• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q26P

Expert-verified
Found in: Page 16

### Physics For Scientists & Engineers

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 ${r}_{Fe}{\left(\frac{{\rho }_{Fe}}{{\rho }_{Al}}\right)}^{1/3}$.

See the step by step solution

## Step 1: Identify given data

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

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

## Step 2: Determine radius of aluminum sphere

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 ${r}_{Fe}{\left(\frac{{\rho }_{Fe}}{{\rho }_{Al}}\right)}^{1/3}$.