JEE Advanced 2018 Paper 1 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A large cylindrical open tank of cross-sectional area 0.50 m^2 stands on level ground and contains water. A small circular hole of area 4.0 cm^2 is…

Numerical answer

A large cylindrical open tank of cross-sectional area

0.50\ \mathrm{m^2}

stands on level ground and contains water. A small circular hole of area

4.0\ \mathrm{cm^2}

is made in the side wall at its very bottom. The water level is initially

1.8\ \mathrm{m}

above the hole. Take

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

and treat the flow as ideal. Find the time, in seconds, taken by the level to fall from

1.8\ \mathrm{m}

0.45\ \mathrm{m}

above the hole.

Show answer and solution

Answer: 375

Apply continuity and Bernoulli's equation (Torricelli): the efflux speed is

\sqrt{2gh}

, because the tank area is much larger than the hole. Volume balance gives

-A\,\dfrac{dh}{dt}=a\sqrt{2gh}

. Separating variables,

t=\frac{A}{a\sqrt{2g}}\int_{h_2}^{h_1}\frac{dh}{\sqrt h}=\frac{A}{a}\sqrt{\frac2g}\left(\sqrt{h_1}-\sqrt{h_2}\right)

. Here

A/a=0.5/(4\times10^{-4})=1250

and

\sqrt{2/g}=\sqrt{0.2}

. With

\sqrt{1.8}-\sqrt{0.45}=0.6708

we get

t=1250\times0.4472\times0.6708\approx375\ \mathrm{s}

. The exact value is

375

seconds because

\sqrt{0.2}(\sqrt{1.8}-\sqrt{0.45})=0.3

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