Question

Water flows out through a circular pipe whose internal diameter is 2cm at the rate of 6 metre per second into a cylindrical tank the radius of whose base is 60 cm, find the rise in the level of water in 30 minutes.
A. 3m
B. -3m
C. 6m
D. -6m

Answer 1

Hint: As both the pipe and tank are cylindrical in shape so we use the formula of Volume of cylinder which is given by \[V=\pi {{r}^{2}}h\], where r is the radius of the base of the cylinder and h is the height of it.

Complete Step-by-Step solution:
We are given that the Water flows out through a circular pipe whose internal diameter is 2 centimetres at the rate of 6 metre per second into a cylindrical tank the radius of whose base is 60 cm.
So, given the internal diameter d is 2cm, then the radius r becomes equal to,
\[r=\dfrac{d}{2}\]
\[\begin{align}
  & \Rightarrow r=\dfrac{2}{2} \\
 & \Rightarrow r=1cm \\
\end{align}\]
Therefore, we have the radius of the circular pipe r = 1cm.
Since the water flows at the rate of 6m/sec. Converting this value in cm/sec we have,
The flow of the water is 600cm/sec.
Now because the pipe is circular at the end then the original shape of the pipe would be cylinder. Therefore, the volume of the water that flows from the pipe in 1 sec would be equal to the volume of the cylinder of height 600 (as the rate is so) and the radius as 1cm.
The formula of volume of cylinder is given by,
\[V=\pi {{r}^{2}}h\], where r is the radius of the base and h is the height.
In our case r = 1cm and h = 600cm/sec.
Substituting the values of r and h in the formula of volume of cylinder we get,
The volume of the water that flows from the pipe in 1 sec is,
\[\begin{align}
  & V=\pi {{(1)}^{2}}600 \\
 & \Rightarrow V=600\pi \\
 & \Rightarrow V=600\left( \dfrac{22}{7} \right) \\
 & \Rightarrow V=1884c{{m}^{3}} \\
\end{align}\]
Then, we have to calculate the volume of water which flows in 30 min. To do so we first convert 30 min in sec which is (30)(60) = 1800sec.
Then, the volume of water which flows in 1800 sec is (1800) (1884cm3) = 33912 cm3.
Now given that the Radius of a cylindrical tank is 60cm.
Let the increase level of water be hcm
So, the volume of the tank \[V=\pi {{r}^{2}}h\], where r is the radius = 60cm and height is assumed to be h.
Substituting the values of Volume = 33912cm3 and radius we get,
The volume of the tank
\[\begin{align}
  & \Rightarrow V'=\pi {{(60)}^{2}}h \\
 & \Rightarrow 33912=\dfrac{22}{7}{{(60)}^{2}}h \\
 & \Rightarrow h=\dfrac{33912(7)}{22(3600)} \\
 & \Rightarrow h=2.99cm \\
\end{align}\]
Therefore, the rise in the water is h = 2.99cm approximately equal to 3cm.
Hence, the rise in the level of water in 30 minutes is equal to 3cm.

Note: The possibility of error in the question arises at the point of confusion in the volume of the tank and the volume of the pipe, because both are cylindrical in shape therefore, the formula of volume remains the same for both but the radius and height differs.