Discussion
A rectangular paper sheet of dimensions 54 cm × 4 cm is taken. The two longer edges of the sheet are joined together to create a cylindrical tube. A cube whose surface area is equal to the area of the sheet is also taken.
Then, the ratio of the volume of the cylindrical tube to the volume of the cube is
Answer: A
The ratio is 1/π.
Joining the two 54 cm edges makes a tube 54 cm long whose circumference is the short side, 4 cm. So its radius is \(r = 4/(2\pi) = 2/\pi\) cm.
\[ \begin{aligned} V_{tube} &= \pi r^{2} h = \pi \cdot \frac{4}{\pi^{2}} \cdot 54 = \frac{216}{\pi} \\ 6a^{2} &= 54 \times 4 = 216 \Rightarrow a = 6 \\ V_{cube} &= 6^{3} = 216 \end{aligned} \]
The ratio is \(\frac{216/\pi}{216} = \frac{1}{\pi}\).
Rolling the sheet the other way (circumference 54 cm, length 4 cm) would give a much larger volume, \(2916/\pi\), and none of the listed ratios; the 2/π, 3/π and 4/π options come from slips such as using the diameter for the radius.
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