A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration along a fixed inclined plane with angle . and are pressures at points 1 and 2, respectively, located at the base of the tube.

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 points 1 and 2, respectively, located at the base of the tube. Let $β = \frac{(P_{1} - P_{2})}{ρ g d},$ 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 [2021]

Smart Achievers · Accepted Answer

(1, 3)

$P_{1} = P_{2} - ρ a \cos 45 ° d + ρ (g - a \sin 45 °) d$

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g d} = 1 - \frac{\sqrt{2} a}{g}$

$∴$ $β = 0 for a = \frac{g}{\sqrt{2}}$

$and β = \frac{\sqrt{2} - 1}{\sqrt{2}} for a = \frac{g}{2}$

Solution

1 $β = 0$ when $a = \frac{g}{\sqrt{2}}$

2 $β > 0$ when $a = \frac{g}{\sqrt{2}}$

3 $β = \frac{\sqrt{2} - 1}{\sqrt{2}}$ when $a = \frac{g}{2}$

4 $β = \frac{1}{\sqrt{2}}$ when $a = \frac{g}{2}$

Ans.
(1, 3)

$P_{1} = P_{2} - ρ a \cos 45 ° d + ρ (g - a \sin 45 °) d$

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g d} = 1 - \frac{\sqrt{2} a}{g}$

$∴$ $β = 0 for a = \frac{g}{\sqrt{2}}$

$and β = \frac{\sqrt{2} - 1}{\sqrt{2}} for a = \frac{g}{2}$

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Solution

1 β=0 when a=g2

2 β>0 when a=g2

3 β=2-12 when a=g2

4 β=12 when a=g2