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13. Surface Areas and Volumes-Exercise 13.3 Mathematics Class 10 In English
Last Updated : 06 March 2026
Get updated solved mathematics for class 10 chapter 13. Surface Areas and Volumes Topic Exercise 13.3 with clear explanations free study material in English medium.
13. Surface Areas and Volumes-Exercise 13.3 Mathematics Class 10 In English
Last Updated : 06 March 2026
13. Surface Areas and Volumes-Exercise 13.3 Mathematics Class 10 In English
13. Surface Areas and Volumes
Exercise 13.3
Exercise 13.1 Chapter 13. Surface Areas and Volumes
NCERT Solution for class 10 maths in English Medium:
Q1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Radius of sphere (r) = 4.2 cm
Radius of cylinder (R) = 6 cm
Let the height of the cylinder = h cm
Since the sphere is cast into a cylinder, therefore
Volume of cylinder = volume of sphere
Q2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Let the radius of large solid sphere = R cm
GIven : r1 = 6 cm, r2 = 8 cm and r3 = 10 cm
The radius of new sphere is 12 cm
Q3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of well = 7 m
Thus, radius of the well (r) = 3.5 cm
Depth of well (h) = 20 m
Length (l) = 22 m and breadth (b) = 14 m of the platform.
Let the height of platform = h m
Volume of platform = Volume of earth taken out from the well
l × b × h = πr2h
The height of platform = 2.5 m
Q4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of the well = 3 m
Radius of well (r) = 3/2 m = 1.5 m
Depth of well (H) = 14 m
Width of circular ring around the well = 4 m
Thus, outer radius of the ring (R) = 4 m + 1.5 = 5.5 m
Let the height of the circular embankment = h m
Volume of circular platform = Volume of earth taken out from the well
⇒ πR2h - πr2h = πr2H
⇒ πh(R2 - r2) = πr2H
⇒ h (R2 - r2) = r2H
⇒ h [(5.5)2 - (1.5)2] = 1.5 × 1.5 × 14
⇒ h (5.5 +1.5) (5.5 - 1.5) = 1.5 × 1.5 × 14 [ a2 - b2 = (a + b) (a - b) ]
⇒ h = 1.125 m
The height of ring embankment = 1.125 m
Q5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
The diameter of cylindrical vessel = 12 cm
Radius of vessel R = 6 cm
Height of vessel H = 15 cm
The number of such cones which can be filled with ice-cream is 10.
Q6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
Hence, Number of coins is 400
Q7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Radius of cylindrical bucket R = 18 cm
and height H = 32 cm
Height of conical pile = 24 cm
Volume of cylindrical bucket = πR2H
Q8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Length of canal in 1 hour l = 10km = 10000 m
The breadth of canal, b = 6 m
Depth of canal, h = 1.5 m
The volume of water in canal in 1 hour = l × b × h
= 10000 × 6 × 1.5 m3
= 90000 m3
Area = 562500 m2
Therefore, 562500 m2 areas are required for irrigation.
Q9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Diameter of tank = 10 m
Radius of tank = 5 m
depth of tank h = 2 m
Diameter of pipe = 20 cm
Radius of pipe = 10 cm = 0.1 m
Length of pipe in 1 hour = 3 km = 3000 m
Now volume of water in pipe in 1 hour = πr2h
= π × 0.1 × 0.1 × 3000
= π × 30 m3
Volume of water that can be filled in the tank = πr2h
= π × 5 × 5 × 2
Taken time to fill the tank
Hence 100 minutes to take to fill the tank.
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NCERT Solutions for Class 10 mathematics Chapter 13. Surface Areas and Volumes Topic Exercise 13.3
NCERT Solutions for Class 10 mathematics Chapter 13. Surface Areas and Volumes Topic Exercise 13.3 are prepared to help students understand important concepts of the chapter in a simple and clear manner. These solutions are based on the latest CBSE syllabus and follow the official NCERT textbooks. Students who are searching for accurate answers and step-by-step explanations can use these solutions to improve their learning and prepare better for school exams as well as board examinations.
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About Chapter 13. Surface Areas and Volumes of Class 10 mathematics
Chapter 13. Surface Areas and Volumes is an important part of the Class 10 mathematics syllabus. This chapter explains several key concepts which are essential for understanding the subject in detail. Students often face difficulties while solving textbook questions related to this chapter. That is why NCERT Solutions for Chapter 13. Surface Areas and Volumes are provided to explain every concept in a simple and structured way.
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Understanding Topic: Exercise 13.3
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13. Surface Areas and Volumes-Exercise 13.3 Mathematics Class 10 In English
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