Question

12. A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm. and height 3 cm. if a full bowl of liquid is filled in the bottles,find how many bottles are required​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm. and height 3 cm. A full bowl of liquid is filled in the bottles.

Now,

Diameter of hemispherical bowl, d = 9 cm

Radius of hemispherical bowl, r = d/2 = 9/2 cm

We know,

Volume of hemisphere of radius r is given by

$\boxed{\sf{ \: \: Volume_{(hemisphere)} \: = \: \frac{2}{3} \: \pi \: {r}^{3} \: \: }}$

So, on substituting the value of r, we get

$\rm \: Volume_{(liquid \: in \: hemispherical \: bowl)}$

$\rm \: = \: \frac{2}{3} \times \pi \times \dfrac{9}{2} \times \dfrac{9}{2} \times \dfrac{9}{2}$

$\rm \: = \: \frac{243}{3} \pi \: {cm}^{3}$

$\rm\implies \:\boxed{\sf{ \:Volume_{(liquid \: in \: hemispherical \: bowl)} = \frac{243\pi}{4} \: {cm}^{3} \: \: }}$

Now,

Diameter of cylindrical bottle, D = 3 cm

So, Radius of cylindrical bottle, R = 3/2 cm

Height of cylindrical bottle, h = 3 cm

We know,

Volume of cylinder of radius R and height h is given by

$\boxed{\sf{ \:Volume_{(cylinder)} \: = \: \pi \: {R}^{2} \: h \: }}$

So,

$\rm \: Volume_{(liquid \: in \: cylindrical \: bottle)}$

$\rm \: = \: \pi \times \frac{3}{2} \times \frac{3}{2} \times 3$

$\rm \: = \: \frac{27\pi}{4} \: {cm}^{3}$

$\rm\implies \:\boxed{\sf{ \:Volume_{(liquid \: in \: cylindrical \: bottle)} = \frac{27\pi}{4} \: \: }}$

Now, Number of bottles required to fill whole liquid of hemispherical bowl are evaluated as

$\rm \: Number_{(cylindrical\:bottles)}$

$\rm \: = \: \frac{Volume_{(liquid \: in \: hemispherical \: bowl)}}{Volume_{(liquid \: in \: cylindrical \: bottle)}}$

$\rm \: = \: \frac{243\pi}{4} \div \frac{27\pi}{4} \:$

$\rm \: = \: 9 \:$

Hence,

$\rm\implies \:\boxed{\sf{ \:Number_{(cylindrical\:bottles)} \: = \: 9 \: \: }}$

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Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Step-by-step explanation:

Given: Diameter =9 cm , hence radius

$\tt(r)=\dfrac{9}{2} cm$

Volume of of a hemispherical bowl

$\tt = \dfrac{1}{2}\left[\dfrac{4}{3} \pi r^{3}\right]$

$\text{Here \( \tt \: r=\dfrac{9}{2} \)}$

$\text{\( \Rightarrow \) Volume of hemispherical bowl \( = \) \( \dfrac{1}{2}\left[\dfrac{4}{3} \pi\left(\dfrac{9}{2}\right)^{3}\right] \)}$

$\text{ \( \Rightarrow \) Volume of the bowl \( \tt=\dfrac{243}{4} \pi \quad c \) \( \tt m^{3} \)}$

$\text{Now to calculate volume of each cylindrical bottle we are given:}$

Diameter: 3 cm ,

$\text{hence \( \tt r=\dfrac{3}{2} \)}$

and height (h)=3 cm

$\text{Volume of the cylinder \( \tt =\pi r^{2} h \)}$

$\text{\( \Rightarrow \) volume of cylinder \( =\pi \left(\dfrac{3}{2}\right)^{2} \times 3 \)}$

$\text{\( \Rightarrow \) Volume of cylinder \( \tt=\dfrac{27}{4} \pi \: \: cm ^{3} \)}$

$\text{ Now a number of bottles \( = \) Volume of hemispherical bowl \( \div \) volume of the cylinder}$

$\text{ \( \Rightarrow \) Number of bottles} = \dfrac{\dfrac{243} {\cancel{4}} \cancel \pi} { \dfrac{27} {\cancel{4}}\cancel\pi } = \dfrac{243}{27}$

$\text{ \( \Rightarrow \) Number of bottles \( =9 \)}$