Question Statement:
The radius as well as the height of a circular cone increases by 10%. The percentage
increase in its volume is
Options:
-
(A) 17.1
-
(B) 21.0
-
(C) 33.1
-
(D) 72.8
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2019 Q.3)**
**1. Governing Concept**:
- **Topic**: Mensuration & Percentage Changes in Solid Volumes
- Volume of a right circular cone of base radius $r$ and height $h$:
$V = \frac{1}{3}\pi r^2 h$
- Both radius and height are increased by $10\%$:
$r' = 1.1 r, \quad h' = 1.1 h$
- New volume $V'$:
$V' = \frac{1}{3}\pi (r')^2 (h') = \frac{1}{3}\pi (1.1 r)^2 (1.1 h) = \frac{1}{3}\pi r^2 h \times (1.1)^3 = V \times 1.331$
- Percentage increase in volume:
$\Delta V\% = \left(\frac{V' - V}{V}\right) \times 100\% = (1.331 - 1) \times 100\% = 33.1\%$
**2. Option Analysis**:
- **Option (A) [10%] is INCORRECT**: Represents linear dimension increase, not volume.
- **Option (B) [21%] is INCORRECT**: Represents area increase ($(1.1)^2 - 1 = 21\%$).
- **Option (C) [33.1%] is CORRECT**: Directly matches $(1.1)^3 - 1 = 33.1\%$.
- **Option (D) [40%] is INCORRECT**: Naive sum of percentages ($10 + 10 + 10 \neq 40$).
**3. Final Answer**:
- **Correct Choice**: **C**
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