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

Solution(By Examveda Team)

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