Question:
A balloon, which always remains spherical, has a variable radius. The rate at which its volume is increasing with respect to its radius $r$ when $r = 5$ cm is:
A balloon, which always remains spherical, has a variable radius. The rate at which its volume is increasing with respect to its radius $r$ when $r = 5$ cm is:
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Differentiating volume with respect to radius always yields the surface area of the sphere: $\frac{dV}{dr} = 4\pi r^2$. Simply plug in the radius to get the area directly!
Updated On: Jun 3, 2026
- $100\pi$
- $50\pi$
- $25\pi$
- $10\pi$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The rate of change of volume $V$ with respect to the radius $r$ is represented by the derivative $\frac{dV}{dr}$.
Step 2: Meaning
Since the balloon is spherical, its volume is $V = \frac{4}{3}\pi r^3$. We find $\frac{dV}{dr}$ and evaluate it at $r = 5$ cm.
Step 3: Analysis
The volume of a sphere is: \[ V = \frac{4}{3}\pi r^3 \] Differentiating both sides with respect to $r$: \[ \frac{dV}{dr} = \frac{4}{3}\pi (3r^2) = 4\pi r^2 \] At $r = 5$ cm: \[ \left. \frac{dV}{dr} \right|_{r=5} = 4\pi (5)^2 = 4\pi (25) = 100\pi \]
Step 4: Conclusion
The rate at which the volume increases with respect to the radius when $r = 5$ cm is $100\pi$ $\text{cm}^3/\text{cm}$.
Final Answer: (A)
The rate of change of volume $V$ with respect to the radius $r$ is represented by the derivative $\frac{dV}{dr}$.
Step 2: Meaning
Since the balloon is spherical, its volume is $V = \frac{4}{3}\pi r^3$. We find $\frac{dV}{dr}$ and evaluate it at $r = 5$ cm.
Step 3: Analysis
The volume of a sphere is: \[ V = \frac{4}{3}\pi r^3 \] Differentiating both sides with respect to $r$: \[ \frac{dV}{dr} = \frac{4}{3}\pi (3r^2) = 4\pi r^2 \] At $r = 5$ cm: \[ \left. \frac{dV}{dr} \right|_{r=5} = 4\pi (5)^2 = 4\pi (25) = 100\pi \]
Step 4: Conclusion
The rate at which the volume increases with respect to the radius when $r = 5$ cm is $100\pi$ $\text{cm}^3/\text{cm}$.
Final Answer: (A)
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