NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 17

EASY

10th CBSE

IMPORTANT

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During conversion of a solid from one shape to another, the volume of the new shape will

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Important Questions on Surface Areas and Volumes

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 18

The diameters of the two circular ends of the bucket are

44 cm

and

24 cm .

The height of the bucket is

35 cm .

The capacity of the bucket is

EASY

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 19

In a right circular cone, the cross-section made by a plane parallel to the base is a

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10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 20

If volumes of two spheres are in the ratio

64 : 27,

then the ratio of their surface areas is

EASY

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 1

Two identical solid hemispheres of equal base radius

r cm

are stuck together along their bases. The total surface area of the combination is

6 {πr}^{2} .

MEDIUM

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 2

A solid cylinder of radius

r

and height

h

is placed over the cylinder of same height and radius. The total surface area of the shape so formed is

4 π r h + 4 π r^{2} .

EASY

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 3

A solid cone of radius

r

and height

h

is placed over a solid cylinder having the same base radius and height as that of a cone. The total surface area of the combined solid is

πr [\sqrt{r^{2} + h^{2}} + 3 r + 2 h]

.

EASY

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 4

A solid ball is exactly fitted inside the cubical box of side

a .

The volume of the ball is

\frac{4}{3} π a^{3} .

HARD

10th CBSE

IMPORTANT

NCERT EXEMPLAR PROBLEMS-SOLUTIONS MATHEMATICS>Chapter 12 - Surface Areas and Volumes>Exercise>Q 5

The volume of the frustum of a cone is

\frac{1}{3} π h [r_{1}^{2} + r_{2}^{2} - r_{1} r_{2}],

where

h

is vertical of frustum and

r_{1}, r_{2}

are the radius of ends.