A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold (given 3 2 = 1.259) is approximately

Answer: Option D

Solution

Answer: Option D

Solution:

$$\begin{array}{rl}& \text{Volume of lead}=\frac{4}{3}\pi r^3\\ & \text{Volume of gold}=\frac{4}{3}\pi R^3-\frac{4}{3}\pi r^3\\ & \text{According to the question,}\\ & \frac{4}{3}\pi R^3-\frac{4}{3}\pi r^3=\frac{4}{3}\pi r^3\\ & \frac{4}{3}\pi R^3=\frac{8}{3}\pi r^3\\ & R^3=2r^3\\ & R^3=2{\left(2\right)}^3\\ & R=\text{root}3\text{of}2\times 2\\ & R=1.259\times 2\\ & R=2.518\\ & \therefore \text{Thickness}=R-r\\ & =2.518-2\\ & =0.518\text{ cm}\end{array}$$