A semi-circular sheet of metal of diameter 28 | vedclass

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Question

(A) The diameter of the semi-circular sheet is $28 \, cm$,so its radius $R = 14 \, cm$. When this sheet is bent to form an open conical cup,the radius

Vedclass · Accepted Answer

$622.16$

Vedclass · Answer

(A) The diameter of the semi-circular sheet is $28 \, cm$,so its radius $R = 14 \, cm$. When this sheet is bent to form an open conical cup,the radius of the sheet becomes the slant height $(l)$ of the cone.Therefore,$l = 14 \, cm$.The circumference of the semi-circular arc becomes the circumference of the base of the cone. The length of the semi-circular arc is $\pi R = \pi 	imes 14$.Let $r$ be the radius of the base of the cone. Then,$2 \pi r = 14 \pi$,which gives $r = 7 \, cm$.Now,we find the height $(h)$ of the cone using the relation $l^2 = r^2 + h^2$:$h = \sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} \approx 12.124 \, cm$.The capacity (volume) of the conical cup is $V = \frac{1}{3} \pi r^2 h$.$V = \frac{1}{3} 	imes \frac{22}{7} 	imes 7 	imes 7 	imes 12.124 \approx 622.38 \, cm^3$. Given the options,the closest value is $622.16 \, cm^3$.

$A$ semi-circular sheet of metal of diameter $28 \, cm$ is bent to form an open conical cup. Find the capacity of the cup. (in $, cm^3$)

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