Question

A semi circular sheet of metal of diameter 28 cm is bend to form a open conical cup. Find the capacity of the cup.

Answer 1

As we know that the circumference of a circle is given as-

Circumference $=\pi r$

Whereas, $r$ is the radius of circle

Diameter of circular sheet $=28cm$

$\therefore$ Radius of circular sheet $=\frac{28}{2}=14cm$

Therefore,

Circumference of circular sheet $=14\pi$

When a semi-circular sheet is bent to form an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.

Slant height of cup $\left(l\right)=$ Radius of circular sheet $=14cm$

Circumference of the base of cone $=$ circumference of circular sheet $=14\pi$

Let $r$ be the radius of the base of cone

$\therefore 2\pi r=14$

$\Rightarrow r=7cm$

Let $h$ be the height of cup.

Therefore,

$l^2=r^2+h^2$

${\left(14\right)}^2={\left(7\right)}^2+h^2$

$\Rightarrow h=\sqrt{196-49}=\sqrt{147}=7\sqrt{2}cm$

Now,

Capacity of cup $=$ Volume of cone

As we know that, volume of cone is given as-

$V=\frac{1}{3}\pi r^{2}h$

Therefore,

Capacity of cup $=\frac{1}{3}\times \frac{22}{7}\times {\left(7\right)}^2\times 7\sqrt{3}=622.4{cm}^3$

Thus the capacity of the cup is $622.4{cm}^3$.

Hence the correct answer is $622.4{cm}^3$.