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:

Volume of the pillar = Volume of the cylindrical part + Volume of conical part

Volume of a Cylinder of Radius "R" and height "h" =πR

2

h

Volume of a cone =

3

1

πr

2

h where r is the radius of the base of the cone and h is the height.

Hence, Volume of the pillar =(

7

22

×8×8×240)+(

3

1

×

7

22

×8

2

×36)=50688cm

3

If one cu cm wieighs 7.8 grams, then 50688cm

3

weighs 50688×7.8=395366.4 grams or 395.37kg

Answer 2

${\sf{\purple{\underline{\overline{Given:-}}}}}$

• Radius = 8 cm
• Height of cylinder = 240 cm
• the conical part in 36 cm high.

${\sf{\pink{\underline{\overline{To \: find:-}}}}}$

Weight of the pillar if one cu. cm of iron weighs 7.8 grams.

${\sf{\purple{\underline{\overline{Solution :-}}}}}$

$\sf \fbox \red {We know that}$

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
• Weight of pillar is 395 kg.

${\sf{\purple{\underline{\overline{❀Tʜᴀɴᴋs!!}}}}}$