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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2020 · Paper 1 · Physics · Mechanical Properties of Fluids

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

An open-ended U-tube of uniform cross-sectional area contains water (density 103kgm−3{10}^{3}kg{m}^{- 3} ). Initially the water level stands at 0.29m0 . 29m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density 800kgm−3800kg{m}^{- 3} is added to the left arm until its length is 0.1m0 . 1m, as shown in the schematic figure below. The ratio (h1h2)\left(\dfrac{{h}_{1}}{{h}_{2}}\right) of the heights of the liquid in the two arms is

An open-ended U-tube of uniform cross-sectional area contains water (density 10 ^ 3 kg m ^ - 3 ). Initially the water

Options

  1. A1514\dfrac{15}{14}
  2. B3533\dfrac{35}{33}
  3. C76\dfrac{7}{6}
  4. D54\dfrac{5}{4}

Correct answer

B. 3533\dfrac{35}{33}

Step-by-step solution

An open-ended U-tube of uniform cross-sectional area contains water (density 10 ^ 3 kg m ^ - 3 ). Initially the water

p1=p2{p}_{1} = {p}_{2}

⇒p0+800g×0.1+1000g(0.29−x)⇒ {p}_{0} + 800g \times 0 . 1 + 1000g ( 0 . 29 - x )=1000g(0.29+x)+p0= 1000g ( 0 . 29 + x ) + {p}_{0}

⇒0.8+2.9−10x=2.9+10x⇒ 0 . 8 + 2 . 9 - 10x = 2 . 9 + 10x

⇒x=0.820=0.04⇒ x = \dfrac{0 . 8}{20} = 0 . 04

h2=0.1+0.29−x=0.35{h}_{2} = 0 . 1 + 0 . 29 - x = 0 . 35

h1=0.29+x=0.33{h}_{1} = 0 . 29 + x = 0 . 33

h2h1=3533=1.06\dfrac{{h}_{2}}{{h}_{1}} = \dfrac{35}{33} = 1 . 06

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