Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Very Short Answer Type Questions: Surface Areas & Volumes

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Q1. What is the longest pole that can be put in a room of dimensions l = 10 cm, b = 10 cm and h = 5 cm?
Ans: 
The longest diagonal of a cuboid =Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ The length  of the required pole (diagonal) = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

 

Q2. The total surface area of a cube is 96 cm2. What is its volume?
Ans: 
Total surface area of the cube = 6l2

∴   Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Thus, the volume of the cube = l3 = 43 = 64 cm


Q3. The radius of a sphere doubled. What percent of its volume is increased?
Ans: 
Original volume = (4/3)πr3

Increased volume = (4/3)π(2r)3 = 32/3πr3

Increase in volume =  32/3πr3 -  4/3πr3 = 28/3πr3

∴ Percent increase in volume = Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

 

Q4. Write ‘True or False’ for the following statements: (i) A right circular cylinder just encloses a sphere of radius r as shown in the figure. The area of the sphere is equal to the curved surface area of the cylinder.
Ans: 
True.

∵ [Radius of the sphere] = [Radius of the cylinder] = r

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

∴ The diameter of the sphere = 2r
⇒ Height of the cylinder (h) = 2r
Now, the surface area of the sphere = 4πr
And curved surface area of the cylinder = 2πrh = 2πr (2r) = 4πr2


(ii) An edge of a cube measures ‘r’ cm. If the largest possible right circular cone is cut out of this cube, then the volume of the cone (in cm3) is 1/6πr3
Ans: False.
∵ Height of the cone = r cm
∴ The diameter of the base of the cone = r cm

⇒ Radius of the base of the cone = (r/2) cm

Now volume of the cone Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

 

Q5. If the total surface area of a sphere is 154 cm2. Find its total volume.
Ans: 
Let ‘r’ be the radius of the sphere
∴ Total S.A. = 4 π r2 = 154 cm2

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

or        Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

Now,                 Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

 

Q6. If the radius of a sphere is 3r then what is its volume?
Ans: 

Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

The document Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 11 Question Answers - Surface Areas and Volumes

1. What is the difference between surface area and volume?
Ans.Surface area measures the total area of all the surfaces of a three-dimensional object, while volume measures the amount of space the object occupies. Surface area is typically measured in square units, while volume is measured in cubic units.
2. How do you calculate the surface area of a cylinder?
Ans.To calculate the surface area of a cylinder, you use the formula: Surface Area = 2πr(h + r), where r is the radius of the base and h is the height of the cylinder. This formula accounts for the area of the two circular bases and the rectangular side that wraps around the cylinder.
3. What formulas are used for finding the volume of common shapes like cubes and spheres?
Ans.For a cube, the volume is found using the formula: Volume = a³, where a is the length of a side. For a sphere, the volume is calculated using the formula: Volume = (4/3)πr³, where r is the radius of the sphere.
4. Why is it important to understand surface areas and volumes in real life?
Ans.Understanding surface areas and volumes is essential in various real-life applications, such as construction, manufacturing, and packaging. It helps in determining materials needed, optimizing space, and ensuring proper design for functionality.
5. How can I apply the concepts of surface area and volume in everyday situations?
Ans.You can apply these concepts in everyday situations such as calculating how much paint is needed to cover a wall (surface area) or determining how much water a swimming pool can hold (volume). Understanding these concepts helps in making informed decisions in home improvement and other practical scenarios.
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