# 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 = πr$$^{2}$$h

= $$\frac{22}{7}$$ × 7$$^{2}$$ × 8 cm$$^{3}$$

= 1232 cm3

The volume of a cube = (edge)$$^{3}$$

= 2$$^{3}$$ cm$$^{3}$$

= 8 cm$$^{3}$$

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

= $$\frac{1232 cm^{3}}{8cm^{3}}$$

= 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 m$$^{2}$$?

Solution:

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

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

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

= 2 × $$\frac{22}{7}$$ × 1.75 × 15 m$$^{2}$$

= 165 m$$^{2}$$

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 m$$^{2}$$, what will be the cost of tin for making the container?

Solution:

Given, diameter of the base is 1 m.

Here, radius = r = $$\frac{1}{2}$$ m and height = h = 1 m.

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

= 2πrh + πr$$^{2}$$

= πr(2h + r)

= π ∙ $$\frac{1}{2}$$ ∙ (2 × 1 + $$\frac{1}{2}$$) m$$^{2}$$

= $$\frac{5π}{4}$$ m$$^{2}$$

= $$\frac{5}{4}$$ ∙  $$\frac{22}{7}$$m$$^{2}$$

= $$\frac{55}{14}$$ m$$^{2}$$

Therefore, the cost of tin = Rs 308 × $$\frac{55}{14}$$ = 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:

When rolled across the breadth

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

Therefore, 2πr = 14 cm

or, r = $$\frac{14}{2π}$$ cm

or, r = $$\frac{14}{2 × \frac{22}{7}}$$ cm

or, r = $$\frac{49}{22}$$ cm

When rolled across the length

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

Therefore, 2πR = 22 cm

or, R = $$\frac{22}{2π}$$ cm

or, r = $$\frac{22}{2 × \frac{22}{7}}$$ cm

or, r = $$\frac{7}{2}$$ cm

Therefore, volume = πR$$^{2}$$h

= $$\frac{22}{7}$$ × ($$\frac{7}{2}$$)$$^{2}$$ × 14 cm$$^{3}$$

= 11 × 49 cm$$^{3}$$

Therefore, the difference in volumes = (11 × 49 - 7 × 49) cm$$^{3}$$

= 4 × 49 cm$$^{3}$$

= 196 cm$$^{3}$$

Therefore, 196 cm$$^{3}$$ is the difference in volumes of the two cylinders.

9th Grade Math

From Problems on Right Circular Cylinder to HOME PAGE

### 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.

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.

 Share this page: What’s this?

## Recent Articles

1. ### Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction

Mar 02, 24 05:31 PM

The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato…

Read More

2. ### Subtraction of Fractions having the Same Denominator | Like Fractions

Mar 02, 24 04:36 PM

To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera…

Read More

3. ### Addition of Like Fractions | Examples | Worksheet | Answer | Fractions

Mar 02, 24 03:32 PM

To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com…

Read More

4. ### Comparison of Unlike Fractions | Compare Unlike Fractions | Examples

Mar 01, 24 01:42 PM

In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…

Read More

5. ### Equivalent Fractions | Fractions |Reduced to the Lowest Term |Examples

Feb 29, 24 05:12 PM

The fractions having the same value are called equivalent fractions. Their numerator and denominator can be different but, they represent the same part of a whole. We can see the shade portion with re…

Read More