Question

5. An iron pillar has some parts in the form of right
circular cylinder and the remaining part in the form of
right circular cone. The radius of the base of both
solid is 8 cm. The cylindrical part of the cylinder 240
cm high and the conical part is 36 cm high. Find the
weight of the pillar if the density of iron is 7.8 g/cm​

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

Answer:

$395 \: kg$

Step-by-step 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.