Question

The bottom of a rectangular tank is 250 cm by 400 cm. Water is pumped in the tank at the rate of 500 cc/min. Find the rate at which the level of water is rising in the tank.

Answer 1

Hint: We here have been given the dimensions of the rectangular bottom of the tank and the rate at which the water inside the tank and we have to calculate the rate at which the water level rises. The rectangular bottom implies that the tank is cuboid in shape. We will then apply the formula given as:
Rate at which the water is pumped inside the tank=rate at which the volume inside the tank rises
Hence, we will get the rate at which the volume rises inside the tank, i.e. $\dfrac{dV}{dt}$. The dimensions of the bottom will be the length, l and the breadth, b of the tank and we will assume the water level inside it at any time t as ‘h’. Thus, our volume ‘V’ will be given as:
$V=l\times b\times h$
We will put this value in the place of V in $\dfrac{dV}{dt}$ as a result to which we will get the value of $\dfrac{dh}{dt}$ which will be our required answer.

Complete step by step answer:
Now, we have been given that the bottom of the rectangular shaped tank has dimensions 250 cm by 400 cm. We have also been given the rate at which the water is pumped in the tank which is equal to 500 cc/min.
The top view of the bottom of the tank would be as below,

Here, the bottom of the tank is given to be rectangular and we know that if a tank has a rectangular bottom then the tank is in a cuboid shape.
Now, we know that:
Rate at which the water is pumped inside the tank=rate at which the volume inside the tank rises.
Hence, the rate at which the volume in the tank rises= 500 cc/min.
Now, we know that the volume of any cuboid is given as:
$\text{volume=length}\times \text{breadth}\times \text{height}$
Here, as the dimensions of the tank are given to us as 250 cm by 400 cm, we can say that:
$\begin{align}
  & length=250cm \\
 & breadth=400cm \\
\end{align}$
Now, as we established above, the rate of rise of volume in the tank=500 cc/min and we also know that the rate of increase in the volume is equal to the derivative of the volume with respect to the time.
Hence, if assume the volume of the tank to be ‘V’, we can say that:
Rate of rise of volume in the tank=$\dfrac{dV}{dt}$
Thus, we can say that:
$\dfrac{dV}{dt}=500c{{m}^{3}}/\min $ .....(i)
Now, the water is constantly being filled in the water, thus its height, i.e. the water level will also keep on rising but the bottom area will remain constant.
At any time t, let us assume the water level to be ‘h’ and the length and breadth to be ‘l’ and ‘b’ respectively.
Thus, we can say that:
$V=l\times b\times h$
Now, we have been asked the rate at which the water level inside the tank rises. This means that we have to calculate the value of the derivative of the height of the water level ‘h’ with respect to time ‘t’, i.e. $\dfrac{dh}{dt}$ .
Now, as we mentioned above, l and b are constants given as:
$\begin{align}
  & l=250cm \\
 & b=400cm \\
\end{align}$
Thus, we get V as:
$\begin{align}
  & V=l\times b\times h \\
 & \Rightarrow V=250\times 400\times h \\
 & \Rightarrow V=100000\times h \\
 & \Rightarrow V={{10}^{5}}\times h \\
\end{align}$
Thus, putting the value of V in equation (i), we get:
$\begin{align}
  & \dfrac{dV}{dt}=500c{{m}^{3}}/\min \\
 & \Rightarrow \dfrac{d}{dt}\left( {{10}^{5}}\times h \right)=500c{{m}^{3}}/\min \\
\end{align}$
Now, we know that $\dfrac{d}{dx}\left( ky \right)=k\dfrac{dy}{dx}$
Thus, putting this into the above obtained equation and solving it we get:
$\begin{align}
  & \Rightarrow {{10}^{5}}.\dfrac{dh}{dt}=500c{{m}^{3}}/\min \\
 & \Rightarrow \dfrac{dh}{dt}=\dfrac{500}{{{10}^{5}}}cm/\min \\
 & \Rightarrow \dfrac{dh}{dt}=\dfrac{1}{200}cm/\min \\
\end{align}$

Hence, the rate at which the water level rises inside the tank is equal to $\dfrac{1}{200}cm/\min $

Note: Here, we have been given the unit of the rate at which the water is pumped as cc/min which means cubic centimetres per minute. We should know that it’s not the SI unit of rate when it comes to the rate of some kind of volume. But since this is the unit mentioned to us in the question, we will calculate the rate of rise in water level corresponding to the unit of rate at which the water is pumped which is given as cm/min, i.e. centimetre per minute. The SI unit of the rate of water pumped will be ${{m}^{3}}/\sec $ and that rate of rise in water level will be $m/\sec $.