Water flows through a cylindrical pipe,whose | vedclass

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(B) Given,radius of the tank,$r_{1} = 40 \,cm$.
Let the rise in the height of the water level in the tank in half an hour be $h_{1} \,cm$.
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Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Given,radius of the tank,$r_{1} = 40 \,cm$.
Let the rise in the height of the water level in the tank in half an hour be $h_{1} \,cm$.
Internal radius of the cylindrical pipe,$r_{2} = 1 \,cm$.
Speed of water flow $= 80 \,cm/s$.
Time $= 30 \,minutes = 30 	imes 60 = 1800 \,seconds$.
Length of water column flowing out of the pipe in $1800 \,seconds$,$h_{2} = 80 	imes 1800 = 144000 \,cm$.
According to the principle of conservation of volume:
Volume of water in the cylindrical tank $=$ Volume of water flowed from the pipe in half an hour.
$\pi r_{1}^{2} h_{1} = \pi r_{2}^{2} h_{2}$
$40^2 	imes h_{1} = 1^2 	imes 144000$
$1600 	imes h_{1} = 144000$
$h_{1} = \frac{144000}{1600} = 90 \,cm$.
Thus,the water level in the tank rises by $90 \,cm$ in half an hour.

Water flows through a cylindrical pipe,whose inner radius is $1 \,cm$,at the rate of $80 \,cm/sec$ into an empty cylindrical tank,the radius of whose base is $40 \,cm$. What is the rise of water level in the tank in half an hour? (in $cm$)

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