Question

A semi-circular sheet of metal of diameter 56 cm is bent to form an open conical cup. The approximate capacity of the cup is

• A. $\frac{2744\sqrt{3}\pi }{3}c{m}^3$
  
• B. $1523\pi c{m}^3$
• C. $\frac{1372\sqrt{3}\pi }{3}c{m}^3$
  
• D. $\frac{1946\sqrt{3}\pi }{3}c{m}^3$
  

Answer 1

The correct option is A $\frac{2744\sqrt{3}\pi }{3}c{m}^3$


Given that the diameter of the semi-circular sheet is 56 cm.
Now, the sheet is bent to form an open conical cup.
Therefore, radius of the sheet becomes the slant height of the conical cup and circumference of the sheet becomes the circumference of the base of cone.
Let $r,h$ and $l$ be the radius, height and slant height of the cone, respectively.

$\therefore l=\frac{56}{2}=28cm$ ....(i)
Also, circumference of base of cone = Circumference of semi-circular sheet
$\Rightarrow 2\pi r=\pi \times 28$
$\Rightarrow r=\frac{28}{2}=14cm$ ....(ii)
Now, $l^2=r^2+h^2$
$\Rightarrow h^2=l^2–r^2=(28{)}^2–(14{)}^2$ [From (i) and (ii)]
$\Rightarrow h^2=(14\times 2{)}^2–(14{)}^2$
$\Rightarrow h^2=(14{)}^2\times (4–1)$
$\Rightarrow h=14\sqrt{3}cm$ …..(iii)
$\therefore \text{Volume of the conical cup}=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}\pi \times (14{)}^2\times 14\sqrt{3}$
$=\frac{2744\sqrt{3}\pi }{3}c{m}^3$
Hence, the correct answer is option (a).