Question:

A cylindrical tube, with its base as shown in the figure, is filled with water It is moving down with a constant acceleration aa along a fixed inclined plane with angle ΞΈ=45∘P1\theta=45^{\circ} P_{1} and P2P_{2} are pressures at points 11 and 22, respectively located at the base of the tube. Let Ξ²=(P1βˆ’P2)/(ρgd)\beta=\left(P_{1}-P_{2}\right) /(\rho g d), where ρ\rho is density of water, dd is the inner diameter of the tube and gg is the acceleration due to gravity. Which of the following statement(s) is(are) correct?
A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a

Updated On: Mar 20, 2024
  • Ξ²=0\beta=0 when a=g2a=\frac {g}{\sqrt{2}}

  • Ξ²>0\beta > 0 when a=g2a=\frac{g}{\sqrt{2}}

  • Ξ²=2βˆ’12\beta=\frac{\sqrt{2}-1}{\sqrt{2}} when a=g2a =\frac{g}{2}

  • Ξ²=12\beta=\frac{1}{\sqrt{2}} when a=g2a = \frac{g}{2}

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A, C

Solution and Explanation

The correct answer is option
(A): Ξ²=0\beta=0 when a=g2a=\frac {g}{\sqrt{2}}
(C): Ξ²=2βˆ’12\beta=\frac{\sqrt{2}-1}{\sqrt{2}} when a=g2a =\frac{g}{2}

Was this answer helpful?
0
0

Questions Asked in JEE Advanced exam

View More Questions