if we recast a sphere into a cylinder what are the speciality of these two solid About the author Maya
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) . Learn more :- from a solid cylinder whose height is 3.6 cm and diameter 2.1 CM a conical cavity of the same height and the same diamet… https://brainly.in/question/24336372 A hemisphere of radius 21 cm is completely filled with milk. There is a hole in the bottom whose radius is 0.1 cm. If ra… https://brainly.in/question/25349591 Reply
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 Reply
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) .
Learn more :-
from a solid cylinder whose height is 3.6 cm and diameter 2.1 CM a conical cavity of the same height and the same diamet…
https://brainly.in/question/24336372
A hemisphere of radius 21 cm is completely filled with milk. There is a hole in
the bottom whose radius is 0.1 cm. If ra…
https://brainly.in/question/25349591
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