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A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

The aim of this question is to learn the volume of a sphere and the density of different materials.

If the radius r is known, the volume V of a sphere is given by:

\[ V \ = \ \dfrac{ 4 }{ 3 } \ \pi r^3 \ … \ … \ … \ (1) \]

Also, for a given material the density $ d $ is defined as:

\[ d \ = \ \dfrac{ m }{ V } \ … \ … \ … \ (2) \]

Where m is the mass of the body. We will manipulate the above two equations to solve the given problem.

Expert Answer

Substituting equation (1) in equation (2):

\[ d \ = \ \dfrac{ m }{ \bigg ( \ \frac{ 4 }{ 3 } \ \pi r^3 \ \bigg ) } \]

\[ \Rightarrow d \ = \ \dfrac{ 4 m }{ 3 \pi r^3 } \]

For lead (say material no. 1 ), the above equation becomes:

\[ d_1 \ = \ \dfrac{ 4 m_1 }{ 3 \pi r_1^3 } \ … \ … \ … \ (3) \]

For Aluminum (say material no. 2 ), the above equation becomes:

\[ d_2 \ = \ \dfrac{ 4 m_2 }{ 3 \pi r_2^3 } \ … \ … \ … \ (4) \]

Dividing and simplifying equation (3) by equation (4):

\[ \dfrac{ d_1 }{ d_2 } \ = \ \dfrac{ m_1 r_2^3 }{ m_2 r_1^3 } \]

Given that:

\[ m_1 = m_2 \]

The above equation further reduces to:

\[ \dfrac{ d_1 }{ d_2 } \ = \ \bigg ( \dfrac{ r_2 }{ r_1 } \bigg )^3 \ … \ … \ … \ (5) \]

\[ \Rightarrow \dfrac{ r_2 }{ r_1 } \ = \ \bigg ( \dfrac{ d_1 }{ d_2 } \bigg )^{ 1/3 } \]

From density tables:

\[ d_1 \ = \ 11.29 \ g/cm^3 \text{ and } d_2 \ = \ 2.7 \ g/cm^3 \]

Substituting these in equation no. (5):

\[ \dfrac{ r_2 }{ r_1 } \ = \ \bigg ( \dfrac{ 11.29 }{ 2.7 } \bigg )^{ 1/3 } \]

\[ \dfrac{ r_2 }{ r_1 } \ = \ \bigg ( 4.1814 \bigg )^{ 1/3 } \]

\[ \Rightarrow \dfrac{ r_2 }{ r_1 } \ = \ 1.61 \]

Numerical Result

\[ \dfrac{ r_2 }{ r_1 } \ = \ 1.61 \]

Example

Find the ratio of the radiuses of two uniform spheres. One is made up of copper and the other one is made of Zinc.

Let copper and zinc be materials no. 1 and 2, respectively. Then from density tables:

\[ d_1 \ = \ 8.96 \ g/cm^3 \text{ and } d_2 \ = \ 7.133 \ g/cm^3 \]

Substituting these in equation no. (5):

\[ \dfrac{ r_2 }{ r_1 } \ = \ \bigg ( \dfrac{ 8.96 }{ 7.133 } \bigg )^{ 1/3 } \]

\[ \dfrac{ r_2 }{ r_1 } \ = \ \bigg ( 1.256 \bigg )^{ 1/3 } \]

\[ \Rightarrow \dfrac{ r_2 }{ r_1 } \ = \ 1.0789 \]

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