B) A hemispherical bowl has diameter 30cm. Find the capacity of the bo

1.	B) A hemispherical bowl has diameter 30cm. Find the capacity of the bowl.Some water is poured into the bowl as shown in the figure. If the radius of thesurface of the water is 9 em. Find the depth of water in the bowl.( Take II = 3.14.)
Answer» Given Diameter of a hemispherical bowl (D) = 30cm , $Radius \: of \:the \: bowl (R) = \frac{D}{2} \\= \frac{30}{2} \\= 15 \:cm$ $Capacity \: of \:the \:Bowl (V) \\= \frac{2}{3} \pi R^{3} \\= \frac{2}{3}\times 3.14 \times 15^{3}\\= 7065 \:cm^{3} \:--(1)$ /* According to the problem GIVEN / Some water is poured into the bowl as shown in the figure. If the radius of the* Some water is poured into the bowl as shown in the figure. If the radius of thesurface of the water is 9 cm. radius of the surface (r) = QD = 9 cm . OD = 15 cm ( Radii OB = OD ) In ∆OQD , <Q = 90° By Pythagoras Theorem : OD² = OQ² + QD² => 15² = OQ² + 9² => 15² - 9² = OQ² => 225 - 81 = OQ² => 144 = OQ² => OQ = 12 cm Now , Depth of the water (H) = OP - OQ = 15 cm - 12 cm = 3 cm THEREFORE., $\red {Capacity \: of \:the \:Bowl} \green { = 7065 \:cm^{3} }$ $\red { Depth\: of \:the \:water }\green {= 3\:cm }$ •••♪

B) A hemispherical bowl has diameter 30cm. Find the capacity of the bowl.Some water is poured into the bowl as shown in the figure. If the radius of thesurface of the water is 9 em. Find the depth of water in the bowl.( Take II = 3.14.)

Answer»

Given Diameter of a hemispherical bowl (D) = 30cm ,

$Radius \: of \:the \: bowl (R) = \frac{D}{2} \\= \frac{30}{2} \\= 15 \:cm$

$Capacity \: of \:the \:Bowl (V) \\= \frac{2}{3} \pi R^{3} \\= \frac{2}{3}\times 3.14 \times 15^{3}\\= 7065 \:cm^{3} \:--(1)$

/* According to the problem GIVEN */

Some water is poured into the bowl as shown in the figure. If the radius of the

Some water is poured into the bowl as shown in the figure. If the radius of thesurface of the water is 9 cm.

radius of the surface (r) = QD = 9 cm .

OD = 15 cm ( Radii OB = OD )

In ∆OQD , <Q = 90°

By Pythagoras Theorem :

OD² = OQ² + QD²

=> 15² = OQ² + 9²

=> 15² - 9² = OQ²

=> 225 - 81 = OQ²

=> 144 = OQ²

=> OQ = 12 cm

Now ,

Depth of the water (H) = OP - OQ

= 15 cm - 12 cm

= 3 cm

$\red {Capacity \: of \:the \:Bowl} \green { = 7065 \:cm^{3} }$

$\red { Depth\: of \:the \:water }\green {= 3\:cm }$

•••♪