Question

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

Answer 1

Answer:

semi circular sheet of metal of diameter 28 cm is bent into an open conical cup find the depth and capacity of cup ...

Answer 2

Answer:

$\bf\underline{\underline{\pink{Question:-}}}$

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

$\bf\underline{\underline{\blue{Solution:-}}}$

When the semicircular sheet is bent into an open conical cup, the radius of the sheet becomes the slant height of the conical cup.

∴ l = 14

Circumference of the base of the cone

= length of arc ABC

= (π×14)cm = $\frac{22}{7}×14$ = 44cm

let the radius of the cone be r cm. Then,

2πr = 44 = $\frac2×{22}{7}×r$ = 44 => r = 7cm.

let the height of the cone be h cm. Then,

h² = l²–r² = (14)²–(7)² = 196–49 = 147.

∴ h = √147 = √7×7×3 = 7√3 cm

= (7×1.732)cm = 12.12cm

Capacity capacity of the conical cup

= volume of the conical cup

= ⅓πr²h

= ½ ×$\frac{22}{7}×7×7×12.12$

= (154×4.04)cm³

=622.16cm³

$\bf\underline{\underline{\green{Extra\:Formulas:-}}}$

★For a right circular cone of radius r units, height h units and land height = l units, we have

☙Slant height of the cone (l) = √h²+r² units

☙Volume of the cone = ⅓πr²h cubic units

☙Area of curved surface = (πrl) sq units = (πr√h²+r²) sq units

☙Total surface area = (area of the curved surface)+(area of the base)

= (πrl+πr²) sq units = πr(l+r) sq units