Question

If the radius of a sphere is doubled, then the increase in percentage of its surface area will be:

• A. $100%$
• B. $200%$
• C. $300%$
• D. $400%$

Hint:

For all 3D shapes whose surface area depends on $r^2$, doubling the radius increases the surface area by 4 times (i.e., a 300% increase).

Answer 1

The Correct Option is C

Step 1: Recall the formula for surface area of a sphere.
The surface area of a sphere is given by: \[ A = 4\pi r^2 \] where \( r \) is the radius of the sphere.

Step 2: When the radius is doubled.
If the new radius is \( 2r \), then the new surface area becomes: \[ A' = 4\pi (2r)^2 = 4\pi \times 4r^2 = 16\pi r^2 \] Actually, the correct simplification should be: \[ A' = 4\pi (2r)^2 = 4\pi \times 4r^2 = 16\pi r^2 \] Oops! Let’s recheck: Since \( (2r)^2 = 4r^2 \), \[ A' = 4\pi \times 4r^2 = 16\pi r^2 \] Wait, that’s incorrect — the coefficient should remain consistent with the definition. Let’s fix that step: Original: \( A = 4\pi r^2 \) When radius doubles: \[ A' = 4\pi (2r)^2 = 4\pi (4r^2) = 16\pi r^2 \] Now correct.

Step 3: Calculate the increase in surface area.
\[ \text{Increase} = A' - A = 16\pi r^2 - 4\pi r^2 = 12\pi r^2 \]
Step 4: Find the percentage increase.
\[ \text{Percentage Increase} = \frac{\text{Increase}}{\text{Original}} \times 100 = \frac{12\pi r^2}{4\pi r^2} \times 100 = 3 \times 100 = 300 \]
Step 5: Conclusion.
When the radius of a sphere is doubled, its surface area increases by $300$.