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

## Surface Areas and Volumes Exercise Ex. 14.1

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*Answer given in the book is incorrect.

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*Answer given in the book is incorrect.

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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.

## Surface Areas and Volumes Exercise Ex. 14.2

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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}

## Surface Areas and Volumes Exercise Ex. 14.3

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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,

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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.

### Solution 23

(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.

## Surface Areas and Volumes Exercise Rev. 14

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## Surface Areas and Volumes Exercise 14.88

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

So the correct option is (c).

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So, the correct option is (b).

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So the correct option is (b).

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So, the correct option is (b).

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So, the correct option is (d).

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So, the correct option is (c).

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So, the correct option is (a).

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So, correct option is (b).

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So, the correct option (d).

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So, the correct option is (c).

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So, the correct option is (b).

## Surface Areas and Volumes Exercise 14.89

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So, the correct option is (a).

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So, the correct option is (b).

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So, the correct option is (c).

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So, the correct option is (d).

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So, the correct option is (a).

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So, the correct option is (b).

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So, the correct option is (d).

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So, the correct option is (d).

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

So, correct option is (b).

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So, correct option is (c).

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So, correct option is (a).

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So, the correct option is (c).

## Surface Areas and Volumes Exercise 14.90

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So, correct option is (d).

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So, the correct option is (b).

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So, the correct option is (b).

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So, the correct option is (a).

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Diameter = 1.6 m = 160 cm

So, radius = 80 cm

So, the correct option is (b).

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So, the correct option is (d).

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So, correct option is (b).

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So, the correct option is (a).

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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.

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So, the correct option is (a).

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So, the correct option is (b).

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So, the correct option is (c).

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So, the correct option is (c).

## Surface Areas and Volumes Exercise 14.91

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So, the correct option is (a).

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So, the correct option is (c)

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So, the correct option is (b).

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So, the correct option is (c).

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So, the correct option is (c).

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So, the correct option is (a).

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So, the correct option is (b).

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Correct option: (b)

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

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Correct option: (a)

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

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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}

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