Question

is some one know how to solve this
please answer me fast it's urgent
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'
​

Answer 1

Answer:

(i) Radius of 1 solid iron sphere =r

⇒Volume =

3

4

​

πr

3

⇒Volume of 27 solid iron spheres =27×

3

4

​

πr

3

⇒

3

4

​

πr

′

 

3

=27×

3

4

​

πr

3

r

′

 

3

=2πr

3

⇒r

′

=3r

(ii) Surface area =4πr

2

SA r

′

=4πr

′

 

2

=4π(3r)

2

=36πr

2

S

′

S

​

=

36πr

2

4πr

2

​

=

9

1

​

=1:9

Step-by-step explanation:

Answer 2

Answer:

For 27 iron spheres,

radius= r , SA=S

For big sphere

radius = r' , SA= S'

Volume of big sphere = volume of 27 spheres

a) 4/3π(r'^3)= 27×4/3×π×(r^3)

r'^3= (3r^3)^3

powers are same

r'= 3r

S = SA of iron sphere = 4π(r^2)

b) S' = SA of big sphere = 4π(r'^2) = 4π( 3r)^2

ratio of S and S' = 4π(r^2)/4π( 3r)^2= 1/9 = 1:9