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Physics · Fluid Mechanics

JEE Advanced 2020 — Paper 1 — Question 5

An open-ended U-tube of uniform cross-sectional area contains water (density 103 kg m−310^{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 kg m−3800 \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 (h1h2)\left(\frac{h_{1}}{h_{2}}\right) of the heights of the liquid in the two arms is

Question figure
  1. Option A:

    1514\frac{15}{14}

  2. Option B:

    3533\frac{35}{33}

    Correct
  3. Option C:

    76\frac{7}{6}

  4. Option D:

    54\frac{5}{4}

Answer: B

Step-by-step solution

800×0.1+1000×x=1000×(0.58−x)800 \times 0.1+1000 \times \mathrm{x}=1000 \times(0.58-\mathrm{x})

2000x=5002000 \mathrm{x}=500

x=0.25 m\mathrm{x}=0.25 \mathrm{~m}

So, h1=0.35 m, h2=0.33 m\mathrm{h}_{1}=0.35 \mathrm{~m}, \mathrm{~h}_{2}=0.33 \mathrm{~m}

h1 h2=3533\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{35}{33}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Physics
Chapter
Fluid Mechanics
Topic
Variation of Static Pressure Inside a Liquid