If the radius of a sphere is increased by $10\%$, then the volume will be increased by

### Concept & Successive Percentage Change The volume of a sphere is given by: $$V = \frac{4}{3} \pi r^3$$ Since $V \propto r^3$, a percentage increase in the radius acts as three successive percentage increases on the volume. ### Step-by-Step Solution * **Given:** * Increase in radius $= 10\%$ * **Calculation (Method 1: Formula):** 1. Let the initial radius be $r=10$ units. Initial Volume $\propto 10^3 = 1000$. 2. New radius $r' = 10 + 10\% \text{ of } 10 = 11$ units. 3. New Volume $\propto 11^3 = 1331$. 4. Increase in volume $= 1331 - 1000 = 331$. 5. Percentage increase $= \left(\frac{331}{1000}\right) \times 100 = 33.1\%$. ### Exam Strategy & Shortcut Use the successive percentage change formula $a + b + \frac{ab}{100}$ for the three factors of $10\%$: First two dimensions: $10 + 10 + \frac{10 \times 10}{100} = 21\%$. Combine this result with the third dimension: $21 + 10 + \frac{21 \times 10}{100} = 31 + 2.1 = 33.1\%$. Alternatively, memorize that $1.1^3 = 1.331$, directly indicating a $33.1\%$ increase. ### Common Pitfall Adding the percentages directly ($10\% \times 3 = 30\%$) is the most common mistake. This ignores the compounding effect of the volume expanding in three dimensions simultaneously. ### Final Answer Therefore, the correct answer is **$33.1\%$**.