A cone of base radius 4 cm is divided into two parts by drawing a plane through the mid-point of its height and parallel to its base. Compare the volume of the two parts.

Given, radius of base of cone = 4 cm The cone is divided in two parts through the height of the cone. Let, the height of the cone is h cm . So, the volume of the cone = 1 3 × π × 4 2 × h cm 3 = 16 πh 3 cm 3 ∴ Height of the smaller cone is h 2 cm and radius is 4 2 cm = 2 cm ∴ Volume of smaller cone = 1 3 π × 2 2 × h 2 cm 3 = 2 πh 3 cm 3 ∴ Volume of lower frustum of the cone = Total volume of cone - Volume of shorter cone = 16 πh 3 - 2 πh 3 cm 3 = 16 πh - 2 πh 3 cm 3 = 14 πh 3 cm 3 ∴ Ratio of two parts of the cone = 2 πh 3 : 14 πh 3 = 2 : 14 = 1 : 7 The ratio of volumes of two parts of the cone is 1 : 7 .

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 c m , and its base is of radius 3.5 c m . Find the total surface area of the article.

From the given diagram it can be clearly observed that radius of hemisphere is same as radius of cylinder. Given radius of cylindrical part = radius of hemispherical part = r = 3 . 5 cm . Also, height of the cylinder h = 10 cm . The total surface area of the article = ( C S A of cylinder) + ( 2 × CSA of hemisphere) = ( 2 π r h ) + ( 2 × 2 π r 2 ) = ( 2 π × 3 .5 × 10 ) + ( 2 × 2 π × 3 .5 × 3 .5 ) = 70 π + 49 π = 119 × 22 7 = 17 × 22 = 374 cm 2

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A spherical ball of radius $r_{1}$ units is melted to make $8$ new identical spherical balls each of radii $r_{2}$ units. Then $r_{1} : r_{2}$ is

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Important Questions on Surface Areas and Volumes

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 8

If the height of a bucket in the shape of frustum of a cone is

16 cm

and the diameters of its two circular ends are

40 cm

and

16 cm

, then its slant height is

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 9

The radii of the circular ends of a bucket of height

40 cm

are

24 cm

and

15 cm

. The slant height

(cm)

of the bucket is :

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 10

The radii of the top and bottom of a bucket of slant height

45 cm

are

28 cm

and

7 cm

respectively. The curved surface area of the bucket is

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 30

The inner diameter of a cylindrical glass pot is $8.4 cm$ and the depth of water in the pot is $10.4 cm$ . What is the area of wet surface of the pot?

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 37

There is some wheat in the form of a right circular cone whose diameter of base is

9

metre and height

3.5

metre. Find the volume of the wheat. To cover the wheat in the conical form how much canvas is required?

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 38

The height of a conical shape tower is

4

metre and the perimeter of the lower base is

11

metre. By pouring sand on the tower a cone of sand is formed whose height is

7

metre and the perimeter of the lower base is

22

metre. Find the amount of sand on the cone.

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 51

Akhil runs a toy manufacturing company. He produces a wide variety of toys suitable for kids of different ages. A wooden toy was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is $10 c m$ , and its base is of radius $3.5 c m$ . Find the total surface area of the toy.
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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Surface Areas and Volumes>Exercise>Q 52

The inside area of four walls of a cubical room is

256

sq. metre. The volume of the room is

A spherical ball of radius r1 units is melted to make 8 new identical spherical balls each of radii r2 units. Then r1 : r2 is

Important Questions on Surface Areas and Volumes

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A spherical ball of radius $r_{1}$ units is melted to make $8$ new identical spherical balls each of radii $r_{2}$ units. Then $r_{1} : r_{2}$ is