Physics · Fluid Mechanics

JEE Advanced 2021 — Paper 1 — Question 8

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 $\theta=45^{\circ} . \mathrm{P}_{1}$ and $\mathrm{P}_{2}$ are pressures at points 1 and 2 ,

respectively located at the base of the tube. Let $\beta=\left(P_{1}-P_{2}\right) /(\rho g d)$ , where $\rho$ 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?

Answer: A, C

$\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right) \mathrm{ds}=\rho d s d \sqrt{2}(\mathrm{~g} \sin 45-\mathrm{a})$

$\left(P_{1}-P_{2}\right)=\rho d(g-a \sqrt{2})$ $\beta=\frac{\left(P_{1}-P_{2}\right)}{\rho g d}=\left(1-\frac{a \sqrt{2}}{g}\right)$

When $\mathrm{a}=\mathrm{g} / \sqrt{2}, \quad \beta=0$

When $\mathrm{a}=\mathrm{g} / 2, \quad \beta=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

Answer key and solution verified before publishing.

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