# A cylindrical tub of radius 12 cm contains water to a depth of 20cm. A spherical ball is dropped into the tub and the level of the water is raised by 6.75 cm, find the radius of the ball.

Last updated date: 26th Mar 2023

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Answer

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Hint: - Volume of spherical ball\[{\text{ = }}\dfrac{4}{3}\pi r_1^3\], (where ${r_1}$is the radius of the ball), and the volume of cylinder\[{{\text{V}}_C} = \pi {r^2}h\], (where r is the radius and h is the height of the cylinder)

Given data:

Radius of cylindrical tube \[\left( r \right) = 12cm\]

Depth of water in cylindrical tube\[\left( h \right) = 20cm\]

Volume of cylinder without rise\[{{\text{V}}_C} = \pi {r^2}h = \pi {\left( {12} \right)^2}\left( {20} \right) = 2880\pi c{m^3}\]

After dropping the spherical ball the level of water is raised by 6.75 cm.

$ \Rightarrow $New depth of water in cylindrical tube\[\left( {{h_1}} \right) = 20 + 6.75 = 26.75cm\]

$ \Rightarrow $New cylindrical volume \[{\text{ = }}\pi {r^2}{h_1} = \pi {\left( {12} \right)^2}\left( {26.75} \right) = 3852\pi c{m^3}\]

Volume of spherical ball\[{\text{ = }}\dfrac{4}{3}\pi r_1^3\], (where ${r_1}$is the radius of the ball)

Therefore new cylindrical volume $ = $ Volume of spherical ball$ + $ Volume of cylinder \[{{\text{V}}_C}\]without rise

\[

\Rightarrow 3852\pi = \dfrac{4}{3}\pi r_1^3 + 2880\pi \\

\Rightarrow \dfrac{4}{3}\pi r_i^3 = 972\pi \\

\Rightarrow r_1^3 = 729 = {\left( 9 \right)^3} \\

\Rightarrow {r_1} = 9cm \\

\]

So, the radius of the ball is 9cm.

Note: - In such types of questions always remember the formula of volume of cylinder and spherical ball which is stated above, then using these formulas, calculate the new volume of the cylindrical tub, when spherical ball is dropped in cylindrical tub and simplify, we will get the required radius of the ball.

Given data:

Radius of cylindrical tube \[\left( r \right) = 12cm\]

Depth of water in cylindrical tube\[\left( h \right) = 20cm\]

Volume of cylinder without rise\[{{\text{V}}_C} = \pi {r^2}h = \pi {\left( {12} \right)^2}\left( {20} \right) = 2880\pi c{m^3}\]

After dropping the spherical ball the level of water is raised by 6.75 cm.

$ \Rightarrow $New depth of water in cylindrical tube\[\left( {{h_1}} \right) = 20 + 6.75 = 26.75cm\]

$ \Rightarrow $New cylindrical volume \[{\text{ = }}\pi {r^2}{h_1} = \pi {\left( {12} \right)^2}\left( {26.75} \right) = 3852\pi c{m^3}\]

Volume of spherical ball\[{\text{ = }}\dfrac{4}{3}\pi r_1^3\], (where ${r_1}$is the radius of the ball)

Therefore new cylindrical volume $ = $ Volume of spherical ball$ + $ Volume of cylinder \[{{\text{V}}_C}\]without rise

\[

\Rightarrow 3852\pi = \dfrac{4}{3}\pi r_1^3 + 2880\pi \\

\Rightarrow \dfrac{4}{3}\pi r_i^3 = 972\pi \\

\Rightarrow r_1^3 = 729 = {\left( 9 \right)^3} \\

\Rightarrow {r_1} = 9cm \\

\]

So, the radius of the ball is 9cm.

Note: - In such types of questions always remember the formula of volume of cylinder and spherical ball which is stated above, then using these formulas, calculate the new volume of the cylindrical tub, when spherical ball is dropped in cylindrical tub and simplify, we will get the required radius of the ball.

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