Question:
The radius of a spherical soap bubble is increasing at a constant rate of \(0.5\) cm/sec. At the instant when the radius is \(5\) cm, the rate at which its volume is increasing is...
The radius of a spherical soap bubble is increasing at a constant rate of \(0.5\) cm/sec. At the instant when the radius is \(5\) cm, the rate at which its volume is increasing is...
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Differentiate the volume of a sphere with respect to time and substitute the given values.
Updated On: Nov 12, 2026
- \(25\pi\) cm\({}^{3}\)/sec
- \(50\pi\) cm\({}^{3}\)/sec
- \(100\pi\) cm\({}^{3}\)/sec
- \(10\pi\) cm\({}^{3}\)/sec
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The volume of a sphere of radius \(r\) is \(V=\frac{4}{3}\pi r^{3}\).
Step 2: Differentiate with respect to time
\(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
Step 3: Substitute the values
\(r=5\) cm and \(\frac{dr}{dt}=0.5\) cm/s.
\(\frac{dV}{dt}=4\pi(25)(0.5)\).
Step 4: Calculate
\(\frac{dV}{dt}=50\pi\ \text{cm}^{3}/\text{s}\).
Final Answer:
The volume is increasing at 50 pi cubic centimetres per second. \[ \boxed{\text{(B) }50\pi \text{ cm}^{3}/\text{sec}} \]
The volume of a sphere of radius \(r\) is \(V=\frac{4}{3}\pi r^{3}\).
Step 2: Differentiate with respect to time
\(\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}\).
Step 3: Substitute the values
\(r=5\) cm and \(\frac{dr}{dt}=0.5\) cm/s.
\(\frac{dV}{dt}=4\pi(25)(0.5)\).
Step 4: Calculate
\(\frac{dV}{dt}=50\pi\ \text{cm}^{3}/\text{s}\).
Final Answer:
The volume is increasing at 50 pi cubic centimetres per second. \[ \boxed{\text{(B) }50\pi \text{ cm}^{3}/\text{sec}} \]
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