Water flows through a cylindrical pipe, whose inner radius is $ 1 \mathrm{~cm} $, at the rate of $ 80 \mathrm{~cm} / \mathrm{sec} $ in an empty cylindrical tank, the radius of whose base is $ 40 \mathrm{~cm} $. What is the rise of water level in tank in half an hour?

Given:

Water flows through a cylindrical pipe, whose inner radius is \( 1 \mathrm{~cm} \), at the rate of \( 80 \mathrm{~cm} / \mathrm{sec} \) in an empty cylindrical tank, the radius of whose base is \( 40 \mathrm{~cm} \).

To do:

We have to find the rise of water level in tank in half an hour.

Solution:

Radius of the cylindrical tank $r = 40\ cm$

Let the height of the raise in water level in tank in half an hour be $h$.

Internal radius of the cylindrical pipe $r_1 = 1\ cm$

Rate of the water flow $= 80\ cm/s$

This implies,

Volume of water flow in 1 sec $=\pi r_1^2 h_1$

$=\pi (1)^2 \times 80$

$=80\pi$

Volume of water flow in 30 minutes $=80\pi \times 30 \times 60$

$= 144000\pi\ cm^3$

According to the question,

Volume of water in the cylindrical tank $=$ Volume of water flown from the circular pipe in half an hour

$\pi r^2 h = 144000\pi$

$40 \times 40 \times h = 144000$

$h = \frac{144000}{40\times40}$

$h= 90\ cm$

The rise of water level in tank in half an hour is $90\ cm$.

Tutorialspoint

e

Updated on: 10-Oct-2022

30 Views

Kickstart Your Career

Get certified by completing the course

Get Started