A semicircular sheet of metal of diameter 28 cm bent into an open conical cup.Find the depth and capacity of the cup .
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Answered by
3
Heya,
Solution to your question is:-
Let r cm and R cm be the radius of the semi-circular sheet and base of the conical cup respectively.
Suppose the depth of the conical cup is H cm.
Given, 2r = 28 cm
⇒ r = 14 cm
When the semi-circular sheet of metal is bent into an open conical cup, then
Slant height of the cone, L = Radius of the semi-circular sheet = 14 cm
Circumference of base of cone = 2πR
∴ 2π R = 14π cm
⇒ 2R = 14 cm
⇒ R = 7 cm
Slant height of the cone, L = 14 cm
√(7 cm²+ H²= 14cm {Since, L=√R²+H²}
⇒ 49 cm²+ H ²= (14 cm)²= 196 cm²
⇒ H ²= 196 cm² – 49 cm² = 147 cm²
⇒ H²=√147cm²=7√3cm
Therefore,
Capacity or volume of the conical cup
= 1/3πR²H
= 1/3×22/7×(7)²×7√3
= 【1078√3/3cm³】 ANSWER...
HOPE THIS HELPS YOU:-))
@Ashu
Solution to your question is:-
Let r cm and R cm be the radius of the semi-circular sheet and base of the conical cup respectively.
Suppose the depth of the conical cup is H cm.
Given, 2r = 28 cm
⇒ r = 14 cm
When the semi-circular sheet of metal is bent into an open conical cup, then
Slant height of the cone, L = Radius of the semi-circular sheet = 14 cm
Circumference of base of cone = 2πR
∴ 2π R = 14π cm
⇒ 2R = 14 cm
⇒ R = 7 cm
Slant height of the cone, L = 14 cm
√(7 cm²+ H²= 14cm {Since, L=√R²+H²}
⇒ 49 cm²+ H ²= (14 cm)²= 196 cm²
⇒ H ²= 196 cm² – 49 cm² = 147 cm²
⇒ H²=√147cm²=7√3cm
Therefore,
Capacity or volume of the conical cup
= 1/3πR²H
= 1/3×22/7×(7)²×7√3
= 【1078√3/3cm³】 ANSWER...
HOPE THIS HELPS YOU:-))
@Ashu
Answered by
0
Answer:
1078 root 3/3
Step-by-step explanation:
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