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Q26P

Expert-verifiedFound in: Page 16

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Let ${{\mathbf{\rho}}}_{{\mathbf{Al}}}$ represent the density of aluminium and ${{\mathbf{\rho}}}_{{\mathbf{Fe}}}$ that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius ${{\mathbf{\rho}}}_{{\mathbf{Fe}}}$ on an equal-arm balance.**

The radius of aluminum sphere is ${r}_{Fe}{\left(\frac{{\rho}_{Fe}}{{\rho}_{Al}}\right)}^{1/3}$.

The density of aluminium is ${\rho}_{Al}$

The mass of iron is ${\rho}_{Fe}$.

**The density is the ratio of the mass of an object tothe unit volume of the space. It describes the heavier or lighter among the different objects.**

The volume of the aluminum sphere is given by

${V}_{Al}=\frac{4}{3}\pi {r}_{Al}^{3}$

The volume of the iron sphere is given by

${V}_{Fe}=\frac{4}{3}\pi {r}_{Fe}^{3}$

The condition for the equal-arm balanceis given by

$\begin{array}{rcl}{m}_{a}& =& {m}_{i}\\ {\rho}_{Al}{V}_{Al}& =& {\rho}_{Fe}{V}_{Fe}\end{array}$

Substitute all the values in the above equation.

role="math" localid="1663665328044" $\begin{array}{rcl}{\rho}_{Al}\left(\frac{4}{3}\pi {r}_{Al}^{3}\right)& =& {\rho}_{Fe}\left(\frac{4}{3}\pi {r}_{Fe}^{3}\right)\\ {r}_{Al}& =& {r}_{Fe}{\left(\frac{{\rho}_{Fe}}{{\rho}_{Al}}\right)}^{1/3}\end{array}$

Therefore, the radius of the aluminum sphere that satisfies the equal-armbalance criterion is ${r}_{Fe}{\left(\frac{{\rho}_{Fe}}{{\rho}_{Al}}\right)}^{1/3}$.

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