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QJEEAdv20Ph0362: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph0362 Solution: An open-ended U-tube of uniform cross-sectional area contains…

An open-ended U-tube of uniform cross-sectional area contains water (density 103 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 (h1h2)\left( {{{{h_1}} \over {{h_2}}}} \right) of the heights of the liquid in the two arms is : JEE Advanced 2020 Paper 1 Offline Physics - Properties of Matter Question 34 English

Multiple Choice Options

  • A 1514{{15} \over {14}}
  • B 3533{{35} \over {33}}
  • C 76{7 \over 6}
  • D 54{5 \over 4}

Step-by-Step Verified Solution

We have, h1 + h2 = 0.29 ×\times 2 + 0.1

h1 + h2 = 0.68 .......(i)

⇒\Rightarrow p0 + ρ\rho kg(0.1) + ρ\rho wg(h1 −- 0.1) −- ρ\rho wgh2 = p0

where, ρ\rho k = density of kerosene and ρ\rho w = density of water.

⇒\Rightarrow ρ\rho kg(0.1) + ρ\rho wgh1 −- ρ\rho wg ×\times (0.1) = ρ\rho wgh2

⇒\Rightarrow 800 ×\times 10 ×\times 0.1 + 1000 ×\times 10 ×\times h1 −- 1000 ×\times 10 ×\times 0.1 = 1000 ×\times 10 ×\times h2

⇒\Rightarrow 10000(h1 −- h2) = 200

⇒\Rightarrow h1 −- h2 = 0.01 .......(ii)

Solving Eqs. (i) and (ii), we get

h1 = 0.35

and h2 = 0.33

So, h1h2=3533{{{h_1}} \over {{h_2}}} = {{35} \over {33}}

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