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 $$\root 3 \of 2 $$  = 1.259) is approximately

Solution (By Examveda Team)

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