A solid brass sphere of radius 2.1 dm is converted into a right circular cylindrical rod of length 7 cm. The ratio of total surface areas of the rod to the sphere is

Answer: Option C

Solution

Answer: Option C

Solution:

$$\begin{array}{rl}& \frac{4}{3}\pi R^3=\pi r^{2}h\\ & \frac{4}{3}\pi \times 21\times 21\times 21=\pi \times R^2\times 7\\ & R^2=21\times 21\times 4\\ & R=42\text{ cm}\\ & \frac{\text{total surface area of rod}}{\text{total surface area of sphere}}\\ & =\frac{2\pi r\left(h+r\right)}{4\pi R^2}\\ & =\frac{42\left(42+7\right)}{2\times 21\times 21}\\ & =\frac{49}{21}\\ & =\frac{7}{3}\\ & =7:3\end{array}$$