## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The sphere's inner diameter is $8.0~cm$
We can find the volume of aluminum used to make the sphere as: $V = \frac{M}{\rho}$ $V = \frac{0.690~kg}{2.70\times 10^3~kg/m^3}$ $V = 2.56\times 10^{-4}~m^3$ This volume is equal to the difference between the volume of the outer sphere $V_o$ and the volume of the inner sphere $V_i$. We can find the radius $R_i$ of the inner sphere. $V_o-V_i = V$ $V_i = V_o-V$ $\frac{4}{3}\pi~R_i^3 = \frac{4}{3}\pi~R_o^3-V$ $R_i^3 = R_o^3-\frac{3V}{4\pi}$ $R_i = (R_o^3-\frac{3V}{4\pi})^{1/3}$ $R_i = [(0.0500~m)^3-\frac{(3)(2.56\times 10^{-4}~m^3)}{4\pi}]^{1/3}$ $R_i = 0.040~m = 4.0~cm$ The sphere's inner diameter is $2R_i$ which is $8.0~cm$.