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 $\theta=45^{\circ} . P_1$ and $P_2$ 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?
$(A)$ $\beta=0$ when $a= g / \sqrt{2}$
$(B)$ $\beta>0$ when $a= g / \sqrt{2}$
$(C)$ $\beta=\frac{\sqrt{2}-1}{\sqrt{2}}$ when $a= g / 2$
$(D)$ $\beta=\frac{1}{\sqrt{2}}$ when $a= g / 2$
IIT 2021, Advanced
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$\therefore P_1-P_3=\rho\left(g-\frac{a}{\sqrt{2}}\right) d$
$P_2-P_3=\rho \frac{a}{\sqrt{2}} d$
$\therefore P_1-P_2=\rho d\left[g-\frac{2 a}{\sqrt{2}}\right]$
$\therefore \frac{P_1-P_2}{\rho g d}=\left[1-\sqrt{2} \frac{a}{g}\right]=\beta$
$\therefore \text { if } \beta=0, a=\frac{g}{\sqrt{2}} \ldots(A)$
$\beta=\frac{\sqrt{2}-1}{2}, a=\frac{g}{2} \ldots(C)$
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