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Math Olympiad Test: Surface Area and Volume- 4 - Class 10 MCQ


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10 Questions MCQ Test - Math Olympiad Test: Surface Area and Volume- 4

Math Olympiad Test: Surface Area and Volume- 4 for Class 10 2024 is part of Class 10 preparation. The Math Olympiad Test: Surface Area and Volume- 4 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Surface Area and Volume- 4 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Surface Area and Volume- 4 below.
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Math Olympiad Test: Surface Area and Volume- 4 - Question 1

A hemispherical bowl is made of steel of 0.25 cm thickness. The inner radius of the bowl is 5 cm. The volume of steel used is _____. (Use π = 3.141)

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 1

Let r a n d R be inner and outer radius of bowl respectively.

Then, r= 5 cm, R = 5 + 0.25 = 5.25 cm.
Volume of steel used = Outer volume - Inner volume

= 41.25 cm3

Math Olympiad Test: Surface Area and Volume- 4 - Question 2

A sector of a circle of radius 12 cm has the angle 120°. It is rolled up so that two bounding radii are joined together to form a cone. Find the volume of the cone.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 2

Given the radius of circle 12 cm
This becomes the slant height of cone 
The angle of sector 120∘



2π × 48π
This become the circumference of base circle = 2πr = 8π2r = 8r = 4
According to Pythagoras theorem
(Slant height)= (height)2 + (radius)2
122 = (height)2 + 42
144 =(height)2 + 16
(height)2 = 144 − 16
(height)2 = 132
height = √132​


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Math Olympiad Test: Surface Area and Volume- 4 - Question 3

The ratio between the volume of two sphere is 8 : 27. What is the ratio between their surface areas?

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 3

Let radius of two spheres be r1, and r2 respectively 
Given, ratio of volumes = 8 : 27
⇒ 
⇒ r1 : r2 = 2 : 3
Ratio between surface areas = 
= 4/9

Math Olympiad Test: Surface Area and Volume- 4 - Question 4

A right circular cone is 4.1 cm high and the radius of its base is 2.1 cm. Another right circular cone is 4.3 cm high and the radius of the base is 2.1 cm. Both the cones are melted and recast into a sphere. Find the diameter of the sphere.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 4

Volume of cone of r = 2.1 cm & h = 4.1cm = 

Volume of cone of r = 2.1 cm & h = 4.3 cm

Let radius of the sphere be r cm.
Now, Volume of sphere = Sum of Volume of both cones

∴ Diameter of sphere = 4.2 cm.

Math Olympiad Test: Surface Area and Volume- 4 - Question 5

A box opened at the top has its outer dimensions 10 cm × 9 cm × 2.5 cm and its thickness is 0.5 cm, find the volume of the metal.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 5

Outer dimension s of the box are 10 cm, 9 cm and 2.5 cm
Thickness of the box is 0.5 cm
So, inner dimensions of the box is (10 - 2 x 0.5) cm,
(9 - 2 x 0.5) cm and (2.5 - 0.5) cm
i.e, 9 cm, 8 cm and 2 c
Volume of the metal = Volume of outer box - Volume of inner box
= 10 x 9 x 2.5 - 9 x 8 x 2 = 225 - 144 = 81 cm3

Math Olympiad Test: Surface Area and Volume- 4 - Question 6

A cuboidal metal of dimensions 44 cm × 30 cm × 15 cm was melted and cast into a cylinder of height 28 cm. Its radius is _____.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 6

Volume of cuboid = 44 x 30 x 15 cm

This volume is equal to volume of cylinder

∴ 44 x 30 x 15 = π x r2 x 28

∴ r = 15 cm

Math Olympiad Test: Surface Area and Volume- 4 - Question 7

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and the height of the cone outside the hemisphere is 5 cm, find the volume of water left in the tub. (Take π = 22/7)

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 7

Volume of water in the cylinder tub = Volume of the tub

Volume of the solid immersed in the tub = Volume of the hemisphere + Volume of the cone

= 154 cm3

Volume of water left in the tub = Volume of the tub - Volume of solid immersed

= ( 770 - 154) cm3 = 616 cm3

Math Olympiad Test: Surface Area and Volume- 4 - Question 8

A cylindrical vessel of diameter 4 cm is partly filled with water. 300 lead balls are dropped in it. The rise in water level is 0.8 cm. The diameter of each ball is ____.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 8

Radius of vessel (R) = 2 cm 
Rise in water level = 0.8 cm
Volume of water displaced by 300 lead balls
= πR2h = π x 4 x 0 .8 = 3.2m
∴ Volume displaced by one ball = 
Let radius of each ball be r cm 
∴ 
∴ Diameter = 2 x r = 0.4 cm

Math Olympiad Test: Surface Area and Volume- 4 - Question 9

The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost to paint 1 cm2 of the surface is ₹ 0.05. Find the total cost of painting the vessel all over.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 9

External radius of hemispherical vessel 

Internal radius of hemispherical vessel 

External curved surface area of hemispherical vessel = 2πr12

Internal curved surface area of hemispherical vessel = 2πr22

Area of top of the hemispherical vessel

Total surface area of the vessel

=

= 1925.78 cm2

Cost of painting the vessel at the rate of ₹ 0.05 per cm2 = 1925.78 x 0.05

= ₹ 96.28 

Math Olympiad Test: Surface Area and Volume- 4 - Question 10

To construct a wall 24 m long, 0. 4 m thick and 6 m high, bricks of diamensions 25 cm × 16 cm × 10 cm each are used. If the mortar occupies 1/10th of the volume of the wall, find the number of bricks used.

Detailed Solution for Math Olympiad Test: Surface Area and Volume- 4 - Question 10

Volume of the wall = 24 x 0.4 x 6 = 57.6m
Volume of mortar = 57.6/10 = 5.76 m3 
Volume of used bricks = (57.6 - 5.76) m= 51.84 m
Volume of each brick = (25 x 16 x 10) cm3 = 4000 cm3

∴ Number o f bricks used = 51.84 / 4/1000
= 12960

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