# A semi-circular sheet of metal of diameter 28cm is bent to form an open conical cup. Find the capacity of the cup.

Answer : $622.38\; cm^{3}$

It is given that a semi-circular sheet is bent to form an open conical cup.

Now, 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.

Diameter of semi-circular sheet $=28\; cm$

So, radius of semi-circular sheet $\left ( R \right )=14\; cm$

Slant height of the conical cup $\left ( l \right )=14\; cm$

Now,

Let the radius of base of the cup = r cm.

Circumference of semi-circular sheet = Circumference of base of conical cup

$\pi R=2\; \pi\; r$

$\Rightarrow \frac{22}{7}\times 14=2.\frac{22}{7}.r$

$\Rightarrow r=7\; cm$

Let the height of cup be h cm.

Then we know that for a right circular cone,

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

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

$\Rightarrow 196-49=h^{2}$

$\Rightarrow h=7\sqrt{3}\; cm$

$\therefore$ Capacity of the cup = Volume of a cone $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}.\frac{22}{7}.\left ( 7 \right )^{2}.7\sqrt{3}$

$=\frac{1078}{3}\sqrt{3}$

$=\frac{1078}{3}\times 1.732$

$=622.38\; cm^{3}$

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