#### 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

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

#### Solution

(i) Radius of 1 solid iron sphere = *r*

`"Volume of 1 solid iron sphere "=4/3pir^3`

`"Volume of 27 solid iron spheres "=27xx4/3pir^3`

27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be *r*'.

`"Volume of new solid iron sphere "=4/3pi(r')^3`

`4/3pi(r')^3=27xx4/3pir^3`

`(r')^3=27r^3`

**r' = 3r**

(ii) Surface area of 1 solid iron sphere of radius* r* = 4π*r*^{2}

Surface area of iron sphere of radius *r*' = 4π (*r*')^{2}

= 4 π (3*r*)^{2} = 36 π*r*^{2}

`S/(S')=(4pir^2)/(36pir^2)=1/9=1:9`