Question

Water flows through a cylindrical pipe, whose inner radius is $1cm$ , at the rate of $80cm$ per second in an empty cylindrical tank, the radius of whose base is $40cm$ . What is the rise of the water level in a tank in $\dfrac{1}{2}hr$ ?

Answer 1

Hint: Start with finding the volume flow rate for one second from the length rate of flow of water from the pipe. Now multiply the time, i.e. half an hour, to the volume flow rate to get the total volume added in that time. You can equate this total volume with the volume expression of the tank, i.e. $\pi {\left( {40} \right)^2}h$ where the unknown ‘h’ will be the height of the water increased.

Complete step-by-step answer:
Here in this problem, it is given that a cylindrical pipe with an inner radius of $1cm$ is filling water into a cylindrical tank at a rate of $80cm/s$ . The cylindrical tank has a base radius of $40cm$. With this given information we need to calculate the rise of the water line in half an hour.

The flow rate of water in this problem i.e. $80cm/s$ represents the length of pipe filled with water flown into the cylindrical tank in one second. So, this means the pipe will deliver $80cm$ the length of pipe filled with water into the tank in one second.
So, let’s first figure out the volume flow rate of the water flowing from the pipe into the tank. For this, we need to calculate the volume of water flowing in one second.
$ \Rightarrow $ The volume of the cylinder with radius $1cm$ and length $80cm$
According to the formula, the volume of a cylinder is given by pi$\left( \pi \right)$ times the square of radius times the length,
Therefore, the volume flow rate of water $ = \pi {r^2}l$
After substituting the values in it, we get:
$ \Rightarrow $ The volume flow rate of water$ = \pi {r^2}l = \pi \times {\left( 1 \right)^2} \times 80{\text{ c}}{{\text{m}}^3}{\text{/s}}$
Now we can put $\pi = 3.14$ in above
$ \Rightarrow $ The volume flow rate of water $ = \pi \times {\left( 1 \right)^2} \times 80 = 3.14 \times 1 \times 80 = 251.2{\text{ c}}{{\text{m}}^3}{\text{/s}}$
Therefore, this is the volume of water that enters in the cylindrical tank in one second. But we need to do calculations for $\dfrac{1}{2}{\text{hour}} = \dfrac{{60}}{2}{\text{minutes}} = \dfrac{{60 \times 60}}{2}{\text{seconds}} = 1800{\text{ seconds}}$
So, the total volume of water added into the cylindrical tank in $1800{\text{ seconds}}$will be ${\text{251}}{\text{.2 c}}{{\text{m}}^3}{\text{/s }} \times 1800s = 452160c{m^3}$
Thus, this is the total volume of water added into the tank in $30$ minutes. And it will be equal to the volume of the water added in the tank.
Volume increased in the tank with height $'h'$ can be given by $\pi {\left( {40} \right)^2}h{\text{ c}}{{\text{m}}^3}$
$ \Rightarrow $ Volume increased in tank $ = \pi {\left( {40} \right)^2}h{\text{ c}}{{\text{m}}^3} = 452160c{m^3}$
We can solve this to find the value for the height ‘h’:
$ \Rightarrow \pi {\left( {40} \right)^2}h = 452160 \Rightarrow h = \dfrac{{452160}}{{\pi \times 1600}}$
This can be further solved using the value $\pi = 3.14$:
$ \Rightarrow h = \dfrac{{452160}}{{3.14 \times 1600}} = \dfrac{{452160}}{{5024}} \Rightarrow h = 90cm$
Thus, we get the height increased in the cylindrical tank after $\dfrac{1}{2}hr$ as $90cm$

Note: In questions like this, the use of the formula of Mensuration. Always be careful with the units used in the quantities. Here we converted the time given in hours to seconds because of the unit of the flow rate of water. Also notice that for calculating the increased height we don’t have to know the height of water already present in the tank.