Question

PLS ANSWER QUICKLY 2MARROW XAM
64 solid iron spheres,each of radius r and surface area s are melted to form a sphere with surface area s` .find
(i)radius R` of the new sphere

Answer 1

radus=r
surface area=s=4πr²
s=12.56 r²
v=$\frac{4}{3} \pi r ^{3}$=s*$\frac{r}{3}$
volume of 64 spheres=64*s*$\frac{r}{3}$=21.33 *rs

Radius of the new sphere=R
volume= 4/3 πR³=21.33*rs
63.99rs= 4πR³
63.99rs/12.56=R³

5.094 rs=R³
5.094*12.56r*r=R³

63.98 r²=R³
R=∛63.98*r²

Answer 2

let radius of small sphere=r
and radius of larger sphere = R
then
for smaller :
surface area = s =4πr²
then volume of it = 4/3πr³
volume of 64 spheres = 64×s×r/3 = 21.333 × rs
for larger i.e., resulted sphere:
volume= 4/3 πR³ = volume of 64 spheres
4/3 πR³ = 21.333 × rs
4πR³ = 21.333
63.999×rs= 4πR³
63.999× rs/12.56=R³
5.094.. × rs = R³
R = ∛5.094 × rs