
A metallic cylinder of diameter $3$ cm and height $3\dfrac{1}{3}$ cm is melted and cast into a sphere. Find the diameter of the sphere.
Answer
473.4k+ views
Hint: The metallic cylinder is converted into a metallic sphere. In this conversion the volume remains the same for both the cylinder and sphere. So equate both the volumes and find the radius and then we can find diameter as twice the radius.
Complete step-by-step answer:
Given information
Cylinder:
Diameter of cylinder, ${D_c} = 5$ cm
Radius of the cylinder is given by,
${r_c} = \dfrac{D}{2}$
${r_c} = \dfrac{5}{2}$ cm
Height of the cylinder, ${H_c} = 3\dfrac{1}{3} = \dfrac{{10}}{3}$ cm
The volume of the cylinder is given by,
$V = \pi {r_c}^2{H_c} \cdots \left( 1 \right)$
Substituting the value of and in equation (1), we get
$\Rightarrow V = \pi {\left( {\dfrac{5}{2}} \right)^2}\left( {\dfrac{{10}}{3}} \right) \cdots \left( 2 \right)$
Sphere
Let the radius and diameter of the sphere be and respectively.
Volume of the sphere is given by,
${V_s} = \dfrac{4}{3}\pi {r_s}^3 \cdots \left( 3 \right)$
According to the question the metallic cylinder is converted into a metallic sphere. In this process of conversion the volume of the cylinder will be the same as that of the sphere.
Therefore, equating (2) and equation (3) , we get
$\Rightarrow \pi {\left( {\dfrac{5}{2}} \right)^2}\left( {\dfrac{{10}}{3}} \right) = \dfrac{4}{3}\pi {r_s}^3$
Cancelling the $\pi $ from both sides,
$\Rightarrow \left( {\dfrac{{25}}{4}} \right)\left( {\dfrac{{10}}{3}} \right) = \dfrac{4}{3}{r_s}^3$
There is only one variable in this equation and i.e. ${r_s}$ . Solving for ${r_s}$ we get,
$
\Rightarrow \left( {\dfrac{{25}}{4}} \right)\left( {\dfrac{{10}}{3}} \right) = \dfrac{4}{3}{r_s}^3 \\
\Rightarrow {r_s}^3 = \dfrac{{25 \times 10 \times 3}}{{4 \times 3 \times 4}} \\
\Rightarrow {r_s}^3 = \dfrac{{125}}{8} \\
\Rightarrow {r_s} = \sqrt[3]{{\dfrac{{125}}{8}}} \\
\Rightarrow {r_s} = 2.5{\text{ cm}} \\
$
The diameter of the sphere is twice its radius.
$
\Rightarrow {D_s} = 2{r_s} \\
\Rightarrow {D_s} = 2\left( {2.5} \right) \\
\Rightarrow {D_s} = 5{\text{ cm}} \\
$
Hence, the diameter of the sphere is, ${D_s} = 5{\text{ cm}}$.
Note: In problems of these types where one shape is converted into another shape, the concept of constant volume should be applied.
For instance, if the cone of radius ${r_c}$ and height ${h_c}$ is converted into a sphere of radius ${r_s}$ . then also the concept of constant volume is used as
Volume of cone = Volume of Sphere.
$\dfrac{1}{3}\pi {r_c}^2{h_c} = \dfrac{4}{3}\pi {r_s}^3$
The only unknown in this equation (i) is ${r_s}$ .So it can be solved for ${r_s}$.
Complete step-by-step answer:

Given information
Cylinder:
Diameter of cylinder, ${D_c} = 5$ cm
Radius of the cylinder is given by,
${r_c} = \dfrac{D}{2}$
${r_c} = \dfrac{5}{2}$ cm
Height of the cylinder, ${H_c} = 3\dfrac{1}{3} = \dfrac{{10}}{3}$ cm
The volume of the cylinder is given by,
$V = \pi {r_c}^2{H_c} \cdots \left( 1 \right)$
Substituting the value of and in equation (1), we get
$\Rightarrow V = \pi {\left( {\dfrac{5}{2}} \right)^2}\left( {\dfrac{{10}}{3}} \right) \cdots \left( 2 \right)$
Sphere
Let the radius and diameter of the sphere be and respectively.
Volume of the sphere is given by,
${V_s} = \dfrac{4}{3}\pi {r_s}^3 \cdots \left( 3 \right)$
According to the question the metallic cylinder is converted into a metallic sphere. In this process of conversion the volume of the cylinder will be the same as that of the sphere.
Therefore, equating (2) and equation (3) , we get
$\Rightarrow \pi {\left( {\dfrac{5}{2}} \right)^2}\left( {\dfrac{{10}}{3}} \right) = \dfrac{4}{3}\pi {r_s}^3$
Cancelling the $\pi $ from both sides,
$\Rightarrow \left( {\dfrac{{25}}{4}} \right)\left( {\dfrac{{10}}{3}} \right) = \dfrac{4}{3}{r_s}^3$
There is only one variable in this equation and i.e. ${r_s}$ . Solving for ${r_s}$ we get,
$
\Rightarrow \left( {\dfrac{{25}}{4}} \right)\left( {\dfrac{{10}}{3}} \right) = \dfrac{4}{3}{r_s}^3 \\
\Rightarrow {r_s}^3 = \dfrac{{25 \times 10 \times 3}}{{4 \times 3 \times 4}} \\
\Rightarrow {r_s}^3 = \dfrac{{125}}{8} \\
\Rightarrow {r_s} = \sqrt[3]{{\dfrac{{125}}{8}}} \\
\Rightarrow {r_s} = 2.5{\text{ cm}} \\
$
The diameter of the sphere is twice its radius.
$
\Rightarrow {D_s} = 2{r_s} \\
\Rightarrow {D_s} = 2\left( {2.5} \right) \\
\Rightarrow {D_s} = 5{\text{ cm}} \\
$
Hence, the diameter of the sphere is, ${D_s} = 5{\text{ cm}}$.
Note: In problems of these types where one shape is converted into another shape, the concept of constant volume should be applied.
For instance, if the cone of radius ${r_c}$ and height ${h_c}$ is converted into a sphere of radius ${r_s}$ . then also the concept of constant volume is used as
Volume of cone = Volume of Sphere.
$\dfrac{1}{3}\pi {r_c}^2{h_c} = \dfrac{4}{3}\pi {r_s}^3$
The only unknown in this equation (i) is ${r_s}$ .So it can be solved for ${r_s}$.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

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

How do you graph the function fx 4x class 9 maths CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Difference Between Plant Cell and Animal Cell

What is pollution? How many types of pollution? Define it

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.
