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QJEEAdv21Ph6c86: JEE Advanced 2021 Physics - Class 11 JEEAdv21Ph6c86 Solution: A cylindrical tube, with its base as shown in the figure, is…

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\theta = 45 ∘^\circ . P1 and P2 are pressures at points 1 and 2, respectively, located at the base of the tube. Let β\beta = (P1 −- P2)/( ρ\rho gd), where ρ\rho 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?

JEE Advanced 2021 Paper 1 Online Physics - Properties of Matter Question 22 English

Multiple Choice Options

  • A β\beta = 0 when a = g/ 2\sqrt 2
  • B β\beta > 0 when a = g/ 2\sqrt 2
  • C β=2−12\beta = {{\sqrt 2 - 1} \over {\sqrt 2 }} when a = g/2
  • D β=12\beta = {1 \over {\sqrt 2 }} when a = g/2

Step-by-Step Verified Solution

(p1−p2)ds=ρdsd2(gsin⁡45∘−a)({p_1} - {p_2})ds = \rho dsd\sqrt 2 (g\sin 45^\circ - a)

(p1−p2)ds=ρd(g−a2)({p_1} - {p_2})ds = \rho d(g - a\sqrt 2 )

β=(p1−p2)ρgd=(1−a2g)\beta = {{({p_1} - {p_2})} \over {\rho gd}} = \left( {1 - {{a\sqrt 2 } \over g}} \right)

When a=g2,β=0a = {g \over {\sqrt 2 }},\beta = 0

When a=g2,β=(2−12)a = {g \over 2},\beta = \left( {{{\sqrt 2 - 1} \over {\sqrt 2 }}} \right)

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