A hollow cylindrical metal pipe, open at both ends, has an external diameter of 28 cm. Its thickness is 4 cm and height 49 cm. Find its volume, if the metal weighs 8g. per cu.cm. What is the weight of the of pipe .
Answers
Step-by-step explanation:
Given, thickness of iron =3cm
external diameter = 56cm
Length of tube = 182cm
We know,
volume of hollow cylinder = πh(R²−r²)
∴ internal diameter = external diameter−2×thickness of iron
= 56−2×3
= 56−6
= 50cm
=> Outer radius(R) = 56÷2 = 28cm
inner radius(r) = 50÷2 = 25cm
∴ Volume of hollow cylinder = πh(R²−r²)
= 3.14×182(28²−25²)
= 3.14×82×159
= 90865.32cm³ Answe
Answer:
Given, thickness of iron =3cm
external diameter = 56cm
Length of tube = 182cm
We know,
volume of hollow cylinder = πh(R²−r²)
∴ internal diameter = external diameter−2×thickness of iron
= 56−2×3
= 56−6
= 50cm
=> Outer radius(R) = 56÷2 = 28cm
inner radius(r) = 50÷2 = 25cm
∴ Volume of hollow cylinder = πh(R²−r²)
= 3.14×182(28²−25²)
= 3.14×82×159
= 90865.32
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