A cylindrical tank of height 0.4,m is open at | vedclass

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(A) The area of the tank $A = \pi R^2$,where $R = 0.08\,m$.The area of the hole $a = \pi r^2$,where $r = 5 	imes 10^{-3}\,m$.According to the equatio

Vedclass · Accepted Answer

$46.26$

Vedclass · Answer

(A) The area of the tank $A = \pi R^2$,where $R = 0.08\,m$.The area of the hole $a = \pi r^2$,where $r = 5 	imes 10^{-3}\,m$.According to the equation of continuity and Torricelli's law,the rate of change of volume is $A \frac{dh}{dt} = -a \sqrt{2gh}$.Rearranging for time $t$: $dt = -\frac{A}{a \sqrt{2g}} h^{-1/2} dh$.Integrating from $h = 0.16\,m$ to $h = 0$:$t = \frac{A}{a \sqrt{2g}} \int_{0}^{0.16} h^{-1/2} dh = \frac{A}{a \sqrt{2g}} [2\sqrt{h}]_{0}^{0.16}$.$t = \frac{\pi (0.08)^2}{\pi (5 	imes 10^{-3})^2 \sqrt{2 	imes 9.8}} 	imes 2 	imes \sqrt{0.16}$.$t = \frac{0.0064}{25 	imes 10^{-6} 	imes 4.427} 	imes 0.8 \approx 46.26\,s$.

$A$ cylindrical tank of height $0.4\,m$ is open at the top and has a diameter $0.16\,m$. Water is filled in it up to a height of $0.16\,m$. How long will it take to empty the tank through a hole of radius $5 \times 10^{-3}\,m$ in its bottom? (in seconds)

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