SELECT * FROM question_mgmt as q WHERE id=1175 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=6 AND chapterId=93 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
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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