JEE Main 2022 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A closed tank standing on level ground contains water up to a height of 6 m above the ground. The air above the water is at a gauge pressure of…

Numerical answer

A closed tank standing on level ground contains water up to a height of

6\ \mathrm{m}

above the ground. The air above the water is at a gauge pressure of

4\times10^{4}\ \mathrm{Pa}

. A small hole is made in the side wall at a depth of

1\ \mathrm{m}

below the water surface. Taking

g=10\ \mathrm{m\,s^{-2}}

and

\rho=1000\ \mathrm{kg\,m^{-3}}

, the horizontal distance (in m) from the tank at which the jet strikes the ground is

Show answer and solution

Answer: 10

Apply Bernoulli between the water surface and the hole:

v^{2}=2\left(\dfrac{P_g}{\rho}+gh\right)=2\left(\dfrac{4\times10^{4}}{1000}+10\times1\right)=2(40+10)=100

, so

v=10\ \mathrm{m\,s^{-1}}

horizontally. The hole is

6-1=5\ \mathrm{m}

above the ground, so the fall time is

t=\sqrt{\dfrac{2\times5}{10}}=1\ \mathrm{s}

. Range

=vt=10\ \mathrm{m}

. The extra gauge pressure is equivalent to

4\ \mathrm{m}

of water head, which must be added to the

1\ \mathrm{m}

depth.

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