Question

"Question 9 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 (i) radius r' of the new sphere, (ii) ratio of S and S'.

Class 9 - Math - Surface Areas and Volumes Page 236"

Answer 1

Surface area:

Surface area of a solid object is a measure of the total area that the surface of the object occupies and it is always measured in square unit.

Sphere:

A sphere is a three dimensional figure which is made up of all points in the space which lie at a constant distance from a fixed point called the centre of the sphere and a constant distance is called its radius.

Surface area of sphere:

For a sphere of radius 'r' ,the  surface area is given by

Surface area of sphere=4πr² sq units.

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Solution:

Given, r be the radius of each solid iron sphere & r' the radius of new solid iron sphere, then

Volume of new sphere= 4/3πr’³

Value of old sphere=4/3πr³

Volume of 27 solid sphere of radius r = 27 × 4/3 πr³ = 36πr³

 

27 solid iron spheres are melted to form a new sphere with radius r'.

A/q,
4/3 πr'3= 36πr³
 r'³= 27r³

r'³= (3r)³

r′ = 3r.............(i)

Radius r' of the new sphere=3r

(ii)

Surface area(S) of solid Iron sphere= 4πr²

Surface area(S') of solid Iron sphere= 4πr’²

S'= 4π(3r)²= 4π×9r²

[ From eq(i)]

S'= 36πr²

Required ratio = S:S′

= 4πr²:36πr² = 1:9

Hence, the ratio of S & S' = 1:9

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 Hope this will help you....

Answer 2

Here is the correct answer