Question

from solid wooden cylinder of height 28 cm and diameter 6 CM two conical cavity is hollowed out the diameters of cones are also of 6 cm and height 10.5 cm find the volume of the remaining solid



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Answer 1

Answer:

Step-by-step explanation:

$\huge{\boxed{\sf{Required \ Answer}}}$

Given that:-

A solid wooden cylinder.

Radius  3 cm [6/2] height 28 cm.

2 cones.

Radius 3 cm [6/2] and height 10.5 cm.

Diameter = 2 × Radius, so in square brackets, values added.

We need to find the remaining volume of the solid, if 2 cones are cut off from the wooden cylinder.

Let's Do!

Volume of Cylinder = $\sf{ \pi \times r^2 \times h}$

Here, radius as 3 and height as 28:-

$\sf{ \dfrac{22}{7}\times (3)^2 \times 28}$

= 792 cm²

Volume of Cone = $\sf{ \dfrac{1}{3} \times \pi \times r^2 \times h}$

Volume of 1 cone =  $\sf{ \dfrac{1}{3} \times \dfrac{22}{7} \times (3)^2 \times 10.5}$ × 2

So, 2 cones volume is 99 × 2 = 198 cm²

Now, subtracting the values:-

$\boxed{\tt{Volume \ of \ Cylinder - Volume \ of \ 2 \ cones}}$

= 792 - 198

$\boxed{\rm{= 594 \ cm^2 \ is \ the \ required \ answer.}}$

Answer 2

Answer :-

$\: \\ \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Volume of Cones and Volume of Cylinder has been used. We know that Volume is the space occupied by matter. So, it can neither be created nor be destroyed. Hence, if we find the volume of Cylinder and subtract the volume of two cones, we can get our required answer.

Let's do it !!

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★ Formula Used :-

$\: \\ \: \large{\boxed{\boxed{\sf{Volume \; of \; Cylinder \; = \; \bf{\pi r^{2} h}}}}}$

$\: \\ \: \large{\boxed{\boxed{\sf{Volume \; of \; Cone \; = \; \bf{\dfrac{1}{3} \: \times \: \pi r^{2} h}}}}}$

$\: \\ \: \large{\boxed{\boxed{\sf{Radius_{(cone \: or \: cylinder)} \; = \; \bf{\dfrac{Diameter}{2}}}}}}$

$\: \: \\ \large{\boxed{\boxed{\sf{Volume \; of \; the \; remaining \; solid \: = \: \bf{Area \: of \: Cylinder \: - \: 2 \: \times \: Area \: of \: Cone}}}}}$

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★ Question :-

From solid wooden cylinder of height 28 cm and diameter 6 CM two conical cavity is hollowed out the diameters of cones are also of 6 cm and height 10.5 cm. Find the volume of the remaining solid

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★ Solution :-

Given,

» Diameter of the Cylinder = 6 cm

» Height of the Cylinder = 28 cm

» Diameter of the Cylinder = 6 cm

» Height of the Cone = 10.5 cm

Here clearly, the Diameter of the Cone and Cylinder are equal. So their radius also will be equal.

~ For the Radius of Cone and Cylinder :-

$\: \\ \: \large{\sf{:\rightarrow \: \: \: Radius_{(cone \: and \: cylinder)} \; = \; \bf{\dfrac{Diameter}{2}}}}$

$\: \\ \: \large{\sf{:\rightarrow \: \: \: Radius_{(cone \: and \: cylinder)} \; = \; \bf{\dfrac{\cancel{6 \: cm}}{\cancel{2}} \: \: = \: \: \underline{\underline{3 \: cm}}}}}$

$\: \: \\ \large{\boxed{\boxed{\tt{Radius_{(cone \:\: and \: \: cylinder)} \: \: = \: \bf{3 \: cm}}}}}$

~ For the Volume of Cylinder :-

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \; of \; Cylinder \; = \; \bf{\pi r^{2} h}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \; of \; Cylinder \; = \; \bf{\dfrac{22}{\cancel{7}} \: \times \: (3)^{2} \: \times \: \cancel{28}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \; of \; Cylinder \; = \; \bf{22 \: \times \: 9 \: \times \: 4 \: \: = \: \: \underline{\underline{792 \: cm^{3}}}}}}$

$\: \\ \: \large{\boxed{\boxed{\tt{Volume \:\; of \:\; Cylinder \: \: = \: \bf{792 \: cm^{3}}}}}}$

~ For the Volume of Cone :-

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \; of \; Cone \; = \; \bf{\dfrac{1}{3} \: \times \: \pi r^{2} h}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \; of \; Cone \; = \; \bf{\dfrac{1}{3} \: \times \: \dfrac{22}{\cancel{7}} \: \times \: (3)^{2} \: \times \: \cancel{10.5}}}}$

$\: \\ \qquad \: \large{\sf{\longrightarrow \: \: \: Volume \; of \; Cone \; = \; \bf{\dfrac{1}{\cancel{3}} \: \times \: 22\: \times \: \cancel{9} \: \times \: 1.5 \: \: = \: \: \underline{\underline{99 \: cm^{3}}}}}}$

$\\ \: \: \large{\boxed{\boxed{\tt{Volume \:\; of \:\; Cone \: \: = \: \bf{99 \: cm^{3}}}}}}$

~ For the Volume of Remaining Solid :-

$\: \\ \qquad \large{\sf{\longrightarrow \: \: \: Volume \; of \; the \; remaining \; solid \: = \: \bf{Area \: of \: Cylinder \: - \: 2 \: \times \: Area \: of \: cone}}}$

$\: \\ \qquad \large{\sf{\longrightarrow \: \: \: Volume \; of \; the \; remaining \; solid \: = \: \bf{792 \: cm^{3} \: - \: 2 \: \times \: 99 \: cm^{3}}}}$

$\: \\ \qquad \large{\sf{\longrightarrow \: \: \: Volume \; of \; the \; remaining \; solid \: = \: \bf{792 \: cm^{3} \: - \: 198 \: cm^{3} \: \: = \: \: \underline{\underline{594 \: cm^{3}}}}}}$

$\: \\ \: \large{\boxed{\boxed{\tt{Volume \:\; of \:\; Remaining \:\; Solid \; \: = \: \bf{594 \: cm^{3}}}}}}$

$\: \: \\ \large{\underline{\underline{\rm{\leadsto \: \: \: Thus, \: the \: area \: of \: remaining \: solid \: is \: \: \boxed{\bf{594 \: cm^{3}}}}}}}$

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$\: \: \qquad \: \large{\underbrace{\underbrace{\rm{\mapsto \: \: \: Let's \: understand \: more \: :-}}}}$

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cube = (Side)³

• Volume of Hemisphere = ⅔ × πr²h

• TSA of Cube = 6 × (Side)²

• TSA of Cylinder = 2πrh + 2πr²

• TSA of Cone = πrl + πr²

• CSA of Cone = 4 × (Side)²

• CSA of Cylinder = 2πrh

• CSA of Cone = πrl