Question

4. A metallic sphere of radius 4.2 CM is melted and recast into the shape of a cylinder of radius 6 CM find the height of the cylinder.​

Answer 1

» Question :

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm then, Find the height of the cylinder.

» To Find :

The height of the cylinder.

» Given :

• Radius of the Sphere = 4.2 cm

• Radius of the cylinder = 6 cm

» We Know :

Volume of a Sphere :

$\sf{\underline{\boxed{V_{s} = \dfrac{4}{3}\pi r^{3}}}}$

Where,

• V = Volume of the Sphere.
• r = Radius of the Sphere.

Volume of the cylinder :

$\sf{\underline{\boxed{V_{c} = \pi r^{2}h}}}$

Where,

• V = Volume of the Sphere.
• r = Radius of the cylinder.

» Concept :

According to the question , as the cylinder was made by recasting the sphere ,the volume of the sphere will be Equal to the volume of the cylinder. i.e,

$\sf{V_{s} = V_{c}}$

Now by this information we can find the height of the cylinder.

» Solution :

• Radius of the Sphere = 4.2 cm

• Radius of the cylinder = 6 cm

Formulae :

For Cylinder :

$\sf{\underline{\boxed{V_{c} = \pi r^{2}h}}}$

For Sphere :

$\sf{\underline{\boxed{V_{s} = \dfrac{4}{3}\pi r^{3}}}}$

Putting the two formulas , as :

$\sf{V_{s} = V_{c}}$

We Get :

$\sf{\Rightarrow \dfrac{4}{3}\pi r^{3} = \pi r^{2}h}$

$\sf{\Rightarrow \dfrac{4}{3}\cancel{\pi} r^{3} = \cancel{\pi}r^{2}h}$

$\sf{\Rightarrow \dfrac{4}{3}r^{3} = r^{2}h}$

Putting the value in the formula ,we get :

$\sf{\Rightarrow \dfrac{4}{3} \times 4.2^{3} = 6^{2}h}$

$\sf{\Rightarrow \dfrac{4}{3} \times 74.088 = 36h}$

$\sf{\Rightarrow \dfrac{4}{\cancel{3}} \times \cancel{74.088} = 36h}$

$\sf{\Rightarrow 4 \times 24.7(approx.) = 36h}$

$\sf{\Rightarrow 98.8 = 36h}$

$\sf{\Rightarrow \cancel{\dfrac{98.8}{36}} = h}$

$\sf{\Rightarrow 2.7(approx.) = h}$

Hence , the height of the cylinder is 2.7 cm.

Additional information :

• Surface area of a Cylinder = 2πr(h + r)

• Curved surface area of Cylinder = 2πrh

• Surface area of a Cuboid = 2(lb + lh + bh)

• Curved surface area of a Cuboid = 2(l + b)h

Answer 2

☯️ Given ☯️

• A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm.

☯️ To Find ☯️

• Height Of Cylinder.

☯️ Solution ☯️

We can say that,

Volume Of Sphere = Volume Of Cylinder.

Because, Sphere is melted into cylinder.

Now,

Let the height be H.

Volume Of Sphere = 4/3πr³

Volume Of Cylinder = πr²h.

We can put them equal.

4/3πr³ = πr²h

Put the values and solve the Equation.

⇛ 4/3 × π × (4.2)³ = π × 6² × H.

Cancel Out π.

⇛ 4/3 × 4.2 × 4.2 × 4.2 = 6 × 6 × H.

⇛ 98.784 = 36H.

⇛ 36H = 98.784.

⇛ H = 98.784/36

⇛ H = 2.744m.

⇛ H = 2.7m (approx.)

ADDITIONAL FORMULAE :-

• CSA Of Cylinder = 2πrh

• Volume Of Cylinder = πr²h

• TSA Of Cylinder = 2πr(l+r).

• CSA Of Sphere = 2πr²

• TSA Of Sphere = 3πr²