Ten spherical shaped gold jewels of radius 'r' are melted and recasted into a cylindrical shaped gold jewel having same radius , then find the curved surface area of the new jewel.
Answers
Question :---- we have to Find CSA of new cylinder .
Given:----- 10 spherical shaped gold jewels of radius r are melted and recast in cylinder .
Formula to be used :-------
- Volume of cylinder = πr²h
- volume of sphere = 4/3 πr³
- CSA of cylinder = 2πrh
- π = 3.14
Solution :---------
since 10 spherical shaped gold of radius r are melted , and recast in a cylinder of same radius r . their volume will be same ..
comparing their volume we will get , Height of cylinder .
so,
10× 4/3 π r³ = π r² × H
40 r = 3H
H = (40r/3)
we get, Height of cylinder and we have its radius same as r , so its CSA will be now ,
CSA of cylinder = 2πrh
2πrh = 2 × π × r × 40r/3
CSA = 80πr²/3
CSA = 80×3.14r²/3
CSA = 83.74cm² (Ans)
So, CSA of formed cylinder will be 83.74cm² ......
Answer:
Radius of 1 sphere = r cm.
Volume = (4/3)πr³
Volume of 10 spheres = 10 * (4/3)πr³
It is melted and recasted into a cylindrical shaped gold jewel having same radius.
Volume of sphere = Volume of cylinder.
==> 10 * (4/3)πr³ = πr²h
==> 40/3 * r = h
==> h = 40r/3
Curved surface area of cylinder = 2πrh
==> 2 * 22/7 * r * 40r/3
==> 83.74r²
Hence, curved surface area of new jewel = 83.74r²