# Water flows at the rate of 15 m per minute through a cylindrical pipe, having the diameter 20 mm. How much time (in min) will it take to fill a conical vessel of base diameter 40 cm and depth 45 cm?

Last updated date: 19th Mar 2023

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Hint- Here, we will be using the formula for the volume of cones.

Given, Rate of flow of water through a cylindrical pipe, ${r_f} = 15{\text{ }}\dfrac{{\text{m}}}{{\min }} = 1500{\text{ }}\dfrac{{{\text{cm}}}}{{{\text{min}}}}$

Diameter of the cylindrical pipe through which water is flowing, $d = 20$mm

Radius of the cylindrical pipe through which water is flowing, $r = \dfrac{d}{2} = \dfrac{{20}}{2} = 10$mm

Base diameter of the conical vessel, $D = 40$cm

Base radius of the conical vessel, $R = \dfrac{D}{2} = \dfrac{{40}}{2} = 20$cm

Height or depth of the conical vessel, $H = 45$cm

As we know that the formula for volume of a cone with radius $R$ and height $H$ is given by $V = \dfrac{1}{3}\pi {R^2}H$

Total volume of conical vessel to be filled, $V = \dfrac{1}{3}\pi {R^2}H = \dfrac{1}{3}\left( {\dfrac{{22}}{7}} \right){\left( {20} \right)^2}\left( {45} \right) = \dfrac{{132000}}{7}{\text{ c}}{{\text{m}}^3}$

Since the cross-sectional area of a cylindrical pipe is circular so the area of the cylindrical pipe having a radius $r$ will be given by $A = \pi {r^2}$

Area of the cylindrical pipe through which water is flowing $A = \pi {r^2} = \dfrac{{22}}{7} \times {\left( {10} \right)^2} = \dfrac{{2200}}{7}{\text{ m}}{{\text{m}}^2}$

Since, $1{\text{ m}}{{\text{m}}^2} = \dfrac{1}{{100}}\;{\text{c}}{{\text{m}}^2}$

Area of the cylindrical pipe through which water is flowing (in cm2) $A = \dfrac{{2200}}{7}{\text{ m}}{{\text{m}}^2} = \left( {\dfrac{{2200}}{7}} \right)\left( {\dfrac{1}{{100}}} \right) = \dfrac{{22}}{7}{\text{ c}}{{\text{m}}^2}$

Volume of water flowing through the cylindrical pipe per minute can be obtained by multiplying the area of cylindrical pipe and rate of flow of water which is given in length per time units.

Volume of water flowing through the cylindrical pipe per minute,

${V_f} = \left( {\dfrac{{22}}{7}} \right) \times \left( {1500} \right) = \dfrac{{33000}}{7}$cm3 per minute

Since we have to fill the conical vessel completely consisting of $\dfrac{{132000}}{7}{\text{ c}}{{\text{m}}^3}$

Therefore, time required by the cylindrical pipe to fill the conical vessel $t = \dfrac{{{\text{Total Volume of the conical vessel}}}}{{{\text{Volume of water flowing through the cylindrical pipe per minute}}}} = \dfrac{V}{{{V_f}}} = \dfrac{{\dfrac{{132000}}{7}}}{{\dfrac{{33000}}{7}}} = 4$min

Hence, time required by the cylindrical pipe to fill the conical vessel completely is 4 minutes.

Note- In these types of problems we find out the total volume of the vessel to be filled and the volumetric rate at which the vessel is getting filled and then by dividing the total volume of the vessel to the volumetric rate, we can obtain the time needed to fill the vessel completely.

Given, Rate of flow of water through a cylindrical pipe, ${r_f} = 15{\text{ }}\dfrac{{\text{m}}}{{\min }} = 1500{\text{ }}\dfrac{{{\text{cm}}}}{{{\text{min}}}}$

Diameter of the cylindrical pipe through which water is flowing, $d = 20$mm

Radius of the cylindrical pipe through which water is flowing, $r = \dfrac{d}{2} = \dfrac{{20}}{2} = 10$mm

Base diameter of the conical vessel, $D = 40$cm

Base radius of the conical vessel, $R = \dfrac{D}{2} = \dfrac{{40}}{2} = 20$cm

Height or depth of the conical vessel, $H = 45$cm

As we know that the formula for volume of a cone with radius $R$ and height $H$ is given by $V = \dfrac{1}{3}\pi {R^2}H$

Total volume of conical vessel to be filled, $V = \dfrac{1}{3}\pi {R^2}H = \dfrac{1}{3}\left( {\dfrac{{22}}{7}} \right){\left( {20} \right)^2}\left( {45} \right) = \dfrac{{132000}}{7}{\text{ c}}{{\text{m}}^3}$

Since the cross-sectional area of a cylindrical pipe is circular so the area of the cylindrical pipe having a radius $r$ will be given by $A = \pi {r^2}$

Area of the cylindrical pipe through which water is flowing $A = \pi {r^2} = \dfrac{{22}}{7} \times {\left( {10} \right)^2} = \dfrac{{2200}}{7}{\text{ m}}{{\text{m}}^2}$

Since, $1{\text{ m}}{{\text{m}}^2} = \dfrac{1}{{100}}\;{\text{c}}{{\text{m}}^2}$

Area of the cylindrical pipe through which water is flowing (in cm2) $A = \dfrac{{2200}}{7}{\text{ m}}{{\text{m}}^2} = \left( {\dfrac{{2200}}{7}} \right)\left( {\dfrac{1}{{100}}} \right) = \dfrac{{22}}{7}{\text{ c}}{{\text{m}}^2}$

Volume of water flowing through the cylindrical pipe per minute can be obtained by multiplying the area of cylindrical pipe and rate of flow of water which is given in length per time units.

Volume of water flowing through the cylindrical pipe per minute,

${V_f} = \left( {\dfrac{{22}}{7}} \right) \times \left( {1500} \right) = \dfrac{{33000}}{7}$cm3 per minute

Since we have to fill the conical vessel completely consisting of $\dfrac{{132000}}{7}{\text{ c}}{{\text{m}}^3}$

Therefore, time required by the cylindrical pipe to fill the conical vessel $t = \dfrac{{{\text{Total Volume of the conical vessel}}}}{{{\text{Volume of water flowing through the cylindrical pipe per minute}}}} = \dfrac{V}{{{V_f}}} = \dfrac{{\dfrac{{132000}}{7}}}{{\dfrac{{33000}}{7}}} = 4$min

Hence, time required by the cylindrical pipe to fill the conical vessel completely is 4 minutes.

Note- In these types of problems we find out the total volume of the vessel to be filled and the volumetric rate at which the vessel is getting filled and then by dividing the total volume of the vessel to the volumetric rate, we can obtain the time needed to fill the vessel completely.

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