Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm

Solution:

We have,

Speed of the water flowing through the pipe, $H=6 \mathrm{~km} / \mathrm{h}=\frac{600000 \mathrm{~cm}}{3600 \mathrm{~s}}=\frac{500}{3} \mathrm{~cm} / \mathrm{s}$,

Radius of the pipe, $R=\frac{14}{2}=7 \mathrm{~cm}$,

Length of the rectangular $\operatorname{tank}, l=60 \mathrm{~m}=6000 \mathrm{~cm}$,

Breadth of the rectangular tank, $b=22 \mathrm{~m}=2200 \mathrm{~cm}$ and

Rise in the level of water in the tank, $h=7 \mathrm{~cm}$

Now,

Volume of the water in the rectangular tank $=l b h$

$=6000 \times 2200 \times 7$

$=92400000 \mathrm{~cm}^{3}$

Also,

Volume of the water flowing through the pipe in $1 \mathrm{~s}=\pi R^{2} H$

$=\frac{22}{7} \times 7 \times 7 \times \frac{500}{3}$

$=\frac{77000}{3} \mathrm{~cm}^{3}$

So,

The time taken $=\frac{\text { Volume of the water in the rectangular tank }}{\text { Volume of the water flowing through the pipe in } 1 \mathrm{~s}}$

$=\frac{92400000}{\left(\frac{77000}{3}\right)}$

$=\frac{92400 \times 3}{77}$

$=3600 \mathrm{~s}$

$=1 \mathrm{hr}$

So, the time in which the level of water in the tank will rise by 7 cm is 1 hour.