Question

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.​

Answer 1

Answer:

SOLUTION-

Volume of the pillar = Volume of the

cylindrical part + Volume of conical part

"h" = πr²h

Volume of a cone = 1/3πr²h where r is the radius of the base of the cone and h is the height.

Hence, Volume of the pillar = (22/7 × 8 × 8 × 24) + 1/3 × 22/7 × 8² × 36 = 50688cm³

If one cu cm wieighs 7.8 grams, then 50688cm³ weighs 50688 ×7.8 = 395366.4 grams or 395.37kg.

Therefore, the weight of the pillar if one cu. cm of iron weighs 7.8 grams is 395.37 kg.

Answer 2

Answer:

$395 \: kg$

Explanation:

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,

W = 48320.4 + 2411.52

W = 50730

Now,

1kg = 1000gm

7.8/1000 × 50730

0.0078 × 50730

395.4 kg

Weight of pillar is 395 kg.