ and $\frac{2 T \cos 	heta}{r} = ho g h_0$:$P_0 - ho g h_0 + ho g h = 1.01 P_0$$ho g h = 0.01 P_0 + ho g h_0$$h = h_0 + \frac{0.01 P_0}{ho g} = 0.15 \,m + \frac{0.01 	imes 10^5}{10^3 	imes 10} = 0.15 \,m + 0.1 \,m = 0.25 \,m = 25 \,cm$

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