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.

$Height of a Cylindrical Pillar$

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

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

= 2 × $\frac{22}{7}$ × 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.

$A Cylindrical Container$

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:

$Dimensions of a Rectangular Piece$

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 = $\frac{14}{2π}$ 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

$Circumference of the Cross Section of Cylinder$

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.

You might like these

Hollow Cylinder | Volume |Inner and Outer Curved Surface Area |Diagram

We will discuss here about the volume and surface area of Hollow Cylinder. The figure below shows a hollow cylinder. A cross section of it perpendicular to the length (or height) is the portion bounded by two concentric circles. Here, AB is the outer diameter and CD is the
Cylinder | Formule for the Volume and the Surface Area of a Cylinder

A solid with uniform cross section perpendicular to its length (or height) is a cylinder. The cross section may be a circle, a triangle, a square, a rectangle or a polygon. A can, a pencil, a book, a glass prism, etc., are examples of cylinders. Each one of the figures shown
Cross Section | Area and Perimeter of the Uniform Cross Section

The cross section of a solid is a plane section resulting from a cut (real or imaginary) perpendicular to the length (or breadth of height) of the solid. If the shape and size of the cross section is the same at every point along the length (or breadth or height) of the
Lateral Surface Area of a Cuboid | Are of the Four Walls of a Room

Here we will learn how to solve the application problems on lateral surface area of a cuboid using the formula. Formula for finding the lateral surface area of a cuboid Area of a Rooms is example of cuboids. Are of the four walls of a room = sum of the four vertical
Volume of Cube | Formula for Finding the Volume of a Cube | Diagram

Here we will learn how to solve the application problems on Volume of cube using the formula. Formula for finding the volume of a cube Volume of a Cube (V) = (edge)3 = a3; where a = edge A cubical wooden box of internal dimensions 1 m × 1 m × 1 m is made of 5 cm thick