Question

1. A solid piece of iron in the form of a cuboid of dimensions 49cm x 33cm x 24cm, is
molded to form a solid sphere. The radius of the sphere is:
(A)21cm
(B) 23cm
(C) 25cm
(D) 19cm prove​

Answer 1

Answer :-

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

Here the concept of equality in Volume of Cuboid and Volume of Sphere has been used. Its given that the solid Cuboidal piece is transformed into a Solid Sphere. So the volume of of Cuboid will be Volume of Sphere. Using this concept, let's do it !!

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★ Formula Used :-

$\: \large{\boxed{\boxed{\sf{Volume \: of \: Cuboid \: = \: \bf{Length(L) \: \times \: Breadth(B) \: \times \: Height(H)}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Volume \: of \: Sphere \: = \: \bf{\dfrac{4}{3} \: \times \: \pi r^{3}}}}}}$

$\large{\boxed{\boxed{\sf{Volume \: of \: Sphere \:, \bf{(\dfrac{4}{3} \: \times \: \pi r^{3})} \: = \: Volume \: of \: Cuboid, \: \bf{(L \: \times \: B \: \times \: H)}}}}}$

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★ Question :-

A solid piece of iron in the form of a cuboid of dimensions 49cm x 33cm x 24cm, is moulded to form a solid sphere. The radius of the sphere is  ?

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★ Solution :-

Given,

» Dimensions of Cuboid = 49 cm × 33 cm × 24 cm

» Solid iron cuboidal piece is moulded in the form of = Solid Sphere

• Let the radius of the sphere be 'r' cm.

Then, according to the question :-

➣ Volume of Sphere  =  Volume of Cuboid

$\: \qquad \qquad \large{\bf{\longrightarrow \: \: \: {\dfrac{4}{3} \: \times \: \pi r^{3} \: = \: Length(L) \: \times \: Breadth(B) \: \times \: Height(H)}}}$

$\: \\ \qquad \qquad \large{\bf{\longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: r^{3} \: = \: 49 \: cm \; \times \; 33 \: cm \; \times \; 24 \: cm}}$

$\: \qquad \qquad \large{\bf{\Longrightarrow \; \; \; r^{3} \: = \: \tt{\dfrac{49 \: cm \; \times \; 33 \: cm \; \times \; 24 \: cm}{4 \; \times \; 22}}}}$

Reducing the like terms from LHS and RHS, we get,

➣ r³ = 9261

$\: \\ \large{\bf{\longmapsto \: \: \: r \: = \: \sqrt[3]{9261} \: = \: \underline{21}}}$

$\: \\ \qquad \large{\boxed{\boxed{\rm{r \: = \: 21 \: cm \; \; \bf{(Option \: A.)}}}}}$

$\: \\ \large{\boxed{\sf{\leadsto \: \: Hence, \: the \: radius \: of \: the \: sphere \: is \: \boxed{\underline{\bf{21 \; cm}}}}}}$

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$\: \large{\underline{\underline{\mapsto \: \: Confused? , \; Don't \: worry \: let's \: verify \: it \: :-}}}$

For verification we need to simply apply the value we got into the equation we formed, then,

$\: \\ \large{\bf{\longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \pi r^{3} \: = \: Length(L) \: \times \: Breadth(B) \: \times \: Height(H)}}$

$\: \\ \large{\bf{\longrightarrow \: \: \: \dfrac{4}{3} \: \times \: \dfrac{22}{7} \: \times \: 21^{3} \: = \: 49 \: cm \; \times \; 33 \: cm \; \times \; 24 \: cm}}$

Reducing the factorial terms, we get,

➣ 4 × 22 × 1 × 21 × 21 = 49 × 24 × 33

➣ 38808 cm³ = 38808 cm³

Clearly, LHS = RHS. So our answer is correct.

Hence, Verified.

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$\: \quad \quad \large{\underbrace{\tt{\mapsto \: \: \: Let's \; understand \; more \; formulas \; :-}}}$

• Volume of Cube = (Side)³

• Volume of Cylinder = πr²h

• Volume of Cone = ⅓ × πr²h

• Volume of Hemisphere = ⅔ × πr³

• TSA of Cube = 6 × (Side)²

• LSA of Cube = 4 × (Side)²

• LSA of Cuboid = 2 × (L + B) × h

• CSA of Cone = πrl

where r is radius and l is the slant height.

• CSA of Cylinder = 2πrh

Answer 2

Given :

• Dimensions of cuboid = 49 cm * 33 cm * 24 cm
• It is molded into a solid sphere.

To find :

• Radius of sphere

Solution :

Since the cuboid iron piece is molded to form a solid sphere, so their volume will be same.

∴ Volume of cuboid = Volume of sphere

⇒  l * b * h = 4/3 πr³

⇒  49 * 33 * 24 = 4/3 * 22/7 * r³

⇒  38808 = 88/21 * r³

⇒  38808 = 4.19 * r³

⇒  r³ = 38808/4.9

⇒  r³ = 9261

⇒  r = ∛9261

⇒  r = 21 cm

Therefore,

Radius of sphere = 21 cm   (Option : A)