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# Question 2

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

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According to the question,

Diameter of semi circular sheet = 28cm

Radius of semi circular sheet (r) = 28/2

= 14cm

Semi Circular sheet is bent to form an open conical cup

Thus, slant height of a conical cup (l)=radius of semicircular sheet(r) = 14cm

We know that,

Circumference of base of a cone = 2πR,

(where, R = radius of a cone and circumference of semi circle = πr)

Thus,

Circumference of base of a cone = Circumference of a Semi circle

⇒ 2πR= πr

\Rightarrow \mathrm{R}=\frac{\pi \mathrm{r}}{2 \pi}=\frac{\mathrm{r}}{2}=\frac{14}{2}=7 \mathrm{cm}

To find height,

R2 + h2 = l2

Where,

h=height of a cone

l=slant height of a cone

⇒ (7)2 + h2 = (14)2

⇒ 49+ h2 = 196

⇒ h2 = 196 – 49 = 147

⇒ h = √147 = 7√3 cm

Thus, volume of a cone = 1/3 πR2h

\begin{array}{l} \Rightarrow \text { Volume of a cone }=\frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 7 \sqrt{3}=\frac{1078 \sqrt{3}}{3} \mathrm{cm}^{3}=\frac{1078}{\sqrt{3}} \mathrm{cm}^{3} \\ \Rightarrow \text { Volume of a cone }=\frac{1078}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \mathrm{cm}^{3}=\frac{1078 \sqrt{3}}{3} \mathrm{cm}^{3} \\ \Rightarrow \text { Volume of a cone }=\frac{1078 \sqrt{3}}{3} \mathrm{cm}^{3}=622.38 \mathrm{cm}^{3} \end{array}

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