The minimum horizontal acceleration (in ${ms}^{- 2}$ ) of the container filled with water upto height of $2 m$ as shown, so that the pressure at point $A$ of the container becomes atmospheric is (the tank is of sufficient height)

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Solution

As the container accelerates, free surface of liquid tilts as shown below
Volume equality gives
$2 \times 3 = \frac{1}{2} \times h \times 3 \Rightarrow h = 4 m$

$∴ tan θ = \frac{4}{3} = \frac{a}{g} \Rightarrow a = \frac{4}{3} g = 13.33 {ms}^{- 2}$

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The minimum horizontal acceleration (in ms−2) of the container filled with water upto height of 2 m as shown, so that the pressure at point A of the container becomes atmospheric is (the tank is of sufficient height)

As the container accelerates, free surface of liquid tilts as shown below Volume equality gives 2×3=12×h×3⇒h=4 m ∴tanθ=43=ag⇒a=43g=13.33 ms−2

The minimum horizontal acceleration (in ${ms}^{- 2}$ ) of the container filled with water upto height of $2 m$ as shown, so that the pressure at point $A$ of the container becomes atmospheric is (the tank is of sufficient height)

As the container accelerates, free surface of liquid tilts as shown below
Volume equality gives
$2 \times 3 = \frac{1}{2} \times h \times 3 \Rightarrow h = 4 m$

$∴ tan θ = \frac{4}{3} = \frac{a}{g} \Rightarrow a = \frac{4}{3} g = 13.33 {ms}^{- 2}$