# A hemispherical tank full of water is emptied by a pipe at the rate of $\dfrac{{25}}{7}$ litres per second. How much time will it take to empty the tank, if the tank is ${\text{3}}$ meters in diameter?

Last updated date: 23rd Mar 2023

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Answer

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Hint: First, let’s calculate the volume of the tank. From that volume let’s take half of it to calculate the time taken to empty that half of the volume of water.

Complete step-by-step answer:

Given,

The diameter of the hemispherical tank is $D = 3m$

Hence the radius will be $R = \dfrac{3}{2}m$

The total volume of the water in the hemispherical tank

$

V{\text{ }} = {\text{ }}\dfrac{2}{3}\pi {R^3} \\

V = \dfrac{2}{3}{\text{x}}\dfrac{{22}}{7}{\text{x(1}}{\text{.5}}{{\text{)}}^3}{m^3} \\

{\text{ then }}V = \dfrac{{148500}}{{21}}litres{\text{ (as }}1{m^3} = 1000{\text{ }}litres{\text{)}} \\

$

This above volume is the volume of the hemispherical tank.

Volume of the half-emptied tank ${\text{ = }}\dfrac{V}{2}{\text{ = }}\dfrac{{74250}}{{21}}litres$

Given that time required to empty $\dfrac{{25}}{7}$ litres of water is $1$ second.

Hence the time required to empty $1$ litres of water is $\dfrac{7}{{25}}$ seconds.

Now, Time taken to half empty the tank will be equal to

$

\dfrac{{74250}}{{21}}litres{\text{ of water = }}\dfrac{7}{{25}}{\text{x}}\dfrac{{74.25}}{{21}}secs{\text{ }} \\

{\text{ }}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\text{ = 990 }}secs. \\

{\text{ = 16}}{\text{.5 }}\min{\text{ ( since 1 minute = 60 seconds)}} \\

$

Hence time taken to half empty the tank is ${\text{16}}{\text{.5}}$ minutes.

Note: In order to solve these surface areas and volume problems, you should have command over formulas related to all types of 3D shapes.

Complete step-by-step answer:

Given,

The diameter of the hemispherical tank is $D = 3m$

Hence the radius will be $R = \dfrac{3}{2}m$

The total volume of the water in the hemispherical tank

$

V{\text{ }} = {\text{ }}\dfrac{2}{3}\pi {R^3} \\

V = \dfrac{2}{3}{\text{x}}\dfrac{{22}}{7}{\text{x(1}}{\text{.5}}{{\text{)}}^3}{m^3} \\

{\text{ then }}V = \dfrac{{148500}}{{21}}litres{\text{ (as }}1{m^3} = 1000{\text{ }}litres{\text{)}} \\

$

This above volume is the volume of the hemispherical tank.

Volume of the half-emptied tank ${\text{ = }}\dfrac{V}{2}{\text{ = }}\dfrac{{74250}}{{21}}litres$

Given that time required to empty $\dfrac{{25}}{7}$ litres of water is $1$ second.

Hence the time required to empty $1$ litres of water is $\dfrac{7}{{25}}$ seconds.

Now, Time taken to half empty the tank will be equal to

$

\dfrac{{74250}}{{21}}litres{\text{ of water = }}\dfrac{7}{{25}}{\text{x}}\dfrac{{74.25}}{{21}}secs{\text{ }} \\

{\text{ }}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\text{ = 990 }}secs. \\

{\text{ = 16}}{\text{.5 }}\min{\text{ ( since 1 minute = 60 seconds)}} \\

$

Hence time taken to half empty the tank is ${\text{16}}{\text{.5}}$ minutes.

Note: In order to solve these surface areas and volume problems, you should have command over formulas related to all types of 3D shapes.

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