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JEE Advanced Physics Mechanical Properties of Fluids 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Physics Question (2021) — Solution

Question

A cylindrical tube, with its base as shown in the figure., is filled with water. It is moving down with a constant acceleration  a  along a fixed inclined plane with angle  θ = 45 ° .  P 1  and  P 2  are pressures at point  1  and  2 , respectively, located at the base of the tube. Let  β = P 1 − P 2 ρ g d , where  ρ  is density of water,  d  is the inner diameter of the tube and  g  is the acceleration due to gravity. Which of the following statement(s) is (are) correct?

Options

  1. A. β = 0  when  a = g 2
  2. B. β > 0  when  a = g 2
  3. C. β = 2 - 1 2  when  a = g 2
  4. D. β = 1 2  when  a = g 2

Answer

C. β = 2 - 1 2  when  a = g 2

Step-by-step solution

Given, β = P 1 − P 2 ρ g d               . . . 1 Here we can write now, P 3 = P 2 + ρ g − a 2 d P 1 = P 3 − ρ a 2 d P 1 = P 2 + ρ g d − 2 ρ a d P 1 = P 2 + ρ g − a 2 d − ρ a 2 d P 1 − P 2 ρ d g = 1 − 2 a g Compare it with given equation 1 , β = 1 − 2 a g Now check the options, If  a = g 2 ,  β = 0 . If  a = g 2 ,  β = 2 - 1 2 . Option 1 and 3  are correct

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