- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per $100\ cm^2$. (Take $\pi = 3.14$)

Given:

A container opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of the height of 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively.

To do:

We have to find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre and the cost of the metal sheet used to make the container, if it costs Rs. 8 per $100\ cm^2$.

Solution:

Radius of the lower end $(r_1) = 8\ cm$

Radius of the upper end $(r_2) = 20\ cm$

Height of the frustum $(h) = 16\ cm$

Volume of the container $=\frac{\pi h}{3}[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}]$

$=3.14 \times \frac{16}{3}[8^{2}+20^{2}+8 \times 20]$

$=3.14 \times \frac{16}{3}[64+400+160]$

$=3.14 \times \frac{16}{3} \times 624$

$=10449.92 \mathrm{~cm}^{3}$

$=10.449$ litres (Since $1 \mathrm{~cm}^{3}=\frac{1}{1000} 1$)

Cost of 1 litre of milk $=Rs.\ 20$

Total cost of milk $=Rs,\ 20 \times 10.449$

$=Rs.\ 208.98$

Slant height of the container $(l)=\sqrt{(r_{2}-r_{1})^{2}+h^{2}}$

$=\sqrt{(20-8)^{2}+16^{2}}$

$=\sqrt{12^{2}+16^{2}}$

$=\sqrt{144+256}$

$=\sqrt{400}$

$=20\ cm$

Curved surface area of the bucket $=\pi l(r_{1}+r_{2})$

$=3.14 \times 20(20+8)$

$=3.14 \times 20 \times 28$

$=1758.4 \mathrm{~cm}^{2}$

The base area of the bucket $=\pi r_{1}^{2}$

$=3.14 \times 8^2$

$=200.96 \mathrm{~cm}^{2}$

The metal sheet used $=$ Curved surface area of the frustum $+$ base area of the bucket

$=1758.4+200.96$

$=1959.36 \mathrm{~cm}^{2}$

The total cost of the metal sheet used $=Rs.\ \frac{8 \times 1959.36}{100}$

$=Rs.\ 156.75$

- Related Questions & Answers
- Program for Volume and Surface area of Frustum of Cone in C++
- Container to create a grid of Bootstrap 4 cards of equal height and width
- Set the height of a line of text with CSS
- Slide up and down of a div animates its content - jQuery?
- Set the height of a box with CSS
- Python Program to Form a New String Made of the First 2 and Last 2 characters From a Given String
- Get the width and height of a three-dimensional array
- Python Program to find the number of operations to pack several metal bars in a container
- Integrate a polynomial and set the lower bound of the integral in Python
- Find the count of palindromic sub-string of a string in its sorted form in Python
- Integrate a Laguerre series and set the lower bound of the integral in Python
- Integrate a Hermite series and set the lower bound of the integral in Python
- Integrate a Chebyshev series and set the lower bound of the integral in Python
- Integrate a Legendre series and set the lower bound of the integral in Python
- Integrate a Hermite_e series and set the lower bound of the integral in Python
- Check if a given string is made up of two alternating characters in C++