# RD SHARMA Solutions for Class 10 Maths Chapter 14 - Surface Areas and Volumes

## Chapter 14 - Surface Areas and Volumes Exercise Ex. 14.1

*Answer given in the book is incorrect.

*Answer given in the book is incorrect.

Height of the conical vessel h = 24 cm

Radius of the conical vessel r =5 cm

Let h be the height of the cylindrical vessel which is filled by water of the conical vessel.

Radius of the cylindrical vessel =10 cm

Volume of the cylindrical vessel = volume of water

π(10)^{2}h=150π

h = 150π¸ 100π

h = 1.5 cm

Thus, the height of the cylindrical vessel is 1.5 cm.

## Chapter 14 - Surface Areas and Volumes Exercise Ex. 14.2

Total area of the canvas = curved surface area of the cone + curved surface area of a cylinder radius = 28 m height (cylinder) = 6 m

height (cone) = 21 m

*l* = slant height of cone

curved surface area of the cone = πrl

=π×28×35

=×28×35 = 3080 m^{2}

curved surface area of the cylinder = 2πrh

=2××28×6

=1056

Total area of the canvas = 3080+1056 =4136 m^{2}

## Chapter 14 - Surface Areas and Volumes Exercise Ex. 14.3

Let the height of the cone be H and the radius be R. This cone is divided into two equal parts.

AQ=1/2 AP

Also,

QP||PC

Therefore,ΔAQD~ΔAPC.

So,

A bucket, made of metal sheet, is in the form of a cone.

R = 15 cm, r = 6 cm and H=35 cm

Now, using the similarity concept, we can writ

Volume of the frustum is

The rate of milk is Rs. 40 per litre.

So, the cost of 51.48 litres is Rs. 2059.20.

(i)

Given:

Radius of lower end (r_{1}) = Diameter/2 = 5
cm

Radius of upper end (r_{2}) = Diameter/2 = 15
cm

Height of the bucket (h) = 24 cm

Area of metal sheet used in making the bucket

= CSA of bucket + Area of smaller circular base

Hence, area of the metal sheet used in making the
bucket is 1711.3 cm^{2}.

(ii)

We should avoid the bucket made by ordinary plastic because it is less strength than metal bucket and also not ecofriendly.

## Chapter 14 - Surface Areas and Volumes Exercise Rev. 14

## Chapter 14 - Surface Areas and Volumes Exercise 14.88

After melting a sphere and converting it into a wire, the volume remains same.

So the correct option is (c).

So, the correct option is (b).

So the correct option is (b).

So, the correct option is (b).

So, the correct option is (d).

So, the correct option is (c).

So, the correct option is (a).

So, correct option is (b).

So, the correct option (d).

So, the correct option is (c).

So, the correct option is (b).

## Chapter 14 - Surface Areas and Volumes Exercise 14.89

So, the correct option is (a).

So, the correct option is (b).

So, the correct option is (c).

So, the correct option is (d).

So, the correct option is (a).

So, the correct option is (b).

So, the correct option is (d).

So, the correct option is (d).

Volume of the sphere = Sum of the volume of the three spheres

So, correct option is (b).

So, correct option is (c).

So, correct option is (a).

So, the correct option is (c).

## Chapter 14 - Surface Areas and Volumes Exercise 14.90

So, correct option is (d).

So, the correct option is (b).

So, the correct option is (b).

So, the correct option is (a).

Diameter = 1.6 m = 160 cm

So, radius = 80 cm

So, the correct option is (b).

So, the correct option is (d).

So, correct option is (b).

So, the correct option is (a).

The cylinder completely encloses the sphere.

Hence, diameter of the sphere = diameter of the cylinder = 2r

Now, h is also given to be 2r.

So, the correct option is (a) or (c).

Note: Both can be the answer since h = 2r.

So, the correct option is (a).

So, the correct option is (b).

So, the correct option is (c).

So, the correct option is (c).

## Chapter 14 - Surface Areas and Volumes Exercise 14.91

So, the correct option is (a).

So, the correct option is (c)

So, the correct option is (b).

So, the correct option is (c).

So, the correct option is (c).

So, the correct option is (a).

So, the correct option is (b).

Correct option: (b)

From the figure, it is clear that diameter of sphere is 2r.

Correct option: (a)

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

Correct option: (a)

When two solid-hemispheres of same base radius r are joined together along their bases, it forms a sphere.

And, CSA of sphere = 4πr^{2}

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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