## College Physics (4th Edition)

Let $d$ be the diameter of a sphere. Then the radius $r$ is $d/2$. Note that $d = 2r$. The volume of the sphere is $V = \frac{4}{3}\pi r^3$ We need to make a new sphere with a volume $V_2$ that is 8 times the original volume. Let $r_2$ be the radius of this new sphere. $V_2 = 8\times V$ $\frac{4}{3}\pi r_2^3 = 8 \times \frac{4}{3}\pi r^3$ $r_2^3 = 8\times r^3$ $r_2 = \sqrt[3] {8\times r^3} = 2r$ The new diameter is $2\times (2r)$ which is $2d$ The correct answer is (b) 2d