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Race of Rollers

A vertical U-...

Question

A vertical U-tube contains a liquid of density $^{'} ρ^{'}$ Now it moves with a constant horizontal acceleration 'a'. Let $h_{1}$ & $h_{2}$ are the heights of liquid column in left and right limbs and $r_{1}$ and $r_{2}$ are the radius of curvature of meniscus are respectively as shown in the figure.
Assume ' $P_{0}$ ' the atmosphere pressure and 'S' is the surface tension of liquid The pressure at point 'A' is /are

ρ g h_{1} + P_{0} - \frac{2 S}{r_{1}}

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ρ g h_{2} + P_{0} - \frac{2 S}{r_{2}} + ρ a l

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ρ g h_{1} + P_{0} + \frac{2 S}{r_{1}}

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ρ g h_{2} + P_{0} + \frac{2 S}{r_{2}} - ρ a l

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Solution

The correct options are
A $ρ g h_{1} + P_{0} - \frac{2 S}{r_{1}}$
D $ρ g h_{2} + P_{0} - \frac{2 S}{r_{2}} + ρ a l$
Radius of curvature of meniscus at C is $r_{1}$
Using $P_{o} - P_{C} = \frac{2 S}{r_{1}}$ $⟹ P_{C} = P_{o} - \frac{2 S}{r_{1}}$
$∴$ Pressure at A $P_{A} = P_{C} + ρ g h_{1} = P_{o} - \frac{2 S}{r_{1}} + ρ g h_{1}$

Radius of curvature of meniscus at D is $r_{2}$
Using $P_{o} - P_{D} = \frac{2 S}{r_{2}}$ $⟹ P_{D} = P_{o} - \frac{2 S}{r_{2}}$
$∴$ Pressure at B $P_{B} = P_{D} + ρ g h_{2} = P_{o} - \frac{2 S}{r_{2}} + ρ g h_{2}$

Using $P_{A} - P_{B} = ρ g l$ $⟹ P_{A} = P_{B} + ρ l a$
$⟹$ $P_{A} = P_{o} - \frac{2 S}{r_{2}} + ρ g h_{2} + ρ l a$

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