Question

The ratio of the diameter of the vessels A and B is 9:4. The vessel A is filled with water till the brim. If half of the water from vessel A would leave quarter of the vessel B empty, what is the ratio of the heights of the vessels?

• A. 6:25
• B. 8:27
• C. 4:29
• D. 2:13

Answer 1

The correct option is B 8:27

Given, the ratio of the diameter of the vessels A and B is 9:4.
Let, the diameter of vessel A and B be $d_1$ and $d_2$.
So, $\frac{d_1}{d_2}=\frac{9}{4}$
According to question,
$\frac{1}{2}\times \mathrm{Volume}\mathrm{of}\mathrm{vessel}A=\frac{3}{4}\times \mathrm{Volume}\mathrm{of}\mathrm{vessel}B$

or Volume of vessel A = $\frac{3}{2}\times$ Volume of vessel B
$\Rightarrow \pi {r_1}^2{h}_1=\frac{3}{2}\times \pi {r_2}^2{h}_2$
where $r_1\mathrm{and}{r}_2\mathrm{are}\mathrm{radius}\mathrm{of}\mathrm{vessel}A\mathrm{and}B,h_1\mathrm{and}{h}_2\mathrm{are}\mathrm{height}\mathrm{of}\mathrm{vessel}A\mathrm{and}B$
$\Rightarrow \frac{2}{3}\times {\left(\frac{r_1}{r_2}\right)}^2=\frac{h_2}{h_1}$$\Rightarrow \frac{2}{3}\times {\left(\frac{d_1}{d_2}\right)}^2=\frac{h_2}{h_1}$
$\Rightarrow \frac{2}{3}\times {\left(\frac{9}{4}\right)}^2=\frac{h_2}{h_1}$
$\Rightarrow \frac{2}{3}\times \frac{81}{16}=\frac{h_2}{h_1}$
$\Rightarrow \frac{h_2}{h_1}=\frac{27}{8}$
$\Rightarrow \frac{h_1}{h_2}=\frac{8}{27}$

Hence, the ratio of the heights of the vessels is 8:27.