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A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is
Explanation
Equate the pressures at the oil–water interface level in both arms.
Detailed Solution
Oil column height = 65 + 65 + 10 = 140 mm; water column above the interface level = 130 mm.
$h_{oil}\rho_{oil}g = h_{water}\rho_{water}g$
$140\times\rho_{oil} = 130\times\rho_{water}$
$\rho_{oil} = \frac{13}{14}\times1000\ kg/m^3$
$\rho_{oil} = 928\ kg\ m^{-3}$