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 2m/s2 in horizontal direction as shown in figure. Neglect the width of the limbs.
A
15cm
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B
18cm
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C
20cm
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D
10cm
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Solution
The correct option is D10cm 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=0.1m=10cm Hence, difference in heights of the liquid in the two limbs is 10cm.