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

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

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)² = (height)² + (radius)²
122 = (height)² + 4²
144 =(height)² + 16
(height)² = 144 − 16
(height)² = 132
height = √132​

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

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.

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

Volume of cuboid = 44 x 30 x 15 cm³ 

This volume is equal to volume of cylinder

∴ 44 x 30 x 15 = π x r² x 28

∴ r = 15 cm

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

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

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 

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) m³ = 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