Math Olympiad Test: Surface Area and Volume- 1 - Class 10 MCQ

Let the no. of marbles = x
Radius of marble = 1.4/2 = 0.7 cm
Radius of cylinder = 7/2 = 3.5 cm

⇒ x = 150

The volume of cone = Volume of metal in the sphere

Total surface area of the remaining solid = curved surface area of the cylinder + curved surface area of the cone + area of upper base of the cylinder 
= 2πrh + πrl + πr²

Let r be the radius of the cylinder Total surface area = 2πr (h + r)

⇒ r² + 20r - 476 = 0
⇒ r² + 34r - 14r - 476 = 0
⇒ r(r + 34) - 14(r + 34) = 0
⇒ (r - 14) (r + 34) = 0
⇒ r = 14; r = -34 (Not possible)
diameter = 2 × 14 = 28 cm

Curved surface area of cylindrical pillar = 264

From (1) and (2)

264r = 924 × 2

The radius of hemisphere = R
∴ Volume of hemisphere = 2/3 πR³
radius of spherical balls = R/4
Let the no. of spherical balls casted be x
∴ Volume of resulting spheres

Now,
Initial volume = Final volume,

Let the number of cones made be n. The volume of solid sphere = volume of resulting cones

Let the thickness of the shell be t
Inner radius = r, outer radius = R
∴ Volume of spherical shell = 4/3 π(R³ - r³)

∴ Weight of spherical shell

= 11.1786 kg
⇒ R³ = 343 ⇒ R = 7 cm
∴ Thickness = t = (R - r) = (7 - 6) = 1 cm

Ratio of volumes of spheres of radii r₁ and r₂ respectively

Similarly,
ratio of surface areas

According to condition

∴ Ratio of surface areas

Surface area of sphere = 4πr² = 616

∴ diameter = 2r = 2 × 7 = 14 cm