Math Olympiad Test: Surface Area and Volume- 2 - Class 10 MCQ

The volume of lead in the shell = volume of a cylinder

⇒ r² = 36 ⇒ r = 6
∴ diameter =  2r = 2 × 6 = 12 cm

Let r be the radius of base of the conical part
Height of the conical part = 4r
Volume of cone with hemispherical top = volume of cone + volume of hemispherical top

Volume of 10 such cones with hemispherical top
= 10 × 2πr³ = 20πr³ cm³
Volume of cylindrical container

20πr³ = 540π
r³ = 27 ⇒ r = 3 cm
diameter = 3  × 2 = 6 cm

Radius of cylinder = 12/2 = 6 cm
height = 15 cm
Volume of cylinder = πr²h
= 22/7 x 6² x 15
= 540π cm³ 
Volume of 12 toys = 540π cm³
Volume of 1 toy = 540π/12 = 45π cm³
Let the radius of the hemisphere be r cm.
Height of the cone = 3r cm
Volume of one toy = Volume of hemisphere + volume of cone

Now

The volume of the cube = 1728
a³ = 1728 ⇒ A = 12 cm
Total surface area = 6a² = 6 × (12)²
= 6 × 144 = 864 cm².

Area of canvas = 2πrh + πrl

= 1320 + 6600 = 7920 cm².

Let the base areas and heights be A and h respectively.
For cylinder, volume = V_C = πr²h = Ah
For cone, volume = V_CO = 1/3 πr²h = 1/3 Ah
For hemisphere, volume = V_h = 2/3 πr³
∴ Ratio of their volumes = 2/3 (πr²)r
= 2/3 Ah [∵ h =r]

Volume of given cylinder = πr²h 
= π(12)² × 16
Now, 

Surface area of sphere = 4πR²
If the radius of sphere is doubled, i.e., R becomes 2R, then
New surface area = 4π(2R)² = 16πR²
= 4 (4πR²)
= 4 (surface area)
∴ Surface area will become 4 times.

Inner surface Area = 4πr² = 324π cm²

∴ Outer radius

∴ thickness of the shell = (12 - 9) = 3 cm

Let the height of the required cone be h cm
∴ Required base area = (16) × 5
= 80 cm² = πr²
Height = h  cm volume = 1/3 (πr²)h
According to given condition
Total volume required = 5 × 100 cm³ = 500 cm³

Ratio of volumes 

Total surface area of cylinder
= 2πr (r + h)
= 2 x 22/7 x 14(14 + h)
= 88 (14 + h)
= 1760 cm².
⇒ h = 1760/88 - 14 = 6 cm
∴ Lateral Surface Area = 2 × π × r × h
= 2 x 22/7 x 14 x 6
= 498 cm².

Let the original radius be r and height be h
Original volume = V = 1/3 πr²h
New radius = 120% of r = 120r/100 = 6r/5
New height = 120% of h = 120h/100 = 6h/5
New volume

Increase in volume 

Increase %

Let the radius of two spheres be R and r

Ratio of their surface area

Let the radii of two cylinders are 2r, 3r and heights be 5h and 3h
Ratio of their volumes