We have 2 rectangular sheets of paper, M and N, of dimensions 6 cm x 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
Options:
(A) 1 : 1
(B) pi : 2
(C) 9 : pi
(D) 3*pi : 1
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2021 Q.6)**
**1. Governing Concept**:
- **Topic**: Mensuration & Solid Geometry (Cylinder vs. Cube Volume Ratio)
- Two sheets $M$ and $N$, each of dimensions $6\text{ cm} \times 1\text{ cm}$ (area $= 6\text{ cm}^2$).
- **Sheet M rolled into open cylinder**:
- Bringing the short edges (length $1\text{ cm}$) together means the circumference of the cylinder base is $C = 6\text{ cm}$, and cylinder height $h = 1\text{ cm}$.
- Base radius: $2\pi r = 6 \implies r = \frac{3}{\pi}\text{ cm}$.
- Assuming closed ends, volume of cylinder:
$V_{\text{cylinder}} = \pi r^2 h = \pi \left(\frac{3}{\pi}\right)^2 (1) = \frac{9}{\pi}\text{ cm}^3$
- **Sheet N cut into square patches to form a closed cube**:
- A closed cube has 6 identical square faces.
- Total surface area of sheet $N = 6\text{ cm}^2$.
- Area of each square face $= \frac{6}{6} = 1\text{ cm}^2 \implies$ edge length of cube $a = 1\text{ cm}$.
- Volume of cube:
$V_{\text{cube}} = a^3 = 1^3 = 1\text{ cm}^3$
- Ratio of the volume of the cylinder to that of the cube:
$\frac{V_{\text{cylinder}}}{V_{\text{cube}}} = \frac{9/\pi}{1} = \frac{9}{\pi}$
**2. Option Analysis**:
- **Option (A) [1 : 1] is INCORRECT**: Fails circular area calculation.
- **Option (B) [pi : 2] is INCORRECT**: Miscalculates cylinder base radius.
- **Option (C) [9 : pi] is CORRECT**: Exactly matches $\frac{9}{\pi} : 1 = 9 : \pi$.
- **Option (D) [3pi : 1] is INCORRECT**: Inverted ratio factor.
**3. Final Answer**:
- **Correct Choice**: **C**
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