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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The weight of the pillar if one cu. Cm of iron weighs 7.8 grams. is 395366.322 g
Step-by-step explanation:
Height of cylinder = 240 cm
Volume of cylinder =
Radius of cylinder = 8 cm
\pi r^2 h
\frac{22}{7} \times (8)^2 (240)
Volume of cylinder =
Volume of cylinder =
Height of cone =36 cm
Radius of cone = 8 cm
Volume of cone =
\frac{1}{3} \pi r^2 h
Volume of cone =
2413.7142 cm^3
Volume of cone =
Total volume of pillar = Volume of cone + Volume of cylinder
Total volume of pillar = 2413.7142 + 48274.2857 = 50687.99
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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
Volume of cylinder
= 3.14 × 8 × 8 × 240
= 48320.4 cm³
Volume of cone
= ⅓ × 3.14 × 8 × 8 × 36
= 1 × 3.14 × 8 × 8 × 12
= 3.14 × 64 ×12
= 2411.52 cm³
Now,
Weight of pillar = Volume of cylinder + volume of cone
W = 48320.4 + 2411.52
W = 50730
1kg = 1000gm
7.8/1000 × 50730
0.0078 × 50730
395.4 kg
Weight of pillar is 395 kg
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