Answer
The ratio of the radius of the aluminum sphere to the radius of the lead sphere is 1.6.
Work Step by Step
Let the mass of each sphere be $M$.
Then: $M =(V_{aluminum})~(\rho_{aluminum})= (V_{lead})~(\rho_{lead})$
$\frac{V_{aluminum}}{V_{lead}} = \frac{\rho_{lead}}{\rho_{aluminum}}$
$\frac{\frac{4}{3}\pi~R_{aluminum}^3}{\frac{4}{3}\pi~R_{lead}^3} = \frac{\rho_{lead}}{\rho_{aluminum}}$
$\frac{R_{aluminum}^3}{R_{lead}^3} = \frac{\rho_{lead}}{\rho_{aluminum}}$
$\frac{R_{aluminum}}{R_{lead}} = (\frac{\rho_{lead}}{\rho_{aluminum}})^{1/3}$
$\frac{R_{aluminum}}{R_{lead}} = (\frac{11.3\times 10^3~kg/m^3}{2.7\times 10^3~kg/m^3})^{1/3}$
$\frac{R_{aluminum}}{R_{lead}} = 1.6$
The ratio of the radius of the aluminum sphere to the radius of the lead sphere is 1.6.
