JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Fluid Mechanics

Consider a water tank as shown in the figure. Its cross-sectional area is

0.4 m^{2}

. The tank has an opening B near the bottom whose cross-sectional area is

1 cm^{2}

. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is

v ms^{- 1}

.

The value of v, to the nearest integer, is ____
[Take value of g to be

10 ms^{- 2}

]

Answer: 3

Solution

The load rests on the water over the whole cross-section of the tank, so the pressure at the surface exceeds atmospheric by

Δ P = \frac{M g}{A} = \frac{24 \times 10}{0.4} = 600 Pa

The opening is four thousand times smaller in area than the tank, so by continuity the surface falls far too slowly for its speed to matter, and Bernoulli's equation between the surface and the opening gives

\frac{1}{2} ρ v^{2} = Δ P + ρ g h = 600 + 1000 \times 10 \times 0.4 = 4600 Pa

Hence

v^{2} = \frac{2 \times 4600}{1000} = 9.2

and

v = 3.03 m s^{- 1}

, which to the nearest integer is

3 m s^{- 1}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Numerical answer

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JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Fluid Mechanics

Solution

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