if we recast a sphere into a cylinder what are the speciality of these two solid
Answers
Step-by-step explanation:
Radius of solid sphere is r
Hight of hollow cylinder h=24 cm
External radius of base of cylinder R
2
=4 cm
Thickness of hollow cylinder t=2 cm
Hence, inner radius of the hollow cylinder = External radius(R) - thickness of cylinder (t)
Inner radius R
1
=(4−2) cm=2 cm
Now,
Volume of solid sphere V=
3
4
πr
3
And, Volume of hollow cylinder V=πh(R
2
2
−R
1
2
)
solid sphere of radius rr is melted and recast into a hollow cylinder of uniform thickness.
→Volume of sphere = Volume of hollow cylinder
3
4
πr
3
=πh(R
2
2
−R
1
2
)
r
3
=
4π
3πh(R
2
2
−R
1
2
)
r
3
=
4
3h(R
2
2
−R
1
2
)
r
3
=
4
3×24×(4
2
−2
2
)
cm
3
r
3
=
4
3×24×(16−4)
cm
3
r
3
=
4
3×24×12
cm
3
r
3
=
4
864
cm
3
r
3
=216 cm
3
r
3
=(6)
3
cm
3
r=6 cm
solution
Given :- if we recast a sphere into a cylinder what are the speciality of these two solid ?
Answer :-
If we recast a sphere into a cylinder , volume of both solids will be equal .
Let us assume that, a sphere o radius R cm is recast into a cylinder of radius r cm and height h cm .
so,
→ Volume of sphere = Volume of cylinder
→ (4/3) π * (radius)³ = π * (radius)² * height
→ (4/3) * π * R³ = π * (r)² * h
→ (4/3)R³ = r²h .
Note :- if radius of cylinder so formed is equal to its height,
then,
→ (4/3)R³ = r² * r
→ (4/3) = r³/R³
→ (4/3) = (r/R)³
→ r/R = (4/3)^(1/3)
→ r : R = (4)^(1/3) : (3)^(1/3) .
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