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Read MoreWelcome to the NCERT Solutions for Class 12 Mathematics - Chapter Application of Derivatives. This page offers a step-by-step solution to the specific question from Exercise 1, Question 9: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
The volume of a sphere (V) with radius (r) is given by
\begin{align} V=\frac{4}{3 }\pi r^3\end{align}
Rate of change of volume (V) with respect to its radius (r) is given by,
\begin{align} \frac{dV}{dr }=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi \left(3r^2\right)=4\pi r^2\end{align}
Therefore, when radius = 10 cm,
\begin{align} \frac{dV}{dr }=4\pi(10)^2=400\pi\end{align}
Hence, the volume of the balloon is increasing at the rate of 400π cm3/s.
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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 9: A balloon, which always remains spherical has a variable radius. Find the rate at which its volume i....
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