An open-ended U-tube of uniform cross-sectional area contains water (density $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800 \mathrm{~kg} \mathrm{~m}^{-3}$ is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio $\left(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}\right)$ of the heights of the liquid in the two arms is-

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Fluid Mechanics

An open-ended U-tube of uniform cross-sectional area contains water (density

1 0^{3} kg m^{- 3}

). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density

800 kg m^{- 3}

is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio

(\frac{h _{1}}{h _{2}})

of the heights of the liquid in the two arms is-

(A) $\frac{15}{14}$
(B) $\frac{35}{33}$ ✓ Correct
(C) $\frac{7}{6}$
(D) $\frac{5}{4}$

Answer: (B)

Solution

From conservation of volume: h₁ + h₂ = 0.29 × 2 + 0.1 = 0.68 m. Applying pressure balance at the bottom level: P₀ + ρ_k g(0.1) + ρ_w g(h₁ - 0.1) = P₀ + ρ_w g h₂. This simplifies to: 800(0.1) + 1000h₁ - 1000(0.1) = 1000h₂, giving h₁ - h₂ = 0.02 m. Solving the two equations: h₁ = 0.35 m and h₂ = 0.33 m. Therefore, h₁/h₂ = 35/33.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Single correct option

Practise more: all JEE Advanced PYQs on Fluid Mechanics →

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Fluid Mechanics

Solution

More JEE Advanced questions on Fluid Mechanics