NCERT Solutions For Class 10 Maths Chapter 12 Surface Area And Volume 2026-27

NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume 2026-27 are designed to help students understand important concepts related to three-dimensional shapes in a clear and simple way. This chapter explains how to calculate the surface area and volume of objects such as cubes, cuboids, cylinders, cones, spheres, hemispheres, and combinations of different solids. These concepts are useful not only for board exams but also in real-life situations, such as finding the capacity of a tank, the space inside a container, or the material needed to cover an object.

The solutions follow the latest NCERT textbook pattern for the 2026-27 academic session and explain every question step by step. Students can use them to learn the correct formulas, understand the method of solving problems, and avoid common calculation mistakes. Each answer is written in simple language so that learners can revise the chapter easily, even before an exam.

Infinity Learn provides NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume 2026-27 to support better understanding and confident exam preparation. The solutions are useful for homework, classroom practice, self-study, and quick revision. They also help students check their answers and improve their problem-solving skills.

Download NCERT Solutions For Class 10 Maths Chapter 12 Surface Area And Volume 2026-27 PDF

The NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume 2026-27 PDF is a helpful study resource for students preparing for school and board examinations. It provides clear, step-by-step solutions to all the questions given in the NCERT textbook. Students can easily understand how to calculate the surface area and volume of different solid shapes, such as cubes, cuboids, cylinders, cones, spheres, and hemispheres. 

Access Class 10 Maths Chapter 12 Surface Area And Volume NCERT Solutions

Exercise 12.1

Question 1: Two cubes, each having a volume of 64 cm³, are joined end to end. Find the surface area of the cuboid formed.

Given:

Step 1: Find the edge of each cube.

Let the edge of each cube be a cm.

Volume of a cube = a³

Therefore, a³ = 64

Since 4 × 4 × 4 = 64, we get:

a = 4 cm

Step 2: Find the dimensions of the resulting cuboid.

When the two cubes are joined end to end, one dimension becomes twice the edge of a cube.

Step 3: Calculate the total surface area of the cuboid.

Total surface area of a cuboid = 2(lb + bh + hl)

= 2[(8 × 4) + (4 × 4) + (4 × 8)]

= 2(32 + 16 + 32)

= 2 × 80

= 160 cm²

Answer: The surface area of the resulting cuboid is 160 cm².

Question 2: A vessel consists of a hollow hemisphere joined to a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find its inner surface area. Use π = 22/7.

Given:

Step 1: Find the common radius.

Radius, r = Diameter/2

= 14/2

= 7 cm

Step 2: Find the height of the cylindrical portion.

The height of a hemisphere is equal to its radius.

Height of the cylinder, h = Total height - Radius of hemisphere

= 13 - 7

= 6 cm

Step 3: Find the inner surface area.

The inner surface consists of the curved surface of the cylinder and the curved surface of the hemisphere.

Inner surface area = 2πrh + 2πr²

= 2 × (22/7) × 7 × 6 + 2 × (22/7) × 7²

= 264 + 308

= 572 cm²

Answer: The inner surface area of the vessel is 572 cm².

Question 3: A toy consists of a cone mounted on a hemisphere. Both parts have a radius of 3.5 cm, and the total height of the toy is 15.5 cm. Find the total surface area of the toy. Use π = 22/7.

Given:

Step 1: Find the vertical height of the cone.

Height of the cone, h = Total height - Radius of hemisphere

= 15.5 - 3.5

= 12 cm

Step 2: Find the slant height of the cone.

Slant height, l = √(r² + h²)

= √[(3.5)² + 12²]

= √(12.25 + 144)

= √156.25

= 12.5 cm

Step 3: Calculate the exposed surface area.

The circular face joining the cone and hemisphere is inside the toy. Therefore, it is not included.

Total surface area = Curved surface area of cone + Curved surface area of hemisphere

= πrl + 2πr²

Curved surface area of cone:

= (22/7) × 3.5 × 12.5

= 137.5 cm²

Curved surface area of hemisphere:

= 2 × (22/7) × (3.5)²

= 77 cm²

Total surface area = 137.5 + 77

= 214.5 cm²

Answer: The total surface area of the toy is 214.5 cm².

Question 4: A hemisphere is placed on a cubical block whose side is 7 cm. What is the greatest possible diameter of the hemisphere? Also find the surface area of the solid. Use π = 22/7.

Step 1: Find the greatest diameter.

The circular base of the hemisphere must fit completely on the square top of the cube. Therefore, its greatest diameter is equal to the side of the cube.

Greatest diameter = 7 cm

Radius, r = 7/2 = 3.5 cm

Step 2: Find the exposed surface area.

The circular part of the cube covered by the hemisphere is not exposed.

Surface area of solid = Surface area of cube - Area of circular base + Curved surface area of hemisphere

= 6a² - πr² + 2πr²

= 6a² + πr²

= 6 × 7² + (22/7) × (3.5)²

= 6 × 49 + 38.5

= 294 + 38.5

= 332.5 cm²

Answer: The greatest diameter of the hemisphere is 7 cm, and the surface area of the solid is 332.5 cm².

Question 5: A hemispherical depression is made in one face of a cubical wooden block. The diameter of the hemisphere is equal to the edge l of the cube. Find the surface area of the remaining solid.

Given:

Edge of cube = Diameter of hemisphere = l

Radius of hemisphere, r = l/2

Step 1: Identify the exposed surfaces.

When the depression is made, the circular portion on the face of the cube is removed. At the same time, the curved inner surface of the hemisphere becomes exposed.

Surface area of remaining solid:

= Surface area of cube - Area of circular opening + Curved surface area of hemisphere

= 6l² - πr² + 2πr²

= 6l² + πr²

Step 2: Substitute r = l/2.

= 6l² + π(l/2)²

= 6l² + (πl²)/4

= (24l² + πl²)/4

= [(24 + π)l²]/4

Answer: The surface area of the remaining solid is [(24 + π)l²]/4 square units.

Question 6: A medicine capsule is shaped like a cylinder with a hemisphere attached at each end. Its total length is 14 mm and its diameter is 5 mm. Find its surface area. Use π = 22/7.

Given:

Step 1: Find the length of the cylindrical part.

The two hemispherical ends together occupy a length equal to 2r.

Length of cylinder, h = 14 - 2(2.5)

= 14 - 5

= 9 mm

Step 2: Calculate the surface area.

The two hemispheres together form a complete sphere.

Surface area of capsule = Curved surface area of cylinder + Surface area of sphere

= 2πrh + 4πr²

= 2π × 2.5 × 9 + 4π × (2.5)²

= 45π + 25π

= 70π

= 70 × (22/7)

= 220 mm²

Answer: The surface area of the capsule is 220 mm².

Question 7: A tent consists of a cylindrical part topped by a cone. The cylindrical portion has a height of 2.1 m and a diameter of 4 m. The slant height of the conical top is 2.8 m. Find the area of canvas required and its cost at Rs. 500 per m². The base is not covered.

Given:

Step 1: Find the canvas area.

Since the base is open, only the curved surfaces of the cylinder and cone are covered.

Canvas area = 2πrh + πrl

= 2 × (22/7) × 2 × 2.1 + (22/7) × 2 × 2.8

= 26.4 + 17.6

= 44 m²

Step 2: Calculate the cost.

Total cost = Area × Rate per square metre

= 44 × 500

= Rs. 22,000

Answer: The tent requires 44 m² of canvas, and its cost is Rs. 22,000.

Question 8: A conical cavity is hollowed out from a solid cylinder. The cylinder and cavity both have a height of 2.4 cm and a diameter of 1.4 cm. Find the total surface area of the remaining solid to the nearest square centimetre. Use π = 22/7.

Given:

Step 1: Find the slant height of the conical cavity.

l = √(r² + h²)

= √[(0.7)² + (2.4)²]

= √(0.49 + 5.76)

= √6.25

= 2.5 cm

Step 2: Find the exposed surface area.

The exposed surfaces are the outer curved surface of the cylinder, the bottom circular base and the inner curved surface of the cone.

Total surface area = 2πrh + πr² + πrl

= 2 × (22/7) × 0.7 × 2.4 + (22/7) × (0.7)² + (22/7) × 0.7 × 2.5

= 10.56 + 1.54 + 5.5

= 17.6 cm²

To the nearest square centimetre, 17.6 cm² becomes 18 cm².

Answer: The total surface area of the remaining solid is approximately 18 cm².

Question 9: A wooden article is formed by scooping out a hemisphere from each end of a solid cylinder. The cylinder is 10 cm high and has a base radius of 3.5 cm. Find the total surface area of the article. Use π = 22/7.

Given:

Step 1: Identify the exposed surfaces.

The surface includes the outer curved surface of the cylinder and the curved inner surfaces of the two hemispherical hollows.

The two hemispherical curved surfaces together have the same area as a complete sphere.

Step 2: Calculate the total surface area.

Total surface area = 2πrh + 4πr²

= 2π × 3.5 × 10 + 4π × (3.5)²

= 70π + 49π

= 119π

= 119 × (22/7)

= 374 cm²

Answer: The total surface area of the wooden article is 374 cm².

Exercise 12.2

Question 1: A solid consists of a cone standing on a hemisphere. Both have a radius of 1 cm, and the height of the cone is also 1 cm. Find the volume of the solid in terms of π.

Given:

Step 1: Write the volume formula.

Volume of solid = Volume of cone + Volume of hemisphere

= (1/3)πr²h + (2/3)πr³

Step 2: Substitute the values.

= (1/3)π × 1² × 1 + (2/3)π × 1³

= π/3 + 2π/3

= 3π/3

= π cm³

Answer: The volume of the solid is π cm³.

Question 2: Rachel makes a model from a thin aluminium sheet. The model consists of a cylinder with a cone attached at each end. Its diameter is 3 cm and its total length is 12 cm. Each conical portion has a height of 2 cm. Find the volume of air inside the model. Assume that its inner and outer measurements are nearly equal. Use π = 22/7.

Given:

Step 1: Find the length of the cylinder.

Length of cylinder = Total length - Heights of the two cones

= 12 - (2 × 2)

= 12 - 4

= 8 cm

Step 2: Find the volume of the cylindrical part.

Volume of cylinder = πr²h

= π × (3/2)² × 8

= π × (9/4) × 8

= 18π cm³

Step 3: Find the volume of the two cones.

Volume of two cones = 2 × (1/3)πr²h

= 2 × (1/3)π × (3/2)² × 2

= 2 × (1/3)π × (9/4) × 2

= 3π cm³

Step 4: Find the total volume of air.

Total volume = 18π + 3π

= 21π

= 21 × (22/7)

= 66 cm³

Answer: The volume of air inside the model is 66 cm³.

Question 3: A gulab jamun contains sugar syrup equal to about 30% of its volume. Each gulab jamun is shaped like a cylinder with two hemispherical ends. Its total length is 5 cm and its diameter is 2.8 cm. Find approximately how much syrup is present in 45 such gulab jamuns. Use π = 22/7.

Given:

Step 1: Find the length of the cylindrical part.

The two hemispherical ends together occupy a length of 2r.

Height of cylinder, h = 5 - 2(1.4)

= 5 - 2.8

= 2.2 cm

Step 2: Find the volume of the cylindrical portion.

Volume = πr²h

= (22/7) × (1.4)² × 2.2

= 13.552 cm³

Step 3: Find the volume of the two hemispherical ends.

Two hemispheres together form one sphere.

Volume = (4/3)πr³

= (4/3) × (22/7) × (1.4)³

= approximately 11.499 cm³

Step 4: Find the volume of one gulab jamun.

Volume of one gulab jamun = 13.552 + 11.499

= approximately 25.051 cm³

Step 5: Find the total volume of 45 gulab jamuns.

Total volume = 45 × 25.051

= approximately 1,127.28 cm³

Step 6: Calculate 30% of the total volume.

Volume of syrup = (30/100) × 1,127.28

= approximately 338.18 cm³

Answer: The 45 gulab jamuns contain approximately 338 cm³ of sugar syrup.

Question 4: A wooden pen stand is shaped like a cuboid and has four conical depressions for holding pens. The cuboid measures 15 cm by 10 cm by 3.5 cm. Each depression has a radius of 0.5 cm and a depth of 1.4 cm. Find the volume of wood in the stand. Use π = 22/7.

Given:

Step 1: Find the volume of the cuboid.

Volume of cuboid = l × b × h

= 15 × 10 × 3.5

= 525 cm³

Step 2: Find the volume removed by the four conical depressions.

Volume of four cones = 4 × (1/3)πr²h

= 4 × (1/3) × (22/7) × (0.5)² × 1.4

= approximately 1.467 cm³

Step 3: Find the remaining volume of wood.

Volume of wood = Volume of cuboid - Volume of four depressions

= 525 - 1.467

= approximately 523.53 cm³

Answer: The volume of wood in the pen stand is approximately 523.53 cm³.

Question 5: An inverted conical vessel has a height of 8 cm and an open top of radius 5 cm. It is filled completely with water. Spherical lead shots, each having a radius of 0.5 cm, are dropped into it. As a result, one-fourth of the water flows out. Find the number of lead shots dropped into the vessel.

Given:

Step 1: Relate the displaced water to the lead shots.

Because the vessel was initially full, the volume of water that flows out is equal to the total volume of the lead shots.

Let the number of lead shots be n.

(1/4) × Volume of cone = n × Volume of one sphere

(1/4) × (1/3)πR²h = n × (4/3)πr³

Step 2: Substitute the measurements.

(1/4) × (1/3)π × 5² × 8 = n × (4/3)π × (0.5)³

After cancelling π and simplifying:

5² × 8 = 16n × (0.5)³

200 = 16n × 0.125

200 = 2n

n = 100

Answer: The number of lead shots dropped into the vessel is 100.

Question 6: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, with another cylinder of height 60 cm and radius 8 cm placed on it. Find the mass of the pole if 1 cm³ of iron has a mass of approximately 8 g. Use π = 3.14.

Given:

Step 1: Find the volume of the larger cylinder.

V₁ = πr₁²h₁

= π × 12² × 220

= 31,680π cm³

Step 2: Find the volume of the smaller cylinder.

V₂ = πr₂²h₂

= π × 8² × 60

= 3,840π cm³

Step 3: Find the total volume.

Total volume = 31,680π + 3,840π

= 35,520π

= 35,520 × 3.14

= 111,532.8 cm³

Step 4: Calculate the mass.

Mass = Volume × Mass per cubic centimetre

= 111,532.8 × 8

= 892,262.4 g

Since 1,000 g = 1 kg:

Mass = 892,262.4/1,000

= 892.2624 kg

Answer: The mass of the iron pole is approximately 892.26 kg.

Question 7: A solid is formed by placing a right circular cone of height 120 cm and radius 60 cm on a hemisphere of radius 60 cm. It is placed upright inside a cylindrical vessel full of water and touches the bottom. The cylinder has a radius of 60 cm and a height of 180 cm. Find the volume of water left in the cylinder. Use π = 22/7.

Given:

Step 1: Find the volume of the cylindrical vessel.

Volume of cylinder = πr²h

= π × 60² × 180

= π × 3,600 × 180

= 648,000π cm³

Step 2: Find the volume of the cone.

Volume of cone = (1/3)πr²h

= (1/3)π × 60² × 120

= 144,000π cm³

Step 3: Find the volume of the hemisphere.

Volume of hemisphere = (2/3)πr³

= (2/3)π × 60³

= 144,000π cm³

Step 4: Find the volume of the complete solid.

Volume of solid = 144,000π + 144,000π

= 288,000π cm³

Step 5: Find the water remaining.

Volume of water left = Volume of cylinder - Volume of solid

= 648,000π - 288,000π

= 360,000π cm³

= 360,000 × (22/7)

= approximately 1,131,428.57 cm³

Since 1 m³ = 1,000,000 cm³:

Volume of water left = 1,131,428.57/1,000,000

= approximately 1.131 m³

Answer: The volume of water left in the cylinder is approximately 1.131 m³.

Question 8: A spherical glass vessel has a cylindrical neck that is 8 cm long and 2 cm in diameter. The diameter of the spherical portion is 8.5 cm. A child measures its capacity as 345 cm³. Check whether this measurement is correct, using the given dimensions as internal measurements and π = 3.14.

Given:

Step 1: Find the volume of the spherical portion.

Volume of sphere = (4/3)πr³

= (4/3) × 3.14 × (4.25)³

= approximately 321.392 cm³

Step 2: Find the volume of the cylindrical neck.

Volume of cylinder = πr²h

= 3.14 × 1² × 8

= 25.12 cm³

Step 3: Find the total capacity of the vessel.

Total volume = 321.392 + 25.12

= 346.512 cm³

The calculated volume is approximately 346.51 cm³, which is not equal to 345 cm³.

Answer: The child's measurement is not correct. The vessel can hold approximately 346.51 cm³ of water.

:contentReference[oaicite:0]{index=0}