Question

Water is flowing out of a cylinder through a horizontal suble of length 50 cm and diameter 0.4 mm. The tube is 50 cm below the water level in the cylinder. How much water will flow out per minute? Coefficient of viscosity of water=4* 10-3 Pa-s​

Answer 1

Explanation:

height of cylinder be h

h=50cm

D=0.4,r=0.4/2=0.2

Volume of cylinder = pie r2 h

Value of pie=22/7

=22/7×0.2×0.2×5

Solve this u can get,

=44/7

V. Of cylinder=6.2

C. Of water=4*10-3

4*7

=28

=6.2/28

4.5

4 hours and 5 min..

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Answer 2

By Bernoulli's Theorem,

$\displaystyle\longrightarrow\sf{P_0+\rho gh=P_A+\dfrac {1}{2}\rho v^2}$

Since no velocity is at the surface and hence no kinetic energy, and no potential energy at the tube as water flows through it.

But here, $\displaystyle\sf {P_0=P_A,}$ because both ends are open to the atmosphere. Thus,

$\displaystyle\longrightarrow\sf{\rho gh=\dfrac {1}{2}\rho v^2}$

$\displaystyle\longrightarrow\sf{v=\sqrt{2gh}}$

This velocity is the so called "velocity of efflux".

Here, $\displaystyle\sf {h=50\ cm=0.5\ m.}$ Then,

$\displaystyle\longrightarrow\sf{v=\sqrt{2\times10\times0.5}}$

$\displaystyle\longrightarrow\sf{v=\sqrt{10}\ m\ s^{-1}}$

$\displaystyle\longrightarrow\sf{v=60\sqrt{10}\ m\ min^{-1}}$

Then, amount of water flown through the tube per minute will be,

$\displaystyle\longrightarrow\sf{\Delta V=Av}$

$\displaystyle\longrightarrow\sf{\Delta V=\pi r^2v}$

$\displaystyle\longrightarrow\sf{\Delta V=3.14\left (\dfrac {0.4\times10^{-3}}{2}\right)^2\times60\sqrt{10}}$

$\displaystyle\longrightarrow\underline {\underline {\sf{\Delta V}=\bf{2.38\times10^{-5}}\ \sf{m^3\ min^{-1}}}}$