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(B) Let the density of the liquid be $\rho_l = 1.2 \ g \ cm^{-3}$ and the density of the oil be $\rho_o = 0.9 \ g \ cm^{-3}$.When the liquid rises by

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(B) Let the density of the liquid be $ho_l = 1.2 \ g \ cm^{-3}$ and the density of the oil be $ho_o = 0.9 \ g \ cm^{-3}$.When the liquid rises by $15 \ cm$ on the right side, it must have fallen by $15 \ cm$ on the left side relative to the initial equilibrium level.Thus, the total difference in the liquid levels between the two arms is $h_l = 15 \ cm + 15 \ cm = 30 \ cm$.Let $h_o$ be the height of the oil column on the left side. The pressure at the interface of the oil and liquid on the left side must equal the pressure at the same horizontal level on the right side.Using the hydrostatic pressure balance: $h_o ho_o g = h_l ho_l g$.Substituting the values: $h_o 	imes 0.9 = 30 	imes 1.2$.$h_o = \frac{30 	imes 1.2}{0.9} = 40 \ cm$.The oil column stands $40 \ cm$ above the initial liquid level on the left. Since the liquid on the left fell by $15 \ cm$, the top of the oil column is $40 - 15 = 25 \ cm$ above the initial level.The liquid on the right is $15 \ cm$ above the initial level.Therefore, the difference in height between the oil level and the right liquid level is $25 \ cm - 15 \ cm = 10 \ cm$.

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