Class 10 Math Chapter 12
Surface Area and Volumes Important Questions

Here are some essential questions for Class 10 Mathematics Chapter 12, Surface Area and Volumes, thoughtfully chosen to aid students in their preparation for the CBSE Class 10 Mathematics Examination in 2023-24. By practicing these varied problems, students can enhance their comprehension of Surface Area and Volumes concepts and refine their problem-solving skills.

Introduction

In Chapter 12 of Class 10 Mathematics,Surface Area and Volumes, we will study the surface areas and volumes of combinations of any two of the following geometric shapes: cubes, cuboids, spheres, hemispheres, and right circular cylinders/cones. We will explore how to calculate the total surface area, lateral surface area, and volume of various combinations involving these shapes.

What is meant by the surface area of a three-dimensional shape?

The surface area of a three-dimensional shape refers to the total measure of all its external surfaces. It encompasses the combined areas of every face, side, and curved surface that make up the object. In essence, it quantifies the extent of the outer covering of the shape, accounting for all its visible and touchable parts.

 Class 10 Surface area and volumes Important Questions and Answers

Q1. If the surface area of a sphere is 616cm², then its diameter (in cm) is :

Options

(a) 7
(b) 14
(c) 28
(d) 56

Ans.(b) 14

Explanation:
Surface area of sphere = 616 cm^2
‍⇒ 4r² = 616 
⇒ 4 ×\(\frac{22}{7}\)× r² = 616
⇒ r² = 616 ×\(\frac{7}{22×4}\)
⇒ r = 7 cm
So, the diameter = 2r = 2 × 7 = 14 cm.

Q2.  The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What is the ratio (approximate) of their volumes ?

Options

(a) 1:16
(b) 1:64
(c) 1:4
(d) 1:128

Ans. (b) 1:64

Explanation:
The diameter of Moon is approximately one fourth of the diameter of Earth.
Let, Radius of Moon = r, 
Then, Radius of Earth = 4r
Required ratio =\(\frac{\text {Volume of Moon}}{\text {Volume of Earth}}\)
\(=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi (4r)^3}\)
=\(frac{r^3}{64r^3}=\frac{1}{64}\)
= 1 : 64

Q3. A housing society collects rain water from the roof of its building of area 22 m × 20 m in a cylindrical vessel of diameter 2 m and height 3.5 m and then pumps this water into the main water tank so that everyone can use it. On a particular day, the rain water collected from the roof just fills the cylindrical vessel to the brim. Calculate the rainfall in cm.

Ans. Rainfall = 2.5 cm

Explanation:
Given, Area of the roof = 22 m × 20 cm 
Height of cylindrical vessel = 3.5 m
Diameter of the vessel = 2 m
∴ Radius of the vessel = 1 m
Let the total rainfall be h m. 
∴ Volume of rain water collected on the roof 
= volume of the cylindrical vessel 
⇒ 22 × 20 × h = π (1)3.5
⇒ 22 × 20 × h =\(\frac{22}{7}\)(3.5)
⇒ 20h = 0.5
⇒ h=\(\frac{1}{40}\)m = 2.5 cm.

Q4. The internal and external diameters of a hollow spherical shell are 6 cm and 10 cm respectively. It is melted and recast into a solid cylinder of base diameter 14 cm. Find the height of the cylinder so formed.

Ans.  Height of the cylinder =\(\frac{8}{3}\)cm

Explanation:
Given, internal diameter of sphere = 6 cm 
External diameter of sphere = 10cm 
Base diameter of cylinder = 14  cm 
Thus, internal radius of sphere ,r₁ = 3 cm 
External radius of sphere r₂ = 5 cm
Base radius of cylinder, r = 7  cm 
Let the height of the cylinder be h cm. 
Thus, volume of cylinder = Volume of spherical shell 
πr²h =\(\frac{4}{3}\) π(r₂³-r₁³)
⇒ π(7)²h =\(\frac{4}{3}\)π[(5)³ - (3)³]
⇒ 3(7)²h = 4 [(5)³ - (3)³]
⇒ 3(49)h = 4 [125 - 27]
⇒ 3(49)h = 4 [98]
⇒ 3h = 4 [2]
⇒ h =\(\frac{8}{3}\)cm.

Q5. A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into 12 toys in the shape of a right-circular cone mounted on a hemisphere. Find the radius of the hemisphere if the total height of the cone is 3 times the radius.

Ans. Given, Diameter of the cylinder = 12 cm
∴ Radius = 6 cm
Height of the cylinder = 15 cm 
Number of toys = 12 
Let the radius of the hemisphere be r cm
Thus,    height of the cone = 3r cm 
Now,   volume of cylinder = π(6)² 15 cm³    
= (36) (15) π cm³    
= 540 π cm^3
‍Now, total volume of 12 toys 
= 12 [volume of hemisphere + volume of cone]
Thus, total volume of 12 toys 
=12[\(\frac{1}{3}\)π3r(r)² +\(\frac{2}{3}\) πr³]cm³
= 4[3πr³ + 2πr³]cm³
=20πr³ cm^3
‍Now,     20πr³ = 540π
⇒ r³ = 27 
⇒ r³ = (3)^3
‍⇒ r = 3 cm 
Thus, radius of the hemisphere = 3 cm and height of the  cone = (3)(3) cm =9 cm
∴   Total  height of the toy 
= (9 + 3) cm = 12 cm.

CBSE Class 10 Maths Chapter wise Important Questions

Conclusion

To improve your understanding of the Surface Area and Volumes chapter, you can explore oswal.io. This platform offers a range of practice questions designed to make learning easier. By practicing these questions, you can strengthen your understanding of various shapes and their measurements, as well as sharpen your skills in solving related problems. 

Frequently Asked Questions

Q1: How is the lateral surface area of a cylinder calculated?

Ans: The lateral surface area of a cylinder is calculated by multiplying its height by its circumference.

Q2: What is a hemisphere, and how is it related to a sphere?

Ans: A hemisphere is half of a sphere. It shares the same base and curved surface, but the hemisphere lacks the bottom half of the sphere.

Q3: How to find the volume of a cylinder?

Ans: The volume of a cylinder is calculated by multiplying the area of its base by its height: V = πr²h, where 'r' is the radius and 'h' is the height.

Q4: What is the formula for the volume of a cone?

Ans: The formula for the volume of a cone is V = (1/3)πr²h, where 'r' is the radius of the base and 'h' is the height.

Q5: How is the total surface area of a sphere calculated?

Ans: The total surface area of a sphere is calculated using the formula A = 4πr², where 'r' represents the radius of the sphere.