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 the pillar = Volume of the cylindrical part + Volume of conical partVolume of a Cylinder of Radius "R" and height "h" =

2

2 h

2 hVolume of a cone =

2 hVolume of a cone = 3

2 hVolume of a cone = 31

2 hVolume of a cone = 31

2 hVolume of a cone = 31 πr

2 hVolume of a cone = 31 πr 2

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =(

2 hVolume of a cone = 31 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

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+(

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 3

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 ×

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 7

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722 ×8

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722 ×8 2

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722 ×8 2 ×36)=50688cm

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722 ×8 2 ×36)=50688cm 3

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722 ×8 2 ×36)=50688cm 3

2 hVolume of a cone = 31 πr 2 h where r is the radius of the base of the cone and h is the height.Hence, Volume of the pillar =( 722 ×8×8×240)+( 31 × 722 ×8 2 ×36)=50688cm 3 If one cu cm w iei ghs 7.8 grams, then 50688cm

ghs 7.8 grams, then 50688cm 3

ghs 7.8 grams, then 50688cm 3 weighs 50688×7.8=395366.4 grams or 395.37kg

Answer 2

Answer:

${\textsf{\textbf{\pink{\underline{Given:- }}}}}$

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

${\textsf{\textbf{\purple{\underline{To find:- }}}}}$

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

${\textsf{\textbf{\orange{\underline{ Solution :- }}}}}$

${\textsf{\textbf{\blue{\underline{ We know that }}}}}$

• 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

${\textsf{\textbf{\pink{\underline{Weight of pillar is 395 kg.}}}}}$