Practice free →
HomeMHT-CETMathematicsApplication of Derivatives › The volume of a sphere is increasing at $8\pi$ c…

The volume of a sphere is increasing at $8\pi$ cm³/s. The rate of change of surface area when the radius is 2 cm is:

A$2\pi$ cm²/s
B$4\pi$ cm²/s
C$8\pi$ cm²/s
D$16\pi$ cm²/s
Answer & Solution
Correct answer: C. $8\pi$ cm²/s
$V = (4/3)\pi r^3$ ⇒ $dV/dt = 4\pi r^2 \cdot dr/dt = 8\pi$. At $r = 2$: $4\pi(4)\cdot dr/dt = 8\pi$ ⇒ $dr/dt = 1/2$. $S = 4\pi r^2$ ⇒ $dS/dt = 8\pi r \cdot dr/dt = 8\pi(2)(1/2) = 8\pi$ cm²/s.
Solve this in the app — MHT-CET practice & 24k+ MCQs →
Related questions