area of 1 solid iron sphere of radius r __twenty seven solid iron spheres each of radius__ (4pir2). (therefore ) Volume of new solid iron sphere (frac43pir'3) (Rightarrow ) (frac43pir'3) 27 (frac43pir3) (Rightarrow ) (r'3 27 r3) (Rightarrow ) r'. Therefore, the radius of new sphere.

Therefore, Volume of the new sphere frac43 pi rprime 3 frac43 pi rprime 336 pi r3 rprime 3frac36 times 3 times r34 27 r3 (3 r)3 rprime3 r, the radius ( rprime ) of the new sphere. Twenty seven solid iron spheres, each of radius r and surface area S are melted to twenty four by seven form a sphere with surface area. Open in App, solution (i radius of 1 solid iron sphere r, volume of 1 solid iron sphere 43r3. Ex.8, 9 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area.

### Twenty seven solid iron spheres, each of radius r and surface area

Therefore, the volume of this iron sphere twenty rs coin will be equal to the volume of 27 solid iron spheres. Surface area of the iron sphere of radius twenty thousend rprime 4 pi rprime. Step - 1: Compare both__twenty seven solid iron spheres each of radius__volume and find the radiusVolume of twenty seven solid spheres 27 34r 336r.(1)Volume of new spheres 34r.(2)Compare (1) & (2)36r 3

*twenty seven solid iron spheres each of radius*34r 327r 3r 3Taking, cube root on both sides3rrr3rStep.

Now, Let the radius of this new sphere be r'. The volume of the solid sphere 4 3.

Volume of each 27 solid iron sphere 27timesfrac43 pi r3 36 pi r3, radius of the new sphere rprime. Hence,

__twenty seven solid iron spheres each of radius__the volume of twenty-seven solid sphere r 3 36.

Question 9 (i twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. New solid iron sphere radius r ' By using the formula, the volume of this new sphere 4 3 r '.

This implies, Volume of each solid iron sphere frac43. I) Radius of 1 solid iron sphere. 4 3 r ' 3 36 r 3 r ' 3 27 r 3 r ' 3.