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Home » A hallow metallic cylindrical tube has an internal radius of 3.5 cm and height 21 cm. The thickness of the metal tube is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place
A hallow metallic cylindrical tube has an internal radius of 3.5 cm and height 21 cm. The thickness of the metal tube is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place
Class 10 | Frank | ICSE | Maths | Mensuration II
A hallow metallic cylindrical tube has an internal radius of 3.5 cm and height 21 cm. The thickness of the metal tube is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place

Solution: –

According to the given question,

The internal radius of hallow metallic cylindrical tube is 3.5 cm.

The height of hallow metallic cylindrical tube is 21 cm.

And the thickness of the metal is 0.5 cm.

Then, the outer radius of the tube is \left( 3.5+0.5 \right)cm=4cm.

We know that, the volume of hallow metallic cylindrical tube is \pi h\left( {{R}^{2}}-{{r}^{2}} \right)

=\left( \frac{22}{7} \right)\times 21\times \left( {{4}^{2}}-{{3.5}^{2}} \right)

=\left( \frac{22}{7} \right)\times 21\times \left( 16-12.25 \right)

=\left( \frac{22}{7} \right)\times 21\times 3.75

=247.5c{{m}^{3}}

Therefore, the volume of metal is used 247.5c{{m}^{3}}

Tube is melted and cast into a right circular cone of height is 7 cm.

Now, let us consider {{r}_{1}} be the radius of right circular cone,

So, the volume is \frac{1}{3}\pi r_{1}^{2}h=247.5

\frac{1}{3}\times \frac{22}{7}r_{1}^{2}\times 7=247.5

Now, by cross multiplication we get,

r_{1}^{2}=\frac{\left( 247.5\times 3\times 7 \right)}{\left( 22\times 7 \right)}

r_{1}^{2}=33.75

{{r}_{1}}=\sqrt{33.75}

{{r}_{1}}=5.8cm. Therefore, the radius of the cone is 5.8 cm.

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