Question

A large solid of sphere 15 m melted and recast into several small sphere of diameter 3 m . what is the percentage increase in the surface area of the smaller spheres over that of a large sphere

Answer 1

this is your answer......

Answer 2

Given :- A large solid sphere of diameter 15 m melted and recast into several small sphere of diameter 3 m . what is the percentage increase in the surface area of the smaller spheres over that of a large sphere ?

Solution :-

Let us assume that, total n number of small sphere were made after recast large sphere .

So,

→ Volume of large sphere = n * volume of small sphere

→ (4/3) * π * (15/2)³ = n * (4/3) * π * (3/2)³

→ 15³ / 8 = n * 3³ / 8

→ 15 * 15 * 15 = n * 3 * 3 * 3

→ n = 5 * 5 * 5 = 125 .

Now,

→ Surface area of large sphere = 4 * π * (radius)² = 4 * π * (15/2)² = 225π cm² .

and,

→ surface area of 125 small spheres = 125 * (4 * π * (3/2)²) = 125 * 9π = 1125π cm².

Therefore,

→ increased in surface area = 1125π - 225π = 900π cm².

Hence,

→ % increased in surface area = (900π * 100) / 225π = 400% (Ans.)

∴ % increase in the surface area of the smaller spheres over that of a large sphere is 400.

Learn more :-

from a solid cylinder whose height is 3.6 cm and diameter 2.1 CM a conical cavity of the same height and the same diamet...

https://brainly.in/question/24336372

A hemisphere of radius 21 cm is completely filled with milk. There is a hole in

the bottom whose radius is 0.1 cm. If ra...

https://brainly.in/question/25349591