Question

twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S' find the (a) radius r' of the new sphere
(b) ratio of S and S'
please answer my question it's urgent
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Answer 1

Answer:

Since 27 solid iron spheres are melted to form a single solid sphere, the volume of the newly formed sphere will be equal to the volume of 27 solid iron spheres together.

The surface area of a sphere = 4πr2

The volume of a sphere = 4/3πr3

Therefore, The volume of 27 solid spheres with radius r = 27 × [4/3πr3] = 36πr3 ---------------- (1)

Also, the volume of the new sphere with radius r' = 4/3πr'3----------------- (2)

(i) Volume of the new sphere = Volume of 27 solid spheres

(4/3) πr'3 = 36πr3 [From equation (1) and (2)]

⇒ r'3 = 36πr³ × 3/4π

⇒ r'3 = 27r³

⇒ r' = ∛27r³

r' = 3r

Radius of the new sphere, r' = 3r

(ii) Ratio of S and S′

Now, surface area of each iron sphere, S = 4πr2

Surface area of the new sphere, S' = 4πr'2 = 4π(3r)2 = 36πr2

The ratio of the S and S’ = 4πr2/36πr2 = 1/9

Hence, the radius of new sphere is 3r and the ratio of S and S’ is 1:9.