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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### 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 m^{3}/min.

**Question 2: The volume of a cube is 729 cm ^{3}. Find its surface area.**

**Solution**:

Volume of a cube is 729 cm^{3} (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— 9^{2}

= 6 Ã— 81

= 486

Total surface area of the cube is 486 cm^{2}.

**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 cm^{3}

Volume of cubes of 10cm edge = 10^{3} = 1000 cm^{3}

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} cm^{3}

Edge of the second cube = 8 cm

Volume of the second cube = (side)^{3} = (8)^{3} cm^{3}

Edge of the third cube = 10 cm

Volume of the third cube = (side)^{3} = (10)^{3} cm^{3}

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

a^{3} = 6^{3} + 8^{3} + 10^{3}

= 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 cm^{3}

**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 = a^{3}

and 27x = b^{3}

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 cm ^{3}. Find the height of the cylinder.**

**Solution**:

Volume of the right circular Cylinder = 176/7 cm^{3}

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

â‡¨ h = r

We know, Volume of the right circular Cylinder = Ï€r^{2}h

Ï€r^{2}h = 176/7

22/7 Ã— h^{2} Ã— h = 176/7

h^{3} = 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 cm ^{3}, then find the radius of the base of the cylinder.**

**Solution**:

Volume of the cylinder = 12936 cm^{3}

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 cm^{3}

Ï€r^{2}h = 12936

22/7 x (2x)^{2}(3x) = 12936

x^{3} = 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 cm^{3}

Ï€r^{2}h = 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