Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: · Marks: · Difficulty:
A water tank open at the top is hemispherical at the bottom and cylindrical above it. If radius of the hemisphere is 12 m and the total capacity of the tank is $3312 \pi \mathrm{m}^{3}$, then the ratio of the surface areas of the hemispherical and the cylindrical portions is
(a)1 : 1
(b)3 : 5
(c)4 : 5
(d)6 : 5
Answer
Answer (as printed): C
Explanation
Let the height of the cylindrical part be $h$ metres. Volume of the tank = Volume of hemispherical part + Volume of cylindrical part $$\begin{aligned} & =\left(\frac{2}{3} \times \pi \times 12 \times 12 \times 12+\pi \times 12 \times 12 \times h\right) \mathrm{m}^{3} \\ & =\pi(1152+144 h) \mathrm{m}^{3} \\ & \therefore \pi(1152+144 h)=3312 \pi \Rightarrow 144 \mathrm{h}=2160 \Rightarrow h=15 \mathrm{m} . \end{aligned}$$ So, ratio of surface areas $=\frac{2 \pi r^{2}}{2 \pi r h}=\frac{r}{h}=\frac{12}{15}=4: 5$.