NCERT Class Solutions
  • Home
  • 9th Solutions
    • Maths Solutions
    • Science Solutions
    • Social Science Solutions
  • 10th Solutions
    • Science Solutions
    • Maths Solutions
    • Social Science Solutions
    • English Solutions
    • Hindi Solutions
    • Sanskrit Solutions
  • NCERT Books
    • Class 10 Books PDF
    • Class 9 Books PDF
  • About Us
    • Write for Us
    • Contact Us
    • Privacy Policy
    • Disclaimer
  • MP Board
    • MP Board Solutions
    • Previous Year Papers
No Result
View All Result
  • Home
  • 9th Solutions
    • Maths Solutions
    • Science Solutions
    • Social Science Solutions
  • 10th Solutions
    • Science Solutions
    • Maths Solutions
    • Social Science Solutions
    • English Solutions
    • Hindi Solutions
    • Sanskrit Solutions
  • NCERT Books
    • Class 10 Books PDF
    • Class 9 Books PDF
  • About Us
    • Write for Us
    • Contact Us
    • Privacy Policy
    • Disclaimer
  • MP Board
    • MP Board Solutions
    • Previous Year Papers
No Result
View All Result
NCERT Class Solutions
No Result
View All Result
ADVERTISEMENT
Home Class 10th Solutions 10th Maths

NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

by Sudhir
December 4, 2021
in 10th Maths, Class 10th Solutions
Reading Time: 6 mins read
0
NCERT Class 10th Maths Solutions
1
VIEWS
Share on FacebookShare on Twitter

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5. These solutions are based on new NCERT Syllabus.

NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 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:
Since, diameter of the cylinder = 10 cm 10
∴ Radius of the cylinder (r) = \(\frac{10}{2}\) cm = 5cm
⇒ Length of wire in one round = 2πr
= 2 × 3.14 × 5 cm = 31.4 cm
∵ Diameter of wire = 3 mm = \(\frac{3}{10}\) cm
∴ The thickness of cylinder covered in one round = \(\frac{3}{10}\) cm
⇒ Number of rounds (turns) of the wire to cover 12 cm = \(\frac{12}{3 / 10}=12 \times \frac{10}{3}\) = 40
∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds
= l = 40 × 31.4 cm = 1256 cm
Now, radius of the wire = \(\frac{3}{2}\) mm = \(\frac{3}{20}\) cm
∴ Volume of wire = πr2l
= 3.14 × \(\frac{3}{20} \times \frac{3}{20}\) × 1256 cm3
∵ Density of wire = 8.88 g/cm3
∴ Mass of the wire = [Volume of the wire] × density
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 1

NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 2.
A right triangle, whose sides are 3cm and 4cm (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 π as found appropriate).
Solution:
Let us consider the right ABAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse BC = \(\sqrt{3^{2}+4^{2}}\) = 5cm
Obviously, we have obtained two cones on the same base AA’ such that radius = DA or DA’.
Now,
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 2
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 3

Question 3.
A cistern, internally measuring 150 cm × 120 cm × 110 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 brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Dimensions of the cistern are 150 cm, 120 cm and 100 cm.
∴ Volume of the cistern = 150 × 120 × 110 cm3 = 1980000 cm3
Volume of water contained in the cistern = 129600 cm3
∴ Free space (volume) which is not filled with water = (1980000 – 129600) cm3
= 1850400 cm3
Now, volume of one brick
= (22.5 × 7.5 × 6.5) cm3 = 1096.875 cm3
∴ Volume of water absorbed by one brick
= \(\frac{1}{17}\) × 1096.875 cm3
Let n bricks can be put in the cistern.
∴ Volume of water absorbed by n bricks
= \(\frac{n}{17}\) × 1096.875 cm3
∴ Volume occupied by n bricks = [free space in the cistern + volume of water absorbed by n bricks]
⇒ [n × 1096.875] = [1850400 + \(\frac{n}{17}\)(1096.875)]
⇒ 1096.875 n – \(\frac{n}{17}\)(1096.875) = 1850400
⇒ (n – \(\frac{n}{17}\)) × 1096.875 = 1850400
⇒ \(\frac{16}{17} n=\frac{1850400}{1096.875} \Rightarrow n=\frac{1850400}{1096.875} \times \frac{17}{16}\)
= 1792.4102 ≈ 1792
Thus, 1792 bricks can be put in the cistern.

NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 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 97280 km2, show that the total rainfall was approximately equivalent to the addition to the norma water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of three rivers = 3 {(Surface area of a river) × Depth}
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 4
Since 0.7236 km3 ≠ 9.728 km3
∴ The additional water in the three rivers is not equivalent to the rainfall.

Question 5.
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, 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 figure).
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 5
Solution:
We have,
For the cylindrical part:
Diameter = 8 cm ⇒ Radius (r) = 4 cm
Height = 10 cm
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 6

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone.
Solution:
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 7
Since, ∆OC1Q ~ ∆OC2S [By AA similarity]
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 8
Now, the total surface area of the frustum = (curved surface area) + (base surface area) + (top surface area)
= πl(r1 + r2) + πr22 + πr12 = π [(r1 + r2)l + r12 + r22]

NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 7.
Derive the formula for the volume of the frustum of a cone.
Solution:
We have,
[Volume of the frustum RPQS] = [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 9
Since, ∆OC1Q ~ ∆OC2S [By AA similarity]
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 10
From (1) and (2), we have
Volume of the frustum RPQS
NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 11

Previous Post

NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Next Post

NCERT Class 10th Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Examples and MCQs

Related

NCERT Class 10th Sanskrit Solutions
10th Sanskrit

Abhyasvan Bhav Sanskrit Class 10 Chapter 3 अनुच्छेदलेखमन्

NCERT Class 10th Sanskrit Solutions
10th Sanskrit

Abhyasvan Bhav Sanskrit Class 10 Chapter 4 चित्रवर्णनम्

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Categories

  • Books
    • Class 10 Books PDF
  • Class 10th Solutions
    • 10th English
    • 10th Hindi
    • 10th Maths
    • 10th Sanskrit
    • 10th Science
    • 10th Social Science
  • Class 9th Solutions
    • 9th Maths
    • 9th Science
    • 9th Social Science
  • MP Board
  • Uncategorized

Recent

NCERT Class 10th Sanskrit Solutions

Abhyasvan Bhav Sanskrit Class 10 Chapter 4 चित्रवर्णनम्

NCERT Class 10th Sanskrit Solutions

Abhyasvan Bhav Sanskrit Class 10 Chapter 3 अनुच्छेदलेखमन्

NCERT Class 10th Sanskrit Solutions

Abhyasvan Bhav Sanskrit Class 10 Chapter 2 पत्रलेखनम्

NCERT Class Solutions

We provide NCERT Solutions

NCERT Class Solutions App Play Store

Follow Us

Browse By Category

  • Books
    • Class 10 Books PDF
  • Class 10th Solutions
    • 10th English
    • 10th Hindi
    • 10th Maths
    • 10th Sanskrit
    • 10th Science
    • 10th Social Science
  • Class 9th Solutions
    • 9th Maths
    • 9th Science
    • 9th Social Science
  • MP Board
  • Uncategorized
  • Write for Us
  • Privacy Policy
  • Contact Us

© 2022 NCERT Class Solutions .

No Result
View All Result
  • Home
  • 9th Solutions
    • Maths Solutions
    • Science Solutions
    • Social Science Solutions
  • 10th Solutions
    • Science Solutions
    • Maths Solutions
    • Social Science Solutions
    • English Solutions
    • Hindi Solutions
    • Sanskrit Solutions
  • NCERT Books
    • Class 10 Books PDF
    • Class 9 Books PDF
  • About Us
    • Write for Us
    • Contact Us
    • Privacy Policy
    • Disclaimer
  • MP Board
    • MP Board Solutions
    • Previous Year Papers

© 2022 NCERT Class Solutions .

This website uses cookies. By continuing to use this website you are giving consent to cookies being used. Visit our Privacy and Cookie Policy.