# 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.

Answer

Verified

364.8k+ views

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.

Last updated date: 23rd Sep 2023

â€¢

Total views: 364.8k

â€¢

Views today: 7.64k

Recently Updated Pages

What do you mean by public facilities

Slogan on Noise Pollution

Paragraph on Friendship

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

What is the Full Form of ILO, UNICEF and UNESCO

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers