Question

Twenty seven solid iron spheres, each of radius r and surface are S are melted to form a sphere with surface area S'.
Fund the
I) radius r' of the new sphere
II) ratio of S and S'.

Answer 1

$\huge\mathfrak{Solution:}$

(i) Volume of the solid sphere = 4/3 × pi × r ^3

Therefore, volume of 27 number of such solid spheres = 27 × 4/3 × pi × r^3

= 36 × pi × r^3

But volume of 27 solid spheres = volume of the sphere having radius r'

=> 36 × pi × r^3 = 4/3 × pi × r'^3

Therefore, (r')^3 = 27 × r^3

Therefore, r' = 3 × r

(ii) Now, S' = 4 × pi × r^3

= 4 × pi × (3 × r )^2

= 36 × pi × r, and S = 4 × pi × r^2

Therefore, S/S' = 4 × pi × r^2 / 26 × pi × r^2
= 1/9

Hence, S : S' = 1 : 9

___________________

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Answer 2

HERE is ur Answer ✍️✍️

(i) Volume of the solid sphere = 4/3 × pi × r ^3

Therefore, volume of 27 number of such solid spheres = 27 × 4/3 × pi × r^3 

= 36 × pi × r^3

But volume of 27 solid spheres = volume of the sphere having radius r'

=> 36 × pi × r^3 = 4/3 × pi × r'^3

Therefore, (r')^3 = 27 × r^3

Therefore, r' = 3 × r

(ii) Now, S' = 4 × pi × r^3

= 4 × pi × (3 × r )^2

= 36 × pi × r, and S = 4 × pi × r^2

Therefore, S/S' = 4 × pi × r^2 / 26 × pi × r^2 

= 1/9

Hence, S : S' = 1 : 9

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