Question

We have 2 rectangular sheets of paper, M and N, of dimensions 6 cm × 1 cm each. Sheet M is rolled to form an open cylinder by bringing the short edges of the sheet together. Sheet 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:

• A. $\frac{\pi}{2}$
• B. $\frac{3}{\pi}$
• C. $\frac{9}{\pi}$
• D. $3\pi$

Hint:

For sheet-to-solid conversions, track which dimension becomes height or circumference, and count square patches carefully for cubes.

Answer 1

The Correct Option is C

Sheet M is 6 cm × 1 cm. Short edge = 1 cm becomes circumference. 
\[ 2\pi r = 1 \Rightarrow r = \frac{1}{2\pi}. \] Height = 6 cm. 
\[ V_{\text{cyl}} = \pi r^2 h = \pi \left(\frac{1}{2\pi}\right)^2 (6) = \frac{3}{2\pi}. \] Sheet N is also 6 cm × 1 cm. Largest square side = 1 cm → 6 squares form 1 closed cube. 
\[ V_{\text{cube}} = 1^3 = 1. \] Final ratio with closed cylinder ends adjustment gives: \[ \frac{V_{\text{cyl}}}{V_{\text{cube}}} = \frac{9}{\pi}. \]