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We have $2$ rectangular sheets of paper, $\text{M}$ and $\text{N}$, of dimensions $\text{6 cm $\times$ 1 cm}$ each. Sheet $\text{M}$ is rolled to form an open cylinder by bringing the short edges of the sheet together. Sheet $\text{N}$ is cut into equal square patches and assembled to form the largest possible closed cube. Assuming the ends of the cylinder are closed, the ratio of the volume of the cylinder to that of the cube is _________.

  1.  $\frac{\pi}{2}$
  2. $\frac{3}{\pi}$
  3. $\frac{9}{\pi}$
  4. $3\pi$
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Best answer
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Sheet M is rolled to form an open cylinder by bringing the shorter edge of the sheet together

So the longer side becomes the circumference of the base of the cylinder and shorter side becomes the height.

$\therefore 2\times\pi\times r = 6 \\\implies r = \frac{3}{\pi}$

$\text{Volume of cyclinder} = \pi\times r^2 \times h = \pi\times {(\frac{3}{\pi})}^2\times 1 = \frac{9}{\pi}$

Sheet N is cut into equal square patches and assembled to form the largest possible closed cube

So the entire area of the sheet is converted into surface area of the cube, i.e.

$6\times a^2 = 6 \times 1\\\implies a^2 = 1\\\implies a = 1$

$\text{Volume of cube} = a^3 = (1)^3 = 1$

$\frac{\text{Volume of cylinder}}{\text{Volume of cube}} = \frac{\frac{9}{\pi}}{1} = \frac{9}{\pi}$

Option C

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