An oil of mass density 800 kg/$$m^3$$​ is contained in a vessel. Calculate the height of water required to develop an equivalent hydrostatic pressure as that developed by oil of height 30 m . Take acceleration due to gravity as

Correct option is A

​$$\text{Mass density of oil } = 800 \; \text{kg/m}^3 \\[6pt]\text{Acceleration due to gravity } g = 9.81 \; \text{m/s}^2 \\[6pt]\text{Height of oil column } = 30 \; \text{m} \\[6pt]\text{Mass density of water } = 1000 \; \text{kg/m}^3 \\[12pt]\text{Using Pascal's Law: Pressure at the same horizontal level is equal.} \\[6pt](P)_{\text{oil}} = (P)_{\text{water}} \\[10pt]\rho_{\text{oil}} g h_{\text{oil}} = \rho_{\text{water}} g h_{\text{water}} \\[10pt]800 \times 9.81 \times 30 = 1000 \times 9.81 \times h \\[10pt]\Rightarrow h = \frac{800 \times 30}{1000} = 24 \; \text{m}$$​​