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The Volume of a Sphere is Increasing at 3 Cm3/Sec. the Rate at Which the Radius Increases When Radius is 2 Cm, is (A) 3 32 π C M / Sec

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Question

The volume of a sphere is increasing at 3 cm3/sec. The rate at which the radius increases when radius is 2 cm, is

Options

  • \[\frac{3}{32\pi}cm/\sec\]

  • \[\frac{3}{16\pi}cm/\sec\]

  • \[\frac{3}{48\pi}cm/\sec\]

  • \[\frac{1}{24\pi}cm/\sec\]

MCQ
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Solution

 \[\frac{3}{16\pi}cm/\sec\]

\[\text { Let  r be the radius and V be the volume of the sphere at any time t.Then },\]

\[V=\frac{4}{3} \pi r^3 \]

\[\Rightarrow\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\]

\[\Rightarrow\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}\]

\[\Rightarrow\frac{dr}{dt}=\frac{3}{4\pi \left( 2 \right)^2}\]

\[\Rightarrow\frac{dr}{dt}=\frac{3}{16\pi}cm/sec\]

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Chapter 12: Derivative as a Rate Measurer - Exercise 13.4 [Page 25]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 12 Derivative as a Rate Measurer
Exercise 13.4 | Q 12 | Page 25

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