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AP Board Class 10 Maths Chapter 10 Mensuration Ex 10.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 10 Mensuration Ex 10.4 Book Answers

AP Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 10 Mensuration Ex 10.4 book solutions pdf online from this page.

Andhra Pradesh Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 10 Mensuration Ex 10.4
Provider	Hsslive

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AP Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Textbooks Solutions with Answer PDF Download

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10th Class Maths 10th Lesson Mensuration Ex 10.4 Textbook Questions and Answers

Question 1.
A metallic sphere of radius 4.2 cm. is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer:
Given, sphere converted into cylinder.
Hence volume of the sphere = volume of the cylinder.
Sphere:
Radius, r = 4.2 cm
Volume V = 43πr³
= 43 × 227 × 4.2 × 4.2 × 4.2
= 4 × 22 × 0.2 × 4.2 × 4.2
= 4 x 22 x 0.2 x 4.2 x 4.2
= 310.464
Cylinder:
Radius, r = 6 cm
Height h = h say
Volume = πr²h
= 227 × 6 × 6 × h
= 22×367ℎ
= 7927ℎ
Hence, 7927ℎ = 310.464
h = 310.464×7792 = 2.744cm
!! π can be cancelled on both sides i.e., sphere = cylinder

Question 2.
Three metallic spheres of radii 6 cm., 8 cm. and 10 cm. respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.
Answer:
Given : Radii of the three spheres r₁ = 6 cm r₂ = 8 cm r₃ = 10 cm
These three are melted to form a single sphere.
Let the radius of the resulting sphere be ‘r’.
Then volume of the resultant sphere = sum of the volumes of the three small spheres.

∴ 1728 = (2 × 2 × 3) × (2 × 2 × 3) × (2 × 2 × 3)
r³ = 12 × 12 × 12
r³ = 12³
∴ r = 12
Thus the radius of the resultant sphere = 12 cm

Question 3.
A 20 m deep well with diameter 7 m. is dug and the earth got by digging is evenly spread out to form a rectangu¬lar platform of base 22 m. × 14 m. Find the height of the platform.
Answer:
Volume of earth taken out = πr²h
= 227 × 72 × 72 × 20
= 770 m
Let height of plot form = H m.
∴ 22 × 14 × H = 227 × 72 × 72 × 20
H = 3514 = 52 = 212 m
∴ The height of the plat form is 212 m

Question 4.
A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment
Answer:
Volume of the well = Volume of the embank
Volume of the cylinder = Volume of the embank
Cylinder :
Radius r = 𝑑2 = 142 = 7 cm
Height/depth, h = 15 m
Volume V = πr²h
= 227 × 7 × 7 × 15
= 22 × 7 × 15
= 2310 m³
Embank:

Let the height of the embank = h m
Inner radius ‘r’ = Radius of well = 7 m
Outer radius, R = inner radius + width
= 7m + 7m = 14 m
Area of the base of the embank = (Area of outer circle) – (Area of inner circle)
= πR² – πr²
= π(R² – r²)
= 227(142−72)
= 227 × (14+7) × (14-7)
= 227 × 21 × 7
= 462 m²
∴ Volume of the embank = Base area × height
= 462 × h = 462 h m³
∴ 462 h m³ = 2310 m³
h = 2310462 = 5 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm. and height 15 cm. is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm. and diameter 6 cm., having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Answer:
Let the number of cones that can be filled with the ice-cream be ‘n’.
Then total volume of all the cones with a hemi spherical top = Volume of the ice-cream

Ice-cream cone = Cone + Hemisphere = πr²h

Cone:
Radius = 𝑑2 = 62 = 3 cm
Height, h = 12 cm
Volume V = 13πr²h
= 13 × 227 × 3 × 3 × 12
= 227 × 36
= 7927
Hemisphere:
Radius = 𝑑2 = 62 = 3 cm
Volume V = 23πr³
= 23 × 227 × 3 × 3 × 3
= 44×97
= 3967
∴ Volume of each cone with ice-cream = 7927 + 3967 = 11887 cm³
Cylinder:
Radius = 𝑑2 = 122 = 6 cm
Height, h = 15 cm
Volume V = πr²h
= 227 × 6 × 6 × 15
= 22×36×157
= 118807
∴ 118807 = n × 118807
⇒ n = 118807 × 71188 = 10
∴ n = 10.

Question 6.
How many silver coins, 1.75 cm in diameter and thickness 2 mm., need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Answer:
Let the number of silver coins needed to melt = n
Then total volume of n coins = volume of the cuboid
n × πr²h = lbh [∵ The shape of the coin is a cylinder and V = πr²h]

∴ 400 silver coins are needed.

Question 7.
A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, 1/4 of the water flows out. Find the number of lead shots dropped into the vessel.
Answer:

Let the number of lead shots dropped = n
Then total volume of n lead shots = 14 volume of the conical vessel.
Lead shots:
Radius, r = 0.5 cm
Volume V = 43πr³
= 43 × 227 × 0.5 × 0.5 × 0.5
Total volume of n – shots
= n × 43 × 227 × 0.125
Cone:
Radius, r = 5 cm;
Height, h = 8 cm
Volume, V = 13 πr²h
= 13 × 227 × 5 × 5 × 8
= 13 × 227 × 200

∴ Number of lead shots = 100.

Question 8.
A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4 𝑑2 cm and height 3 cm. Find the number of cones so formed.
Answer:
Let the no. of small cones = n Then,
total volume of n cones = Volume of sphere Diameter = 28 cm.
Cones:
Radius r = 𝑑2

Height, h = 3 cm

Total volume of n-cones = n . 1549 cm³
Sphere:
Radius = 𝑑2 = 282 = 14 cm

No. of cones formed = 672.

AP Board Textbook Solutions PDF for Class 10th Maths

AP Board Class 10 Maths Chapter 1 Real Numbers Textbook Solutions PDF

AP Board Class 10 Maths Chapter 2 Sets Textbook Solutions PDF

AP Board Class 10 Maths Chapter 3 Polynomials Textbook Solutions PDF

AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Textbook Solutions PDF

AP Board Class 10 Maths Chapter 5 Quadratic Equations Textbook Solutions PDF

AP Board Class 10 Maths Chapter 6 Progressions Textbook Solutions PDF

AP Board Class 10 Maths Chapter 7 Coordinate Geometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 8 Similar Triangles Textbook Solutions PDF

AP Board Class 10 Maths Chapter 9 Tangents and Secants to a Circle Textbook Solutions PDF

AP Board Class 10 Maths Chapter 10 Mensuration Textbook Solutions PDF

AP Board Class 10 Maths Chapter 11 Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 12 Applications of Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 13 Probability Textbook Solutions PDF

AP Board Class 10 Maths Chapter 14 Statistics Textbook Solutions PDF

Andhra Pradesh Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 10 Mensuration Ex 10.4 exam preparation. The AP Board STD 10th Maths Chapter 10 Mensuration Ex 10.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 10 Mensuration Ex 10.4 Books State Board syllabus with maximum efficiency.

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Thursday, June 9, 2022