A hollow iron pipe 25 cm long has an exterior diameter equal to 32 cm. If the thickness of the pipe is 1cm, find the whole surface area of the pipe [ Take pi= 22/7 ]
Answers
Given :
A hollow iron pipe 25 cm long has an exterior diameter equal to 32 cm. If the thickness of the pipe is 1cm.
To find :
The whole surface area of the pipe
Solution :
- Height of a hollow iron pipe = 25 cm
Exterior diameter = 32 cm
Exterior radius (R) = 32/2 = 16 cm
- Thickness of a pipe = 1 cm
Interior diameter = 32 - 1 = 31 cm
Interior radius (r) = 31/2 cm
Whole surface area of hollow iron pipe
★ Exterior curved surface area of pipe + interior curved surface area of pipe + area of two rings
→ 2πRh + 2πrh + 2πR² - 2πr²
→ 2πRh + 2πrh + 2π(R² - r²)
- Apply identity
- a² - b² = (a + b)(a - b)
→ 2πRh + 2πrh + 2π[(R + r)(R - r)]
- Take 2π common
→ 2π[Rh + rh + (R + r)(R - r)]
→ 2 × 22/7[16 × 25 + 31/2 × 25 + (16 + 31/2)(16 - 31/2)
→ 44/7[400 + 15.5 × 25 + (16 + 15.5)(16 - 15.5)]
→ 44/7[400 + 387.5 + 31.5 × 0.5]
→ 44/7[787.5 + 15.75]
→ 44/7 × 803.25
→ 44 × 114.75
→ 5049 cm²
•°• Whole surface area of a hollow pipe is 5049cm²
_______________________________
Answer:
Given :-
- Length of pipe = 25 cm
- External Diameter = 32 cm
- Thickness = 1 cm
To Find :-
Whole SA
Solution :-
At first we will divide the External Diameter to make radius
Radius = 32/2
Radius = 16 cm
Internal Diameter = External Diameter - Thickness
Internal Diameter = 32 - 1
Internal Diameter = 31 cm
Internal Radius = 31/2
Internal Radius = r
External Radius = R
Identity to be applied :-
a² - b² = (a + b)(a - b)
2πRh + 2πrh + 2πR² - 2πr²
- π is common
2πRh + 2πrh + 2π(R² - r²)
Let take 2π as common
2π [Rh + rh + (R + r)(R - r)]
2 × 22/7[16 × 25 + 31/2 × 25 + (16 + 31/2)(16 - 31/2)
44/7[400 + 15.5 × 25 + (16 + 15.5)(16 - 15.5)]
44/7 [400 + 15.5 × 25 + (31.5)(0.5)]
- Removing Bracket
44/7[400 + 387.5 + 31.5 × 0.5]
44/7[787.5 + 31.5 × 0.5]
44/7 [787.5 + 15.75]
44/7 × 803.25
44 × 114.75
5049 cm²