A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is 3.5 cm and the height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub. (Take $\pi = \dfrac{{22}}{7}$)
Answer
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Hint- Here, we will be using the formulas for finding the volume of a cylinder, a hemisphere and a cone.
Given, radius of cylindrical tub $R = 5\,{\text{cm}}$
Length or height of cylindrical tub
Radius of hemisphere of the solid toy $r = 3.5{\text{ cm}}$
Height of cone outside the hemisphere of the solid toy $h = 5{\text{ cm}}$
As we know that volume of a cylinder with radius $R$ and height $H$ is given by ${{\text{V}}_{{\text{cy}}}} = \pi {R^2}H$
Also, volume of a hemisphere with a radius is given by ${{\text{V}}_{\text{h}}} = \left( {\dfrac{2}{3}} \right)\pi {r^3}$
Also, volume of the cone with a base radius $r$ and height $h$ is given by ${{\text{V}}_{{\text{co}}}} = \left( {\dfrac{1}{3}} \right)\pi {r^2}h$
Also given that the cylindrical tub is full of water which means that the complete volume of the cylindrical tub consists of the volume occupied by the solid toy and the volume occupied by the water.
Therefore, Volume of the water left in tubTotal volume of the cylindrical tub$ - $Volume of the solid toy
$ \Rightarrow $ Volume of the water left in tub$ = $Total volume of the cylindrical tub(Volume of hemisphere$ + $Volume of the cone)
$ \Rightarrow $ Volume of the water left in tub$ = \pi {R^2}H - \left[ {\left( {\dfrac{2}{3}} \right)\pi {r^3} + \left( {\dfrac{1}{3}} \right)\pi {r^2}h} \right] = \pi {R^2}H - \left[ {\dfrac{{2\pi {r^3} + \pi {r^2}h}}{3}} \right]$
Volume of the water left in tub $ = \dfrac{{22}}{7} \times {\left( 5 \right)^2} \times 9.8 - \left[ {\dfrac{{2\left( {\dfrac{{22}}{7}} \right) \times {{\left( {3.5} \right)}^3} + \left( {\dfrac{{22}}{7}} \right) \times {{\left( {3.5} \right)}^2} \times 5}}{3}} \right] = 770 - 154 = 616{\text{ c}}{{\text{m}}^3}$
Hence, the volume of the water left in the tub is 616 ${\text{ c}}{{\text{m}}^2}$.
Note- In this particular problem, we have assumed that the hemisphere with a cone is fully immersed in the water present in the cylindrical tub. Here, when a solid toy with the shape of a hemisphere having a cone at the top is immersed in a cylindrical tub full of water, some of the water present in the cylindrical tub gets out.
Given, radius of cylindrical tub $R = 5\,{\text{cm}}$
Length or height of cylindrical tub
Radius of hemisphere of the solid toy $r = 3.5{\text{ cm}}$
Height of cone outside the hemisphere of the solid toy $h = 5{\text{ cm}}$
As we know that volume of a cylinder with radius $R$ and height $H$ is given by ${{\text{V}}_{{\text{cy}}}} = \pi {R^2}H$
Also, volume of a hemisphere with a radius is given by ${{\text{V}}_{\text{h}}} = \left( {\dfrac{2}{3}} \right)\pi {r^3}$
Also, volume of the cone with a base radius $r$ and height $h$ is given by ${{\text{V}}_{{\text{co}}}} = \left( {\dfrac{1}{3}} \right)\pi {r^2}h$
Also given that the cylindrical tub is full of water which means that the complete volume of the cylindrical tub consists of the volume occupied by the solid toy and the volume occupied by the water.
Therefore, Volume of the water left in tubTotal volume of the cylindrical tub$ - $Volume of the solid toy
$ \Rightarrow $ Volume of the water left in tub$ = $Total volume of the cylindrical tub(Volume of hemisphere$ + $Volume of the cone)
$ \Rightarrow $ Volume of the water left in tub$ = \pi {R^2}H - \left[ {\left( {\dfrac{2}{3}} \right)\pi {r^3} + \left( {\dfrac{1}{3}} \right)\pi {r^2}h} \right] = \pi {R^2}H - \left[ {\dfrac{{2\pi {r^3} + \pi {r^2}h}}{3}} \right]$
Volume of the water left in tub $ = \dfrac{{22}}{7} \times {\left( 5 \right)^2} \times 9.8 - \left[ {\dfrac{{2\left( {\dfrac{{22}}{7}} \right) \times {{\left( {3.5} \right)}^3} + \left( {\dfrac{{22}}{7}} \right) \times {{\left( {3.5} \right)}^2} \times 5}}{3}} \right] = 770 - 154 = 616{\text{ c}}{{\text{m}}^3}$
Hence, the volume of the water left in the tub is 616 ${\text{ c}}{{\text{m}}^2}$.
Note- In this particular problem, we have assumed that the hemisphere with a cone is fully immersed in the water present in the cylindrical tub. Here, when a solid toy with the shape of a hemisphere having a cone at the top is immersed in a cylindrical tub full of water, some of the water present in the cylindrical tub gets out.
Last updated date: 17th Sep 2023
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