Question

The heights of two right circular cones are in the ratio \(1 : 2\) and the circumferences of their bases are in the ratio \(3: 4.\) Find the ratio of their volumes.

• A. \(32:9\)
• B. \(9:32\)
• C. \(7:23\)
• D. \(23:7\)

Answer 1

The Correct Option is B

The ratio of the heights of the two right circular cones is \(1:2\), meaning if the height of the first cone is \(h_1\), then the height of the second cone is \(h_2 = 2h_1\).

The ratio of the circumferences of the bases of the cones is \(3:4\). If the circumference of the first cone's base is \(C_1 = 2\pi r_1\) and the second's is \(C_2 = 2\pi r_2\), then:

\(\frac{C_1}{C_2} = \frac{3}{4}\)

This gives:

\(\frac{2\pi r_1}{2\pi r_2} = \frac{3}{4}\)

So, \(\frac{r_1}{r_2} = \frac{3}{4}\)

The volume \(V\) of a cone is given by \(\frac{1}{3}\pi r^2 h\). Therefore, the volumes of the first and second cones are:

\(V_1 = \frac{1}{3}\pi r_1^2 h_1\)

\(V_2 = \frac{1}{3}\pi r_2^2 h_2\)

Substitute \(h_2 = 2h_1\) and \(\frac{r_1}{r_2} = \frac{3}{4}\) into the expression of \(V_2\):

\(V_2 = \frac{1}{3}\pi r_2^2 (2h_1) = \frac{2}{3}\pi r_2^2 h_1\)

Now find \(\frac{V_1}{V_2}\):

\(\frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{2}{3}\pi r_2^2 h_1}\)

= \(\frac{r_1^2}{2r_2^2}\)

Substitute \(r_1 = \frac{3}{4}r_2\):

= \(\frac{\left(\frac{3}{4}r_2\right)^2}{2r_2^2}\)

= \(\frac{\frac{9}{16}r_2^2}{2r_2^2}\)

= \(\frac{9}{32}\)

Thus, the ratio of their volumes is \(9:32\).