An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylinderical part is 240 cm high and the conical part in 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
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45
Answer:
- Radius = 8 cm
- Height of cylinder = 240 cm
- the conical part in 36 cm high.
Weight of the pillar if one cu. cm of iron weighs 7.8 grams.
- Volume of cylinder = πr²h
- Volume of cone = ⅓πr²h
Now,
- Volume of cylinder = 3.14 × 8 × 8 × 240
- => 48320.4 cm^3
Now,
- ⅓ × 3.14 × 8 × 8 × 36
- 1 × 3.14 × 8 × 8 × 12
- 3.14 × 64 ×12
- 2411.52 cm^3
Now,
Weight of pillar = Volume of cylinder + volume of cone
- W = 48320.4 + 2411.52
- W = 50730
Now,
- 1kg = 1000gm
- 7.8/1000 × 50730
- 0.0078 × 50730
- 395.4 kg
Answered by
2
Answer:
Answer:
Given :-
Radius = 8 cm
Height of cylinder = 240 cm
the conical part in 36 cm high.
To Find :-
Weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Solution :-
We know that
Volume of cylinder = πr²h
Volume of cone = ⅓πr²h
Now,
Volume of cylinder = 3.14 × 8 × 8 × 240
=> 48320.4 cm^3
Now
⅓ × 3.14 × 8 × 8 × 36
1 × 3.14 × 8 × 8 × 12
3.14 × 64 ×12
2411.52 cm^3
Now
Weight of pillar = Volume of cylinder + volume of cone
W = 48320.4 + 2411.52
W = 50730
Now
1kg = 1000gm
7.8/1000 × 50730
0.0078 × 50730
395.4 kg
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