Question

Water flows out through a circular pipe, whose internal diameter is $1\frac{1}{3}cm$, at the rate of $0.63$ m per second into a cylindrical tank, the radius of whose base is $0.2$ m. By how much will the level of water rise in one hour?

• A. $2.32m$
• B. $2.52m$
• C. $2.72m$
• D. $2.92m$

Answer 1

Answer: (B)

$2.52m$

In one second, if $0.63m$ of water is flown.

Then in one hour, length or distance covered by the water $=0.63\times 60\times 60=2268m$
Radius of pipe $=\frac{\frac{4}{3}}{2}=\frac{2}{3}cm=\frac{2}{300}m$
Since the pipe is cylindrical in shape, Volume $=\pi {Radius}^2\times height$
Here, height is considered as the distance traveled.
Hence, Volume of water which flows in one hour $=\frac{22}{7}\times {\frac{2}{300}}^2\times 2268m^3=0.3168m^3$

Now, this $0.3168m^3$ of water will fill the cylindrical tank to a height say "h".

=> Volume of the tank till height "h" $=0.3168m^3$

$=>\frac{22}{7}\times {0.2}^2\times h=0.3168$
$=>h=2.52m$