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NCERT Solutions for Exercise 13.5 Class 10 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Exercise 13.5 Class 10 Maths Chapter 13 - Surface Area and Volumes

Edited By Sumit Saini | Updated on Jul 12, 2022 02:36 PM IST | #CBSE Class 10th
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NCERT Solutions for Class 10 Maths exercise 13.5- The region covered by its six rectangular faces is called a cuboid. The sum of the areas of a cuboid's six rectangular faces equals its surface area. The area of all sides apart from the top and bottom faces is known as lateral surface area (LSA). The space occupied by a cuboid's six rectangular faces is known as its volume.

This Story also Contains
  1. Surface Areas and Volumes Class 10 Chapter 13 Excercise: 13.5
  2. More About NCERT Solutions for Class 10 Maths Exercise 13.5
  3. Benefits of NCERT Solutions for Class 10 Maths Exercise 13.5
  4. NCERT Solutions Subject Wise

The formula for cuboid with length l, breadth b, and height h :

Total surface area 1639997164625

Lateral surface area1639997176508 1639997171529

Volume 1639997166122

NCERT solutions Class 10 Maths exercise 13.5- A cylinder is a solid shape with two circular bases that are joined together by a lateral surface. As a result, a cylinder has three faces: two circular and one lateral.

Formulas of cylinder having radius r and height h:

Curved surface area1639997180198

Total surface area of Cylinder1639997177898

Volume of cylinder 16399971787851639997169958

For find the area of complex solid: Complex figure areas can be disassembled and analysed as simpler known shapes. We can determine the required area of the unknown figure by calculating the areas of these known shapes.

Along with NCERT book Class 10 Maths chapter 13 exercise 13.5 the following exercises are also present.


Surface Areas and Volumes Class 10 Chapter 13 Excercise: 13.5

Q1 A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm 3 .

Answer:

A number of rounds are calculated by :

=\ \frac{Height\ of\ cylinder }{Diameter\ of\ wire}

=\ \frac{12 }{0.3}\ =\ 40\ rounds

Thus the length of wire in 40 rounds will be =\ 40\times 2\pi \times 5\ =\ 400 \pi\ cm\ =\ 12.57\ m

And the volume of wire is: Area of cross-section \times Length of wire

=\pi \times (0.15)^2\times 1257.14

=\ 88.89\ cm^3

Hence the mass of wire is. =\ 88.89\ \times 8.88\ =\ 789.41\ gm

Q2 A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \pi as found appropriate.)

Answer:

The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.

=\ \frac{1}{3}\ \pi r^2h_1\ +\ \frac{1}{3}\ \pi r^2h_2

=\ \frac{1}{3}\ \pi \times 2.4^2\times 5 (Note that sum of heights of both the cone is 5 cm - hypotenuse).

=\ 30.14\ cm^3

Now the surface area of a double cone is :

=\ \pi rl_1\ +\ \pi rl_2

=\ \pi \times 2.4 \left ( 4\ +\ 3 \right )

=\ 52.8\ cm^2

Q3 A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm 3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Answer:

The total volume of the cistern is : =\ 150\times 120\times 110\ =\ 1980000\ cm^3

And the volume to be filled in it is = 1980000 - 129600\ =\ 1850400\ cm^3

Now let the number of bricks be n.

Then the volume of bricks : = n\times 22.5\times 7.5\times 6.5 \ =\ 1096.87n\ cm^3

Further, it is given that brick absorbs one-seventeenth of its own volume of water.

Thus water absorbed :

=\ \frac{1}{17}\times 1096.87n\ cm^3

Hence we write :

1850400\ +\ \frac{1}{17}( 1096.87n)\ =\ 1096.87n

n\ =\ 1792.41

Thus the total number of bricks is 1792.

Q4 In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km 2 , show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Answer:

Firstly we will calculate the volume of rainfall :

Volume of rainfall :

=\ 7280\times \frac{10}{100\times 1000}

=\ 0.7280\ Km ^3

And the volume of the three rivers is :

=\ 3\left ( 1072\times \frac{75}{1000}\times \frac{3}{1000} \right )\ Km^3

=\ 0.7236\ Km ^3

It can be seen that both volumes are approximately equal to each other.

Q5 An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig.).

1656422189091

Answer:

From this, we can write the values of both the radius (upper and lower) and the height of the frustum.

Thus slant height of frustum is :

=\ \sqrt{\left ( r_1\ -\ r_2 \right )^2\ +\ h^2}

=\ \sqrt{\left ( 9\ -\ 4 \right )^2\ +\ 12^2}

=\ 13\ cm

Now, the area of the tin shed required :

= Area of frustum + Area of the cylinder

=\ \pi \left ( r_1\ +\ r_2 \right )l\ +\ 2\pi r_2h

=\ \pi \left ( 9\ +\ 4 \right )13\ +\ 2\pi \times 4\times 10

=\ 782.57\ cm^2

Q6 Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

In the case of the frustum, we can consider:- removing a smaller cone (upper part) from a larger cone.

So the CSA of frustum becomes:- CSA of bigger cone - CSA of the smaller cone

\\=\ \pi r_1l_1\ -\ \pi r_2 (l_1\ -\ l)\\\\=\ \pi r_1\left ( \frac{lr_1}{r_1\ -\ r_2} \right )\ -\ \pi r_2\left ( \frac{r_1l}{r_1\ -\ r_2}\ -\ l \right )\\\\=\ \left ( \frac{\pi r_1^2 l}{r_1\ -\ r_2} \right )\ -\ \left ( \frac{\pi r_2^2 l}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( \frac{r_1^2\ -\ r_2^2}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( r_1\ +\ r_2 \right )

And the total surface area of the frustum is = CSA of frustum + Area of upper circle and area of lower circle.

=\ \pi l\left ( r_1\ +\ r_2 \right )\ +\ \pi r_1^2\ +\ \pi r_2^2

Q7 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Answer:

Similar to how we find the surface area of the frustum.

The volume of the frustum is given by-

=Volume of the bigger cone - Volume of the smaller cone

=\ \frac{1}{3} \pi r^2_1 h_1\ -\ \frac{1}{3} \pi r^2_2 (h_1\ -\ h)

=\ \frac{\pi}{3}\left ( r_1^2 h_1\ -\ r_2^2 (h_1\ -\ h) \right )

=\ \frac{\pi}{3}\left ( r_1^2 \left ( \frac{hr_1}{r_1\ -\ r_2} \right )\ -\ r_2^2 \left ( \frac{hr_1}{r_1\ -\ r_2}\ -\ h \right ) \right )

=\ \frac{\pi}{3}\left ( \frac{hr_1^3}{r_1\ -\ r_2}\ -\ \frac{hr_2^3}{r_1\ -\ r_2} \right )

=\ \frac{\pi}{3}h\left ( \frac{r_1^3\ -\ r_2^3}{r_1\ -\ r_2} \right )

=\ \frac{1}{3} \pi h\left ( r_1^2\ +\ r_2^2\ +\ r_1r_2 \right )

More About NCERT Solutions for Class 10 Maths Exercise 13.5

Surface area and volume of Frustum of a cone: When a solid is cut with a plane, a new type of solid is created. The frustum of a cone, which is formed when a plane cuts a cone parallel to the base, is an example of such a solid. A Frustum is the part of a right circular cone with two circular bases that are sliced by a plane parallel to its base.

Geometry includes a wide range of shapes and sizes, including the sphere, cube, cuboid, cone, cylinder, and so on. There is a volume and a surface area for each shape.

Also Read| Surface Areas and Volumes Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 13.5

  • Conversion of solid from one shape to another in exercise 13.5 Class 10 Maths teaches us about shape conversion due to melting and recasting by posing mischievous difficulties.

  • NCERT syllabus Class 10 Maths chapter 13 exercise 13.5 also covers the important concepts of exercise 13.2. and the previous exercise.

  • Class 10 Maths chapter 13 exercise 13.5 also contains concepts related to curved surface area and volume

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NCERT Solutions Subject Wise


Frequently Asked Questions (FAQs)

1. What the NCERT Solution for Class 10 Maths exercise 13.5 deals with?

This exercise deals with surface area, volume of different types of solids, combinations of solids. Definition of frustum, volume and its surface area.

2. What is the basic step we should take to find out the volumes of complex shapes?

We need to simply divide the complex shape into simpler ones like cube cones etc. then we need to either add their volume or subtract in order to get the desired shape.

3. What quality of the object remains unchanged when it's melted and casted into a new figure?

The volume of the object remains unchanged.

4. We have a cube and we remove a hemisphere from its bottom and add a cone. The total volume of the new object is?

The volume of the object remains unchanged.

5. A sphere of radius ‘r’ and a cone of radius ‘r’ and height ‘r’ also is melted and casted into a cube of side ‘x’ find the value of ‘x’?

The volume of the object remains unchanged.

6. How many cubes of volume 10 is needed to be melted to make a sphere pf radius 10 m?

The volume of the object remains unchanged.

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