RD Sharma Solutions for Class 10 Maths Chapter- 16 Surface Areas and Volumes
By Karan Singh Bisht
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Updated on 25 Apr 2025, 09:22 IST
RD Sharma Solutions for Class 10 Maths Chapter 16 – Surface Areas and Volumes are designed to deepen students’ understanding of key geometric concepts and help them score well in exams. This chapter plays a crucial role in applying mathematics to real-world problems, particularly in calculating the surface areas and volumes of various solid shapes and their combinations.
Covering one of the most practical and important topics in Class 10 Mathematics, these RD Sharma solutions are prepared by experienced educators to strengthen conceptual clarity and problem-solving skills. The chapter includes three well-structured exercises, each focusing on specific geometric solids.
Students will learn to solve problems involving cuboids, cubes, right circular cylinders, cones, spheres, hemispheres, and frustums of cones. The RD Sharma Solutions Class 10 offer clear, step-by-step explanations for each question, making it easier for learners of all levels to grasp the methods and apply them confidently.
These solutions serve as a reliable study guide for both classroom learning and board exam preparation, ensuring thorough practice and improved mathematical proficiency.
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RD Sharma Class 10 Chapter 16 PDF includes detailed solutions, examples, and extra questions to help you master Surface Areas and Volumes and other topics. Click here to download the RD Sharma Class 10 Chapter 6 PDF.
Access the RD Sharma Solutions for Class 10 Maths Chapter 16 – Surface Areas and Volumes
Q. How many spherical bullets, each of 5 cm in diameter, can be cast from a rectangular block of metal 11dm x 1 m x 5 dm?
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Solution: Given,
A metallic block of dimension 11dm x 1m x 5dm.
The diameter of each bullet = 5 cm
We know that,
Volume of the sphere = 4/3 πr3
Since, 1 dm = 10-1m = 0.1 m
The volume of the rectangular block = 1.1 x 1 x 0.5 = 0.55 m3
Radius of the bullet = 5/2 = 2.5 cm
Let the number of bullets made from the rectangular block be n.
Then from the question,
The volume of the rectangular block = The sum of the volumes of the n spherical bullets
0.55 = n x 4/3 π(2.5)3
Solving for n, we have
n = 8400
Q. 2.2 cubic dm of brass is to be drawn into a cylindrical wire of 0.25 cm in diameter. Find the length of the wire.
Solution:
Given,
2.2 dm3 of brass is to be drawn into a cylindrical wire of Diameter = 0.25 cm
So, the radius of the wire (r) = d/2
= 0.25/2 = 0.125*10-2 cm
Now, 1 cm = 0.01 m
So, 0.1cm = 0.001 m
Let the length of the wire be (h)
We know that,
The volume of the cylinder = πr2h
It’s understood that,
The volume of cylindrical wire = The volume of brass of 2.2 dm3
22/7 (0.125 x 10-2)2 x h
= 2.2 x 10-3
Q. How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm?
Solution: Given,
A solid sphere of radius, R = 8 cm
With this sphere, we have to make spherical balls of radius r = 1 cm
Let’s assume that the number of balls made as n.
Then, we know that
Volume of the sphere = 4/3 πr3
The volume of the solid sphere = The sum of the volumes of n spherical balls
n x 4/3 πr3 = 4/3 πR3
n x 4/3 π(1)3 = 4/3 π(8)3
n = 83 = 512
Q. A cylindrical vessel having a diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm, which are filled completely. Find the diameter of the cylindrical vessel.
Solution: Given,
The diameter of the cylinder = The height of the cylinder
⇒ h = 2r, where h – the height of the cylinder and r – the radius of the cylinder
We know that,
Volume of a cylinder = πr2h
So, volume of the cylindrical vessel = πr22r = 2πr3 (as h = 2r)….. (i)
Now,
Volume of each identical vessel = πr2h
Diameter = 42 cm, so the radius = 21 cm
Height = 21 cm
So, the volume of two identical vessels = 2 x π 212 × 21 ….. (ii)
Since the volumes on equations (i) and (ii) are equal,
On equating both equations, we have
2πr3= 2 x π 212 × 21
r3 = (21)3
r = 21 cm
So, d = 42 cm
Q. What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm?
Solution: Given,
Diameter of the solid cylinder = 2 cm
Length of hollow cylinder = 16 cm
The solid cylinder is recast into a hollow cylinder of length 16 cm, the external diameter of 20 cm and thickness of 2.5 mm = 0.25 cm
We know that,
Volume of a cylinder = πr2h
Radius of the solid cylinder = 1 cm
So,
Volume of the solid cylinder = π12h = πh cm3
Let’s assume the length of the solid cylinder as h.
And,
Volume of the hollow cylinder = πh(R2– r2)
Thickness of the cylinder = (R – r)
0.25 = 10 – r
So, the internal radius of the cylinder is 9.75 cm.
Volume of the hollow cylinder = π × 16 (100 – 95.0625)
Hence, it’s understood that
Volume of the solid cylinder = Volume of the hollow cylinder
πh = π × 16(100 – 95.06)
h = 79.04 cm
Q. 25 circular plates, each of radius 10.5 cm and thickness 1.6 cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.
Solution:
Given,
250 circular plates, each with radius 10.5 cm and thickness of 1.6 cm.
As the plates are placed one above the other, the total height becomes = 1.6 x 25 = 40 cm
We know that,
Curved surface area of a cylinder = 2πrh
= 2π × 10.5 × 40 = 2640 cm2
And, volume of the cylinder = πr2h
= π × 10.52 × 40 = 13860 cm3
Q. 50 circular plates, each of diameter 14 cm and thickness 0.5 cm, are placed one above the other to form a right circular cylinder. Find its total surface area.
Solution: Given,
50 circular plates, each with diameter 14 cm.
Radius of the circular plates = 7cm
Thickness of the plates = 0.5 cm
As these plates are one above the other, the total thickness of all the plates = 0.5 x 50 = 25 cm
So, the total surface area of the right circular cylinder formed = 2πr × h + 2πr2
= 2πr (h + r)
= 2(22/7) x 7 x (25 + 7)
= 2 x 22 x 32 = 1408 cm2
RD Sharma Solutions for Class 10 Maths Chapter- 16 FAQs
Where can I find RD Sharma Solutions for Class 10 Chapter 16 – Surface Areas and Volumes?
You can find RD Sharma Solutions for Class 10 Chapter 16 on Infinity Learn platform. These expert-curated solutions offer step-by-step answers to help students understand and solve problems related to surface areas and volumes of solid figures with ease.
What topics are covered in RD Sharma Chapter 16 – Surface Areas and Volumes?
Chapter 16 covers the surface areas and volumes of 3D shapes, including cubes, cuboids, cylinders, cones, spheres, hemispheres, and frustums of cones. Infinity Learn provides detailed explanations and solved examples to help students master these topics effectively.
Are RD Sharma Solutions for Chapter 16 helpful for board exam preparation?
Yes, the RD Sharma Solutions for Class 10 Chapter 16 available at Infinity Learn are extremely helpful for board exam preparation. They are designed by subject experts to align with CBSE standards and provide clarity on real-world mathematical applications.
How do Infinity Learn RD Sharma Solutions help in understanding volumes and surface areas?
Infinity Learn RD Sharma Solutions simplify the process of understanding complex geometrical shapes by offering precise, step-by-step solutions. These make it easier for students to visualize and calculate surface areas and volumes of various solid shapes and their combinations.