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

Answer:

Let R cm and r cm denote the radii of the base of the cylinder and cone respectively. Then,

R = r = 8 cm

Let H and h cm be the height of the cylinder and the cone respectively. Then

H = 240 cm and h = 36 cm

Now, volume of the cylinder = πR²H cm³

= (π × 8 × 8 × 240)cm3

= (π × 64 × 240) cm3

$\sf Volume \: of \: the \: cone = \frac{1}{3} πr {}^{2} h \: cm {}^{3} \\ = \sf \:( \frac{1}{3} π × 8 × 8 × 36) cm {}^{3} \\ </p><p></p><p>= \sf ( \frac{1}{3} π × 64 × 36) cm {}^{3} </p><p>$

∴ Total volume of iron = Volume of the cylinder + Volume of the cone

$= \sf (π × 64 × 240 + \frac{1}{3} π × 64 × 36) cm {}^{3} \\ </p><p></p><p> \sf= π × 64 × (240 + 12) cm {}^{3} \\ </p><p></p><p>= \sf \: 22/7 × 64 × 252 cm {}^{3} \\ </p><p></p><p> \sf= 22 × 64 × 36 cm {}^{3} \\ </p><p>$

Hence, total weight of the pillar = Volume Weight per cm3

= (22 × 64 × 36) × 7.8 gms

= 395366.4 gms

= 395.3664 kg.