Question:
The radius of a sphere is increased by 20%. What is the percentage increase in its volume?
The radius of a sphere is increased by 20%. What is the percentage increase in its volume?
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When a dimension of a 3D shape changes by \(x%\), the volume changes approximately by \((1+x/100)^3 - 1\) in decimal, multiply by 100 for percentage.
Updated On: May 27, 2025
- 72.8%
- 66.4%
- 48.8%
- 62.4%
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The Correct Option is A
Solution and Explanation
Step 1: Let the original radius be \(r\).
Step 2: Volume of sphere is \[ V = \frac{4}{3} \pi r^3. \] Step 3: New radius = \[ r' = r + 0.20r = 1.2r. \] Step 4: New volume, \[ V' = \frac{4}{3} \pi (r')^3 = \frac{4}{3} \pi (1.2r)^3 = \frac{4}{3} \pi r^3 (1.2)^3 = V \times (1.728). \] Step 5: Percentage increase in volume, \[ \frac{V' - V}{V} \times 100 = (1.728 - 1) \times 100 = 0.728 \times 100 = 72.8%. \]
Step 2: Volume of sphere is \[ V = \frac{4}{3} \pi r^3. \] Step 3: New radius = \[ r' = r + 0.20r = 1.2r. \] Step 4: New volume, \[ V' = \frac{4}{3} \pi (r')^3 = \frac{4}{3} \pi (1.2r)^3 = \frac{4}{3} \pi r^3 (1.2)^3 = V \times (1.728). \] Step 5: Percentage increase in volume, \[ \frac{V' - V}{V} \times 100 = (1.728 - 1) \times 100 = 0.728 \times 100 = 72.8%. \]
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