**According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 12 in NCERT Class 10 Maths Textbook.**

**CBSE Class 10 Chapter 13 Surface Areas and Volumes Objective Questions **help students to get acquainted with how to solve questions based on surface areas and volumes of different shapes, such as cones, spheres, cylinders and more. Other concepts discussed in this chapter include the conversion of a solid from one shape to another shape and then finding the surface area and volume of the new shape formed. In this article, we provide the solutions to the CBSE Class 10 Maths Chapter 13 – Surface Areas and Volumes Objective Questions, framed as per the textbook.

Based on the changed exam pattern that includes more MCQs, we have covered all the chapter-wise CBSE Class 10 Maths Objective Questions that are expected to be included in the board exams.

## Sub-topics Covered in Chapter 13

Find below the main topics discussed in this chapter.

13.1 Basics (6 MCQs)

13.2 Combination of Solids (6 MCQs)

13.3 Shape Conversion of Solids (7 MCQs)

## Download CBSE Class 10 Maths Chapter 13 – Surface Areas and Volumes Objective Questions Free PDF

** Basics **

1. The lateral surface area of a right circular cone of height 28 cm and base radius 21 cm (in sq. cm) is:

- 2310
- 2110
- 1055
- 1155

** Answer: **(A) 2310

**Solution: **h = 28 cm; r = 21 cm

Therefore, slant height (i) =

= 35 cm

Lateral surface area =Â Ï€rl

=Â 22/7Â Ã— 21Â Ã— 35

= 2310Â cm^{2Â }

2. If the ratio of the radius of a cone and a cylinder of equal volume is 3:5, find the ratio of their heights.

- 25/3
- 28/3
- 23/3
- 7

** Answer:** (A) 25/3

**Solution:** LetÂ r_{1}Â andÂ h_{1} be the radius and height of the cone, andÂ r_{2}Â andÂ h_{2}Â be the radius and height of theÂ cylinder.

It is given that the volume of the cone is equal to the volume of the cylinder.

Thus, the ratio of their height is 25:3.

3. An iron rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

- 0.09 cm
- 0.08 cm
- 0.06 cm
- 0.05 cm

**Answer:** (C) 0.06 cm

**Solution: **Volume of the rod =Â Ï€r^{2}h= (Ï€) Ã— (1/2)^{2} Ã— 8= 2Ï€cm^{3}

Volume of the wire =Â Ï€r^{2}h= (Ï€) Ã— (r)^{2 }Ã— 1800 = 1800Ï€r^{2}cm^{3}

Volume of the rod (old solid shape) = Volume of the wire (New solid shape)

2Â Ï€=1800Ï€r^{2}

r^{2}=1/900

r =Â 1/30

Therefore, the thickness of the wire, i.e. diameter = 1/15 = 0.06 cm

4. What do you understand by the quantity called â€˜areaâ€™?

- It is the height of an object
- It is the quantity that expresses the extent of a planar 2-D surface
- It is the length of an object
- It is the quantity of an object

** Answer: **(B) It is the quantity that expresses the extent of a planar 2-D surface

** Solutions: **The first thing that needs to be understood is that area is a 2-dimensional quantity. The area is the quantity that expresses the extent of a two-dimensional figure or shape, or planar lamina, in the plane. It is only possible to measure the area for 2-D surfaces. There is a different quantity to deal with areas for 3-D surfaces, which we will look at slightly later in the upcoming questions.

Examples of 2-D surfaces are rectangles, circles, ellipses etc. It is possible to find the area for all these 2-D surfaces. But the area of objects such as cubes, cylinders, spheres etc., cannot be defined. But intuitively, we know that there is some area associated with such 3-D objects. How come? The answer is coming soon!

5. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones of diameter 3.5 cm and height 3 cm. The number of cones so formed is:

- 254
- 504
- 540
- 405

**Answer: **(B) 504

**Solution:** Radius of the sphere= 21/2 cm

Volume of the sphere= (4/3) Ï€ (21/2)^{3}cm^{3}

Radius of the cone=7/4Â cmÂ and height=3Â cm

Volume of cone=1/3Ï€r^{2}h=1/3Ï€ (7/4)^{2}Ã—3Â cm^{3}

Let the number of cones formed beÂ n.

Then,

6. How many dimensions are required to make a cuboid?

- 3
- 1
- 15
- 100

**Answer: **(A) 3

**Combination of Solids **

7. There are 2 identical cubes, each having a total surface area equal to â€˜Aâ€™. Let â€˜Sâ€™ be the surface area of the solid obtained by joining these 2 cubes end to end. Which of the following statements is true?

- Â It cannot be determined
- S < 2A
- S > 2A
- S = 2A

** Answer: **(B) S < 2A

**Solution: **When 2 cubes are joined end to end, it can be easily figured out that the ends of the 2 cubes which are joined together will not be visible after joining. But all the other faces will remain visible.

The total surface area of the combined solid can be obtained by adding the surface areas of individual cubes and then subtracting the surface areas of one face of each cube which have become hidden due to joining. The combined solid will be a cuboid whose height and breadth will be the same as the cubeâ€™s side. The length of the cuboid will be the summation of the lengths sides of each cube.

8. The figure consists of 2 cylinders, the inner cylinder is a solid cylinder whose radius is r, and the outer cylinder is a hollow cylinder whose radius is R and height is h; then the volume of fluid it can hold is:

- Ï€r
^{2}h - Ï€R
^{2}h - Ï€(R
^{2}âˆ’r^{2})h - Ï€(R
^{2}+r^{2})h

** Answer:** (C) Ï€ (R^{2}âˆ’r^{2}) h

** Solution: **The inner cylinder is solid, whereas the outer is hollow. If the inner cylinder was not there, then the volume of fluid the outer cylinder can hold would be Ï€R^{2}h, but since the inner cylinder is solid and is occupying some space, it is limiting the volume of the outer cylinder.

So, the volume of fluid the given shape can hold is the difference in the volume of the outer and inner cylinders.

Volume =Â Ï€ (R^{2}âˆ’r^{2}) h

9. Find the volume of the figure.

- 3181.2
- 5162.5
- 7142.8
- 8527.2

** Answer:** (D) 8527.2

**Solution:** Volume of the figure = volume of cone + volume of cylinder + volume of the frustum

= 594 + 5346 + 2587.2

= 8527.2

10. A piece of cloth is required to cover a solid object completely. The solid object is composed of a hemisphere and a cone surmounting it. If the common radius is 7 m and the height of the cone is 1 m, what is the area of cloth required?

- 262.39 m
^{2} - 463.39 m
^{2} - 662.39 m
^{2} - 563 m
^{2}

**Answer**: (B) 463.39 m^{2}

**Solution: **Surface area of hemisphere = 2Ï€r^{2}

=2Ã—22/7Ã— (7)^{2}Â = 308Â m^{2}.

For calculating the surface area of a cone, we need to calculate its slant height,

So, the area of cloth required = (308 + 155.39) m^{2}Â = 463.39Â m^{2 }

11. An oil funnel made of a tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm, and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

- 525.25Â cm
^{2} - 724.25 cm
^{2} - 781.86Â cm
^{2} - 700Â cm
^{2 }

**Answer:** (c) 781.86Â cm^{2}

**Solution: **Curved surface area of the cylinder

2Ï€rh=Ï€Ã—8Ã—10=80Ã—Ï€

The slant height of the frustum can be calculated as follows:

The curved surface area of the frustum

Total curved surface area

=169Ï€+80Ï€

=249Ã—3.14

=781.86 cm^{2}

Hence, the area of the tin sheet required to make the funnel is 781.86 cm^{2}.

12. Ram has a semicircular disc. He rotates it about its diameter by 360 degrees. When he rotates the disc, a volume of air in his room gets swept. What is the name of the object/shape that exactly occupies this volume?

- Cylinder
- Hemisphere
- Sphere
- Cuboid

**Answer:** (C) Sphere

**Solution: **

By seeing 2nd diagram, we can know how the semicircle sweeps and what shape it obtained. The shape obtained is a sphere. If rotation had been done for only 180 degrees instead of 360 degrees, we would get a hemisphere.Â The line segment AB, which acted as the axis of rotation, will also be the diameter of the sphere formed.

**Shape Conversion of Solids **

13. A bucket is in the form of a frustum of a cone, its depth is 15 cm, and the diameters of the top and the bottom are 56 cm and 42 cm, respectively. How many litres of water can the bucket hold?

- 28.49
- 7.5
- 2.5
- 10

**Answer:** (A) 28.49

**Solution: **

R = 28 cm

r = 21 cm

h = 15 cm

Capacity of the bucket =Â 1/3Ï€h (R^{2}+r^{2}+Rr)

= 28.49 litres

So, the bucket holds 28.49 litres of water.

14. A 20 m deep well of diameter 7 m is dug, and the earth taken out is evenly spread out to form a platform of 22 m by 14 m. Find the height of the platform (in m).

- 7.5 m
- 2.5 m
- 10 m
- 5 m

**Answer:** (B) 2.5 m

**Solution: **Diameter of the well = d = 7 m

â‡’Â Radius of the well = r =Â 7/2=3.5m

Height of the well = h =20 m

Length of the platform = 1 = 22 m

Breadth of the platform = b = 14 m

Let the height of the platform = h_{1Â }

According to the given condition, we have:

The volume of the well = Volume of earth dug out

â‡’Â Ï€. r^{2}.h= l Ã— b Ã— h_{1}

â‡’22/7 Ã— 3.5 Ã— 3.5 Ã— 20=22Ã—14Ã—h_{1}

â‡’h_{1}=22/7 Ã— 3.5 Ã— 3.5 Ã— 20 Ã— 122 Ã— 114

â‡’h1 = 2.5 m

Therefore, the height of the platform is 2.5 m.

15. A cylindrical tank is filled by pumping water from a cuboidal tank of dimensions 200cm Ã— 150cm Ã— 95 cm. The radius of the cylindrical tank is 60 cm, and the height is 95 cm. Find the height (in m) of the water left in the cuboidal tank after the cylindrical tank is completely filled. (Take Ï€ = 3.14)

- 0.76 m
- 0.69 m
- 0.59 m
- 0.45 m

**Answer:** (C) 0.59 m

**Solution:** Volume of the cylindrical tank =Â Ï€r^{2}h= (3.14) Ã— (0.6)^{2}Ã—0.95m=1.07m^{3}

Volume of the Cuboidal tank when full = l Ã—Â b Ã—Â h = (2m Ã—Â 1.5m Ã—Â 0.95m) = 2.85Â m^{3}

The volume of water left in the cuboidal tank after completely filling the cylindrical tank = 2.85 â€“ 1.07) = 1.78 m^{3}

Height of water left in cuboidal tank = Volume of water left in the cuboidal tank / (l Ã— b)

= 1.78/ (2 Ã— 1.5) = 0.59 m

So, the height of the water left in the cuboidal tank is 0.59 m.

16. A cylinder is moulded into the shape of a sphere. Which of the following factors will be the same for both shapes?

- Curved surface area
- Surface area
- Volume
- None of these

**Answer:** (C) Volume

**Solution:** Volume is a factor which does not differ with the change of shape. A cylinder can be moulded into a sphere or a cube, or a cuboid of varying dimensions keeping the volume constant.

17. Water in a canal, 6 m wide and 1.5 m deep, is flowing at a speed of 10 km/h. How much area will it irrigate (in m^{2}) in 30 minutes if 8 cm of standing water is needed?

- 256500
- 526500
- 625500
- 562500

** Answer: **(D) 562500

** Solution: **Speed of water flowing through canal =10 km/h =10, 000 m/h

Volume of water flowing through canal in 1 hour = 6Â Ã—Â 1.5Â Ã—Â 10,000 = 90,000Â m^{3}

The volume of water flowing through the canal in 30 minutes = 90000/2=45,000m^{3}

Standing water = 8 cm =0.08 m

According to the given condition, we have:

The area which can be irrigated Ã— 0.08 = 45000

â‡’ Area which can be irrigated = 45000/ 0.08 = 562500 m^{2}.

18. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is:

- 66 m
- 36 m
- 18 m
- 12 m

** Answer:** (B) 36 m

** Solution:** Diameter of metallic sphere=6Â cm

âˆ´ The radius of the metallic sphere=3 cm

Also, we have the diameterÂ of the cross-section of cylindrical wire=0.2 cm

âˆ´ The radius of the cross-section of cylindrical wire=0.1 cm

Let the length of the wire beÂ hÂ cm

Since the metallic sphere is converted into a cylinder-shaped wire of length h cm,

The volume of the metal used in the wire = Volume of the sphere

Hence, the length of the wire is 36 m.

19**. **How many gold coins of 1.75 cm in diameter and 2 mm in thickness can be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?

- 400
- 500
- 350
- 550

** Answer: **(A) 400

**Solution:** Radius of the coin = 0.875 cm

Height of coin = 0.2 cm

Volume of the cylinder =

Volume of cuboid = l Ã—Â b Ã—Â h = 5.5 Ã—Â 10 Ã—Â 3.5

= 192.5Â cm^{3}

Let x be the number of coins that can be made

So, x = volume of cuboid / volume of the cylinder (coin) = 192.5/ 0.481 = 400 coins

So, the number of gold coins required is 400 coins.

The CBSE Class 10 Maths Chapter 13 – Surface Areas and Volumes Objective Questions are available for students who want to excel in the board exam. With the help of these solved questions, students can get practice and thus prepare well for the board exams.

There are also some extra questions from CBSE Class 10 Maths Chapter 13 here, which the students can solve for better practice.

### CBSE Class 10 Maths Chapter 13 Extra Questions

**1. A cylindrical pencil sharpened at one edge is the combination of ______.**

(a) a cone and a cylinder

(b) a hemisphere and a cylinder

(c) two cylinders

(d) frustum of a cone and a cylinder

**2. Given that a right circular cylinder is of radius r cm and height h cm (h>2r); _______ encloses a sphere of diameter.**

(a) 2r cm

(b) r cm

(c) 2h cm

(d) h cm

3.Â **Twelve solid spheres of the same size are made by melting a solid metallic cylinder with a base diameter of 2 cm and height of 16 cm. Calculate the diameter of each sphere.**

(a) 4 cm

(b) 6 cm

(c) 3 cm

(d) 2 cm

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