# RS Aggarwal Solutions for Class 10 Maths Chapter 17 Volume and Surface Area of Solids Exercise 17D

R S Aggarwal solutions are given here contain answers to all the questions enlisted in the exercise 17D. Students should go through the R S Aggarwal Solutions for Class 10 Maths exercise 17d to learn how to find volume and surface areas of geometrical figures. These solutions are prepared by subject experts at BYJUâ€™S as per CBSE latest syllabus. Practice R S Aggarwal Maths Chapter 17 for Class 10 effectively to score good marks in the examinations.

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Exercise 17 A Solutions

Exercise 17 B Solutions

Exercise 17 C Solutions

### Access Solutions of Maths RS Aggarwal Class 10 Chapter 17 – Volume and Surface Area of Solids Exercise 17D

Question 1: A river 1.5 m deep and 36 m wide is flowing at the rate of 3.5 km/hr. Find the amount of water (in cubic metres) that runs into the sea per minute.

Solution:

Depth of the river = 1.5 m

Width of the river = 36 m

Using units:

1 hour = 60 minutes

1 km = 1000 m

Flow rate of river = 3.5 km/hr

= (3.5) x 1000)/1×60) m/min

= 350/6 m/min

The amount of water that runs into sea per minute =

350/6 Ã— 1.5 Ã— 36 = 350 Ã— 1.5 Ã— 6 = 3150

The amount of water that runs into sea per minute is 3150 m3/min.

Question 2: The volume of a cube is 729 cm3. Find its surface area.

Solution:

Volume of a cube is 729 cm3 (given)

we know, Volume of the cube = (side)3

(side)3 = 729

or Side = 9

Each side measure of a cube is 9 cm.

Now,

Total surface area of the cube = 6(side)2

= 6 Ã— 92

= 6 Ã— 81

= 486

Total surface area of the cube is 486 cm2.

Question 3: How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?

Solution:

Let ‘a’ be edge if a cube, so edge = 1m or 100 cm

Volume of cube of 100 cm edge = (side)3

= (100)3

= 1000000 cm3

Volume of cubes of 10cm edge = 103 = 1000 cm3

Now,

Number of required cubes = (Volume of cube of 100 cm edge)/(Volume of cubes of 10cm edge)

= 1000000/1000000

= 1000

Question 4: Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm, respectively are melted and formed into a single cube. Find the edge of the new cube formed.

Solution:

Edge of the first cube = 6 cm

Volume of the first cube = (side)3 = (6)3 cm3

Edge of the second cube = 8 cm

Volume of the second cube = (side)3 = (8)3 cm3

Edge of the third cube = 10 cm

Volume of the third cube = (side)3 = (10)3 cm3

Let “a” be side edge of the new formed cube.

Volume of the formed cube = Volume of the First cube + Volume of the Second cube + Volume of the Third Cube

a3 = 63 + 83 + 103

= 216 + 512 + 1000

= 1728

0r a = 12

Edge of the new cube is 12 cm.

Question 5: Five identical cubes, each of edge 5 cm, are placed adjacent to each other. Find the volume of the resulting cuboid.

Solution:

Edge of the given Cube = 5 cm

When 5 identical cubes are placed adjacent to each other, the length of cuboid formed is 5 x 5 = 25 cm

Now, Volume of the resulting cuboid = lbh

= 25 x 5 x 5

= 625 cm3

Question 6: The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their surface areas.

Solution:

Ratio of two cube = 8:27

Let volumes of the two cubes be 8x and 27x

Let a and b be the sides of first cube and second cube respectively.

Volume of cube = (side)3

8x = a3

and 27x = b3

Formula to find surface area = 6(side)2

The required ratio is 4:9.

Question 7: The volume of a right circular cylinder with its height equal to the radius is 25 1/7 cm3. Find the height of the cylinder.

Solution:

Volume of the right circular Cylinder = 176/7 cm3

Height of the right circular cylinder = Radius of the right circular cylinder

â‡¨ h = r

We know, Volume of the right circular Cylinder = Ï€r2h

Ï€r2h = 176/7

22/7 Ã— h2 Ã— h = 176/7

h3 = 8

or h = 2 cm

Height of the cylinder is 2 cm.

Question 8: The ratio between the radius of the base and the height of a cylinder is 2 : 3. If the volume of the cylinder is 12936 cm3, then find the radius of the base of the cylinder.

Solution:

Volume of the cylinder = 12936 cm3

Ratio of the base and the height of a cylinder is 2:3

Let 2x be the radius and 3x be the height.

Volume of the cylinder = 12936 cm3

Ï€r2h = 12936

22/7 x (2x)2(3x) = 12936

x3 = 343

or x = 7

Radius of the base is 14 cm.

Question 9: The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. Find the ratio of their volumes.

Solution:

The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3.

Let 2r and 3r are the radii and 5h and 3h be the heights.

= 20/27

Ratio of their volumes is 20:27

Question 10: 66 cubic cm of silver is drawn into a wire 1 mm in diameter. Calculate the length of the wire in metres.

Solution:

Diameter of the wire = 1 mm

Radius of wire = r = 1/2 mm = 0.5 mm = 0.05 cm

Let the length of the wire be h

Volume of the wire = 66 cm3

Ï€r2h = 66

22/7 x 0.05 Ã— 0.05 Ã— h = 66

h = (66x7x400)/22

h = 8400

Length of the wire is 8400 cm or 84 m.

## RS Aggarwal Solutions for Class 10 Maths Chapter 17 Volume and Surface Area of Solids Exercise 17D Topics:

Class 10 Maths Chapter 17 Volume and Surface Area of Solids Exercise 17D is based on the topics:

• Volume and Surface Area of Cuboid
• Volume and Surface Area of Cube
• Volume and Surface Area of cone
• Volume and Surface Area of cylinder and hollow cylinder
• Volume and Surface Area of a frustum of a cone