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Question

# A closed cubical vessel is given a horizontal acceleration a. Find the value of a such that pressure at the mid point M of AC is equal to pressure at N.

A
2g
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B
3g
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C
g
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D
4g
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Solution

## The correct option is A 2gFor the given accelerated container, the pressure difference in the liquid along horizontal direction is given by, ΔP=ρax ...(i) The pressure will increase in a direction opposite to that of acceleration of the container. From figure, applying Eq.(i) PM=PC+(ρa×2.5) ...(1) Similarly, PA=PC+(ρa×5) ...(2) The pressure difference in the liquid due to vertical depth is given as: ΔP=ρgh Since PA>PN ⇒PN=PA−ρg×5 ...(3) [∵h=5 m] From eq.(1), (2), (3) PN=(PC+ρa×5)−ρg×5 ...(4) Equating pressure at M & N as given in question, PM=PN⇒PC+ρa×2.5=Pc+ρa×5−ρg×5 ⇒5ρg=2.5ρa ∴a=2g Acceleration of the closed cubical vessel is 2g

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