Import Question JSON

Let the length of the glycerin column be $h_g = 10 \text{ cm}$ and its density be $\rho_g = 1.3 \text{ g/cm}^3$. Let the length of the oil column be $h_{oil}$ and its density be $\rho_{oil} = 0.8 \text{ g/cm}^3$. Let the density of mercury be $\rho_{Hg} = 13.6 \text{ g/cm}^3$. The upper surfaces of the oil and glycerin are at the same horizontal level. At the condition of equilibrium, the pressure at the same horizontal level in the mercury is equal. Let's consider two points, A and B, at the interface of mercury and the other liquids in the two arms. The pressure at point A must be equal to the pressure at point B. Pressure at point A ($P_A$) is the pressure due to the glycerin column. $P_A = \rho_g h_g g = (1.3 \text{ g/cm}^3) \times (10 \text{ cm}) \times g$ Pressure at point B ($P_B$) is the pressure due to the oil column plus the pressure due to the mercury column of height $(h_g - h_{oil})$ below the oil. $P_B = \rho_{oil} h_{oil} g + \rho_{Hg} (h_g - h_{oil}) g$ Equating $P_A$ and $P_B$ and cancelling $g$ from both sides: $1.3 \times 10 = 0.8 \times h_{oil} + 13.6 \times (10 - h_{oil})$ $13 = 0.8h_{oil} + 136 - 13.6h_{oil}$ $13 = 136 - 12.8h_{oil}$ $12.8h_{oil} = 136 - 13$ $12.8h_{oil} = 123$ $h_{oil} = \frac{123}{12.8} \approx 9.609 \text{ cm}$ By solving, we get the length of the oil column as $9.6 \text{ cm}$.