Loading [MathJax]/jax/output/HTML-CSS/jax.js

Problems on Right Circular Cylinder

Here we will learn how to solve different types of problems on right circular cylinder.

1. A solid, metallic, right circular cylindrical block of radius 7 cm and height 8 cm is melted and small cubes of edge 2 cm are made from it. How many such cubes can be made from the block?

Solution:

For the right circular cylinder, we have radius (r) = 7 cm, height (h) = 8 cm.

Therefore, its volume = πr2h

                               = 227 × 72 × 8 cm3

                               = 1232 cm3

The volume of a cube = (edge)3

                                = 23 cm3

                                = 8 cm3

Therefore, the number of cubes that can be made = volume of the cylinder/volume of a cube

                                                                        = 1232cm38cm3

                                                                        = 154

Therefore, 154 cubes can be made from the block.


2. The height of a cylindrical pillar is 15 m. The diameter of its base is 350 cm. What will be the cost of painting the curved surface of the pillar at Rs 25 per m2?

Solution:

The base is circular and so the pillar is a right circular cylinder.

Height of a Cylindrical Pillar

Here, radius = 175 cm = 1.75 m and height = 15 m

Therefore, the curved surface area of the pillar = 2πrh

                                                                    = 2 × 227 × 1.75 × 15 m2

                                                                    = 165 m2

Therefore, the cost of painting this area = Rs 25 × 165 = Rs 4125.


3. A cylindrical container is to be made of tin. The height of the container is 1 m and the diameter of the base is 1 m. If the container is open at the top and tin sheet costs Rs 308 per m2, what will be the cost of tin for making the container?

Solution:

Given, diameter of the base is 1 m.

A Cylindrical Container

Here, radius = r = 12 m and height = h = 1 m.

Total area of tin sheet required = curved surface area + area of the base

                                              = 2πrh + πr2

                                              = πr(2h + r)

                                              = π ∙ 12 ∙ (2 × 1 + 12) m2

                                              = 5π4 m2

                                              = 54 ∙  227m2

                                              = 5514 m2

Therefore, the cost of tin = Rs 308 × 5514 = Rs 1210.


4. The dimensions of a rectangular piece of paper are 22 cm × 14 cm. It is rolled once across the breadth and once across the length to form right circular cylinders of biggest possible surface areas. Find the difference in volumes of the two cylinders that will be formed.

Solution:

Dimensions of a Rectangular Piece

When rolled across the breadth

Circumference of the cross section = 14 cm and height = 22 cm

Circumference of the Cross Section

Therefore, 2πr = 14 cm

or, r = 142π cm

or, r = 142×227 cm

or, r = 4922 cm


When rolled across the length

Circumference of the cross section = 22 cm and height = 14 cm

Circumference of the Cross Section of Cylinder

Therefore, 2πR = 22 cm

or, R = 222π cm

or, r = 222×227 cm

or, r = 72 cm

Therefore, volume = πR2h

                           = 227 × (72)2 × 14 cm3

                           = 11 × 49 cm3

Therefore, the difference in volumes = (11 × 49 - 7 × 49) cm3

                                                     = 4 × 49 cm3

                                                     = 196 cm3

Therefore, 196 cm3 is the difference in volumes of the two cylinders.







9th Grade Math

From Problems on Right Circular Cylinder to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. Divide by Repeated Subtraction | Division as Repeated Subtraction

    Jan 20, 25 03:34 PM

    Repeated Subtraction
    How to divide by repeated subtraction? We will learn how to find the quotient and remainder by the method of repeated subtraction a division problem may be solved.

    Read More

  2. 3rd Grade Multiplication Worksheet | Grade 3 Multiplication Questions

    Jan 20, 25 02:31 PM

    3rd Grade Multiplication Riddle
    In 3rd Grade Multiplication Worksheet we will solve how to multiply 2-digit number by 1-digit number without regrouping, multiply 2-digit number by 1-digit number with regrouping, multiply 3-digit num…

    Read More

  3. 3rd Grade Math Worksheets |3rd Grade Math Sheets|3rd Grade Math Lesson

    Jan 20, 25 12:28 AM

    3rd Grade Math Worksheets
    3rd grade math worksheets is carefully planned and thoughtfully presented on mathematics for the students. Teachers and parents can also follow the worksheets to guide the students.

    Read More

  4. 3rd Grade Multiplication Word Problems Worksheet With Answers | Math

    Jan 19, 25 11:29 PM

    In 3rd Grade Multiplication Word Problems Worksheet we will solve different types of problems on multiplication, multiplication word problems on 3-digits number by 1-digit number and multiplication wo…

    Read More

  5. Multiplying 2-Digit Numbers by 2-Digit Numbers |Multiplying by 2-Digit

    Jan 19, 25 02:34 AM

    Multiplying 2-Digit Numbers by 2-Digit Numbers
    We will learn how to multiply 2-digit numbers by 2-digit numbers.

    Read More