Question

Hola Brainlians!!!

A hemispherical tank full of water is emptied by a pipe at the rate of 25/7 litres per second. How much time it will take to empty half the tank, if it is 3m in diameter?

Use π = 22/7 ​

Answer 1

★ Concept :-

Here the concept of Volume of Hemisphere has been used. We see that we are given the rate of flow of water and diameter of the tank. So firstly we can find the radius of the tank and then find the volume of the tank. Then after that we can find out the volume of water which is to be emptied. Finally after that we can find the time taken to empty that amount of water.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\;\times\;\pi r^{3}}}}$

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★ Solution :-

Given,

» Rate of flow of water = 25/7 literes per second

» Diameter of hemispherical tank = d = 3 m

» Radius of hemispherical tank = r = ½ × 3 = 1.5 m

~ For the Volume of the Hemispherical Tank ::

We know that,

$\;\sf{\rightarrow\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\;\times\;\pi r^{3}}$

By applying values, we get

$\;\sf{\rightarrow\;\;Volume\;of\;Tank\;=\;\dfrac{2}{3}\;\times\;\dfrac{22}{7}\;\times\:(1.5)^{3}}$

$\;\sf{\rightarrow\;\;Volume\;of\;Tank\;=\;\dfrac{2}{3}\;\times\;\dfrac{22}{7}\;\times\:3.375}$

$\;\bf{\rightarrow\;\;\blue{Volume\;of\;Tank\;=\;\dfrac{148.5}{21}\;\;m^{3}}}$

~ For the Volume of Water that is to be emptied ::

It's given that half of the tank is being emptied. This means half the volume of total water is being removed. We know that water takes the shape of the container. So the volume of tank will be equal to total volume of water.

So volume of water to be displaced is ::

$\;\bf{\rightarrow\;\;Volume\;of\;water\;to\;be\;emptied\;=\;\dfrac{1}{2}\:\times\:\dfrac{148.5}{21}\;m^{3}}$

$\;\bf{\rightarrow\;\;\orange{Volume\;of\;water\;to\;be\;emptied\;=\;\dfrac{148.5}{42}\;m^{3}}}$

Now we know that,

→ 1 m³ = 1000 Litres

→ 148.5 / 42 m³ = 148.5/42 × 1000 = 148500/42 L

$\;\bf{\rightarrow\;\;\red{Volume\;of\;water\;to\;be\;emptied\;=\;\dfrac{148500}{42}\;L}}$

~ For the Time taken to empty half of the tank ::

It's given that,

>> Time taken to empty 25/7 L of water = 1 second

(since rate of flow of water = 25/7 L per second)

From this we get,

>> Time taken to empty 1 L of water = 7/25 second

>> Time taken to empty 148500/42 L of water = 7/25 × 148500/42 seconds

$\;\bf{\mapsto\;\;Time\;Taken\;to\;empty\;half\;of\;the\;tank\;=\;\dfrac{1039500}{1050}\;seconds}$

$\;\bf{\mapsto\;\;\green{Time\;Taken\;to\;empty\;half\;of\;the\;tank\;=\;990\;seconds}}$

Now if we need this time in minutes, then,

→ 1 minute = 60 seconds

→ 990 seconds = 990/60 minutes = 16.5 minutes

This is the required answer.

$\;\underline{\boxed{\tt{Required\;\;Time\;=\;\bf{\purple{990\;seconds}}\;\tt{or}\;\bf{\purple{16.5\;minutes}}}}}$

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★ More to know :-

$\;\tt{\leadsto\;\;CSA\;of\; Hemisphere\;=\;2\pi r^{2}}$

$\;\tt{\leadsto\;\;TSA\;of\; Hemisphere\;=\;3\pi r^{2}}$

$\;\tt{\leadsto\;\;CSA\;of\;Sphere\;=\;4\pi r^{2}}$

$\;\tt{\leadsto\;\;TSA\;of\;Sphere\;=\;4\pi r^{2}}$

$\;\tt{\leadsto\;\;Volume\;of\;Sphere\;=\;\dfrac{4}{3}\pi r^{3}}$

Answer 2

$\huge\mathfrak\green{☟ \: \: \: answer \: \: \: \: ✎}$

Volume Of Hemispeherical Tank,

$V = \frac{2}{3} \pi {r}^{3}$

Here,

$Diameter, \: d = 3m$

$\sf \colorbox{pink} {\red★So Redius,}$

$r = 1.5m$

$V = \frac{2}{3} * \frac{22}{7} *(1.5 {)}^{3} = \frac{2}{3} * \frac{22}{7} * \frac{27}{28}$

$= \frac{90}{14}$

Let t in the time taken in seconds to empty half of the tank

As the tank is emptied at the rate of

$\frac{25}{7}$

liters per second then

$\frac{25}{7} * t = 99 \: * \frac{1000}{25}$(Multiplied by 1000 to convert it into liters)

$⇒t = \frac{99000}{25} = 990 \: seconds$

$= \frac{990}{60} = 16.5 \: minutes$

$\\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸} \\ and\\ \sf \colorbox{lightgreen} {\red❤ANSWER ᵇʸ ᶠˡⁱʳᵗʸ ᵇᵒʸ}$

btw good night ji and Allah Afiz