Question

A solid metal cylinder with radius 6cm and height 14 cm is given. On one side of cylinder a hemispherical shape of solid and on the other side a conical shape of solid are removed. If base radius of hemisphere is equal to the radius of cylinder and base radius of cone is 3 cm and it's heights is 5 cm find the volume and total surface area of the remaining solid.

Answer 1

Hey mate...

here's the answer....

A solid metal cylinder with radius 6cm and height 14 cm is given. On one side of cylinder a hemispherical shape of solid and on the other side a conical shape of solid are removed. If base radius of hemisphere is equal to the radius of cylinder and base radius of cone is 3 cm and it's heights is 5 cm find the volume and total surface area of the remaining solid.

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Answer 2

The volume of Remaining part of Sol id is 1130cm³.

Given:-

Radius of solid metal cylinder = 6cm

Height of solid metal cylinder = 14cm

Base radius of hemisphere = 6cm

Base radius of cone = 3cm

Height of cone = 5cm

To Find:-

The volume of the remaining solid.

Solution:-

We can easily find out the volume and total surface area of the remaining solid by using these simple steps.

As

Radius of solid metal cylinder (r1) = 6cm

Height of solid metal cylinder (h1) = 14cm

Base radius of hemisphere (r2) = 6cm

Base radius of cone (r3) = 3cm

Height of cone (h2) = 5cm

Now,

Volume of remaining part,

Volume of cylinder = πr²h

$v1 = \pi {r1}^{2} h1$

$v1 = \frac{22}{7} \times {6}^{2} \times 14$

$v1 = 22 \times 6 \times 6 \times 2$

$v1 = 36 \times 44$

$v1 = 1584 {cm}^{3}$

Volume of Cone = 1/3πr²h

$v2 = \frac{1}{3} \pi {r2}^{2} h2$

$v2 = \frac{1}{3} \times \frac{22}{7} \times {3}^{2} \times 5$

$v2 = 3.14 \times 15$

$v2 = 47.12 {cm}^{3}$

Volume of Hemisphere = 2/3πr³

$v3 = \frac{2}{3} \times \frac{22}{7} \times {6}^{3}$

$v3 = 0.6 \times 3.14 \times 216$

$v3 = 406.94 {cm}^{3}$

Volume of Remaining part = v1 - (v2 + v3)

$v = v1 - (v2 + v3)$

$v = 1584 - (47.12 + 406.94)$

$v = 1584 - 454$

$v = 1130 {cm}^{3}$

Hence, the volume of Remaining part of Sol id is 1130cm³.

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