37. The radius of solid iron sphere is 8cm. Eight rings of iron
plate of external radius 6=cm and thickness 3cm are
3
made by melting this sphere. Find the internal diameter
of each ring
Answers
Answer:
Solution :-
Radius of the iron sphere = 8 cm
External radius of the 8 rings of iron plates = 20/3 cm
Thickness of the 8 rings of iron plates 3 cm
Let internal radius of the ring be r cm
Volume of the sphere = 4/3πr³
⇒ 4/3*22/7*8*8*8
⇒ 45056/21
= 2145.52 cm³
Now, it can be assumed that each ring of iron plate is a hollow cylindrical shell having internal radius 'r' cm and external radius 20/3 cm along with the height 3 cm
Volume of each ring = π*(R² - r²)h
⇒ 3π*[(20/3)² - (r)²]
⇒ Volume of 8 rings = 8*3π[(20/3)² - (r)²]
⇒ 24π*[(20/3)² - (r)²]
Now, volume of sphere = Volume of 8 iron rings
⇒ 2145.52 = 24π[(20/3)² - (r)²]
⇒ 2145.52 = 24*22/7(400/9 - r²)
⇒ 2145.52*7 = 528(400/9 - r²)
⇒ 15018.64/528 = 400/9 - r²
⇒ r² = 44.44 - 28.44
⇒ r² = 16
⇒ r = √16
⇒ r = 4 cm
So, internal radius of the each ring is 4 cm.
Then, diameter = 4*2
= 8 cm
Answer.
Step-by-step explanation:
Answer:
Answer
We have
Volume of solid iron sphere
3
4
π×8
3
cm
3
=
2
2048
πcm
3
External radius of each iron ring =6
3
2
cm=
3
20
cm
Let the internal radius of each ring be r cm
Since each ring forms a hollow cylindrical shell of external and internal radii
3
20
cm and r cm respectively and height 3cm
Volume of each ring =π
⎩
⎪
⎨
⎪
⎧
(
3
20
)
2
−r
2
⎭
⎪
⎬
⎪
⎫
×3cm
3
volume of 8 such rings
=8π(
9
400
−r
2
)×3cm
3
=24π(
9
400
−r
2
)cm
3
Clearly volume of 8 rings= volume of the sphere
⇒24π(
9
400
−r
2
)=
2
2048
π
⇒
9
400
−r
2
=
2
2048
π×
24π
1
⇒r
2
=16
⇒r=4cm