A liquid is filled in a rectangular vessel of dimension 4 m x 3 m X 2m. Now the container starts to move with uniform acceleration 1.25 m/s² at t = 0. The volume of liquid in vessel is 18m3. The speed of liquid coming out from a very small orifice made at bottom of right side wall at t = 4s is

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zero

$\sqrt{30} m / s$

5 m/s

10 m/s

answer is D.

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Detailed Solution

$\begin{aligned} \tan θ = \frac{a}{g} = \frac{y_{1} - y_{2}}{4} \\ ∴ y_{1} - y_{2} = 0 .5 \dots \dots . (i) \end{aligned}$
Volume of water is
$\begin{array}{l} \frac{4 (y_{1} + y_{2})}{2} \times 3 = 18 \\ ∴ y_{1} + y_{2} = 3 \dots \dots . . (ii) \\ ∴ y_{2} = \frac{2 .5}{2} \end{array}$
$∴$ speed of liquid coming from orifice is
$= at + {\sqrt{2 gy}}_{2} = 10 m / s$