A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immisci

Question

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 $20 \mathrm{~mm}$ above the water level on the other side. Meanwhile, the water rises by $40 \mathrm{~mm}$ from its original level. The density of the oil is (Density of water $= 1000 \mathrm{~kg} \mathrm{~m}^{-3}$)

Professor PC Thomas Classes · Answer

(c) : The water level dropped by $40 \mathrm{~mm}$ on the oil side and rose by $40 \mathrm{~mm}$ on the other side. Thus, the total height of the water column balancing the oil is $h_w = 40 + 40 = 80 \mathrm{~mm}$. The oil level is $20 \mathrm{~mm}$ above the new water level, so the height of the oil column is $h_{oil} = 80 + 20 = 100 \mathrm{~mm}$. Equating the pressures at the interface level: $ho_{oil} g h_{oil} = ho_w g h_w \implies ho_{oil} 	imes 100 = 1000 	imes 80$. Solving this gives $ho_{oil} = 800 \mathrm{~kg} \mathrm{~m}^{-3}$.