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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 12.

NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.5, is available here according to the NCERT textbook questions. The solutions are available in PDF and can be downloaded easily. These NCERT Class 10 solutions are designed by expert teachers having vast knowledge of Maths subject, keeping in mind students’ understanding skills.

The solutions provided here for Class 10 Exercise 13.5 are as per the CBSE syllabus and guidelines. It is helpful for students to practise the Solutions for Class 10 Maths, which will help them prepare well and score good marks in the board exam.

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.5

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Access Other Exercise Solutions of Class 10 Maths Chapter 13 – Surface Areas and Volumes

Exercise 13.1 Solutions 9 Question (7 long, 2 short)

Exercise 13.2 Solutions 8 Question (7 long, 1 short)

Exercise 13.3 Solutions 9 Question (9 long)

Exercise 13.4 Solutions 5 Questions (5 long)

Access Answers to NCERT Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.5

1. 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 cm³.

Solution:

Given that,

Diameter of cylinder = 10 cm

So, the radius of the cylinder (r) = 10/2 cm = 5 cm

∴ Length of wire in completely one round = 2πr = 3.14 × 5 cm = 31.4 cm

It is given that diameter of wire = 3 mm = 3/10 cm

∴ The thickness of the cylinder covered in one round = 3/10 m

Hence, the number of turns (rounds) of the wire to cover 12 cm will be

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds

40 x 31.4 cm = 1256 cm

Radius of the wire = 0.3/2 = 0.15 cm

Volume of wire = Area of cross-section of wire × Length of wire

= π(0.15)²×1257.14

= 88.898 cm³

We know,

Mass = Volume × Density

= 88.898×8.88

= 789.41 gm

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

Solution:

Draw the diagram as follows:

Let us consider the ABA

Here,

AS = 3 cm, AC = 4 cm

So, hypotenuse BC = 5 cm

We have got 2 cones on the same base AA’ where the radius = DA or DA’

Now, AD/CA = AB/CB

By putting the value of CA, AB and CB, we get

AD = 2/5 cm

We know,

DB/AB = AB/CB

So, DB = 9/5 cm

As, CD = BC-DB,

CD = 16/5 cm

Now, the volume of the double cone will be

Solving this, we get

V = 30.14 cm³

The surface area of the double cone will be

= 52.75 cm²

3. A cistern, internally measuring 150 cm × 120 cm × 100 cm, has 129600 cm³ 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 being 22.5 cm × 7.5 cm × 6.5 cm?

Solution:

Given that the dimension of the cistern = 150 × 120 × 110

So, volume = 1980000 cm³

Volume to be filled in cistern = 1980000 – 129600

= 1850400 cm³

Now, let the number of bricks placed be “n”

So, the volume of n bricks will be = n×22.5×7.5×6.5

Now, as each brick absorbs one-seventeenth of its volume, the volume will be

= n/(17)×(22.5×7.5×6.5)

For the condition given in the question,

The volume of n bricks has to be equal to the volume absorbed by n bricks + the volume to be filled in the cistern

Or, n×22.5×7.5×6.5 = 1850400+n/(17)×(22.5×7.5×6.5)

Solving this, we get

n = 1792.41

4. 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 km², 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.

Solution:

From the question, it is clear that

Total volume of 3 rivers = 3×[(Surface area of a river)×Depth]

Given,

Surface area of a river = [1072×(75/1000)] km

And,

Depth = (3/1000) km

Now, volume of 3 rivers = 3×[1072×(75/1000)]×(3/1000)

= 0.7236 km³

Now, the volume of rainfall = total surface area × total height of rain

= 0.7280 km³

For the total rainfall to be approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to the volume of 3 rivers.

Since 0.7280 km³ is approximately equivalent to 0.7236 km³,

The question statement is true.

5. An oil funnel made of a 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.).

Solution:

Given,

Diameter of the upper circular end of the frustum part = 18 cm

So, radius (r₁) = 9 cm

Now, the radius of the lower circular end of the frustum (r₂) will be equal to the radius of the circular end of the cylinder.

So, r₂ = 8/2 = 4 cm

Now, height (h₁) of the frustum section = 22 – 10 = 12 cm

And,

Height (h₂) of cylindrical section = 10 cm (given)

Now, the slant height will be-

Or, l = 13 cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

= π(r₁+r₂)l+2πr₂h₂

Solving this, we get

Area of tin sheet required = 782 4/7 cm².

6. 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.

Solution:

Consider the diagram

Let ABC be a cone. From the cone, the frustum DECB is cut by a plane parallel to its base. Here, r₁ and r₂ are the radii of the frustum ends of the cone, and h is the frustum height.

Now, consider the ΔABG and ΔADF,

Here, DF||BG

So, ΔABG ~ ΔADF

Now, by rearranging, we get

The total surface area of the frustum will be equal to the total CSA of the frustum + the area of the upper circular end + the area of the lower circular end

= π(r₁+r₂)l+πr₂²+πr₁²

∴ Surface area of frustum = π[(r₁+r₂)l+r₁²+r₂²].

7. Derive the formula for the volume of the frustum of a cone.

Solution:

Consider the same diagram as the previous question.

Now, approach the question in the same way as the previous one and prove that,

ΔABG ~ ΔADF.

Again,

Now, rearrange them in terms of h and h₁

The total volume of the frustum of the cone will be = Volume of cone ABC – Volume of cone ADE

= (⅓)πr₁²h₁ -(⅓)πr₂²(h₁ – h)

= (π/3)[r₁²h₁-r₂²(h₁ – h)]

Solving this, we get (⅓)πh(r₁²+r₂²+r₁r₂)

∴ Volume of the frustum of the cone = (⅓)πh(r₁²+r₂²+r₁r₂).

Exercise 13.5 of Class 10 Maths has problems based on the topics covered in the chapter, surface areas and volumes. This will include finding the surface area and volumes of the given object, based on different situations, finding the length and height of the given object and also questions related to the frustum of the cone and deriving formulas for the combined shapes.

Learn the entire Solutions for Chapter 13 of Class 10 Maths here in this article. Besides, BYJU’S provides other learning materials, such as notes, books, and previous years’ question papers, along with easy tips and methods, which will help students to do their best in the board exam.

The questions in Exercise 13.5 are prepared as per the topics covered in the complete chapter using the important formulas. The NCERT Solutions will help students to solve problems in a detailed way, following each and every step and method.