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Question

If area of cross-section of both the limbs are equal, find the difference in heights of the liquid in the two limbs when the system is given an acceleration of 2 m/s2 in horizontal direction as shown in figure. Neglect the width of the limbs.


A
15 cm
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B
18 cm
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C
20 cm
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D
10 cm
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Solution

The correct option is D 10 cm
Let the difference in heights of the liquid in the two limbs be x and P0 be the atmospheric pressure.


Since fluid is accelerating horizontally in rightward direction, hence PA>PB.
For pressure at A,
PA=PB+ρaL
PA=P0+(ρ×2×0.5)
[PB=P0, as the limb is open to atmosphere]
PA=P0+ρ ...(1)
Due to vertical height, the pressure difference can also be written as,
PA=P0+ρgx
PA=P0+10ρx ...(2)
From Eq.(1) & (2),
P0+ρ=P0+10ρx
x=110
x=0.1 m=10 cm
Hence, difference in heights of the liquid in the two limbs is 10 cm.

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