Question

An iron pillar consists of a cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1cm^3 of iron weighs 7.5 g.

Answer 1

Answer:

693 kg

Step-by-step explanation:

Given, diameter = 20 cm.

Then, radius = 10 cm.

Given, height of cylinder(h) = 2.8 m = 280 cm.

Height of cone(H) = 42 cm.

∴ Volume of pillar = (πr²h + 1/3πr²H)

= πr²(h + 1/3H)

= (22/7) * (10)² * (280 + 1/3 * 42)

= (2200/7) * (280 + 14)

= (2200/7) * 294

= 2200 * 42

= 92400 cm³.

∴ Weight of the pillar = (92400 * 7.5)/1000

                                   = 693000/1000

                                   = 693 kg.

Therefore, weight of the pillar = 693 kg.

Hope it helps!

Answer 2

HEY

HERE IS ANSWER



Question-


An iron pillar consists of a cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1cm^3 of iron weighs 7.5 g.


Now

Solution -


given

height of cylinder= 2.8 m

= 100*2.8

= 280 cm

diameter= 20 cm

hence radius= 20/2

= 10 cm

height of cone= 42 cm

and radius= 10 cm

hence

$volume \: of \: cylinder = \pi \: { r}^{2} h$

hence

$\frac{22}{7} \times 10 \times 10 \times 280$

$22 \times 10 \times 10 \times 40$
$\bf{88000cm^3}$
now

$volume \: of \: cone = \frac{1}{3} \times \pi \: r ^{2} h$


$\frac{1}{3} \times \frac{22}{7} \times 10 \times 10 \times 42$


$\frac{1}{3} \times 22 \times 10 \times 10 \times 6$

$22 \times 10 \times 10 \times 2$
$\bf{4400cm ^{3}}$

volume of the pillar= volume of cylinder+ volume of cone

volume of pillar=88000+4400cm^3


92400cm^3

now

weight of pillar=7.5*92400

=693000 g

hence

$\bf{weight of pillar = 693kg}$


HOPE IT HELPS

THANKS