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A vessel of height $H$ and length $L$ contains a liquid of density $ρ$ upto height $H / 2$ . The vessel starts accelerating horizontally with acceleration ‘ $a$ ’ towards right. If $A$ is the point at the surface of the liquid at right end while the vessel is accelerating and $B$ is the point at bottom of the vessel on the other end, the difference of pressures at $B$ and $A$ will be

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$2 ρ (g H - a L)$

$\frac{ρ}{2} (g H + a L)$

$\frac{3 ρ}{2} (g H + a L)$

$\frac{ρ}{2} (g H - a L +)$

answer is A.

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Detailed Solution

Let force $(△ P)$ acts on the surface of the liquid while vessel is accelerating.

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