SELECT * FROM question_mgmt as q WHERE id=1174 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 on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
The volume of a sphere (V) with radius (r) is given by,
\begin{align} V=\frac{4}{3}\pi r^2\end{align}
∴Rate of change of volume (V) with respect to time (t) is given by,
\begin{align} \frac{dV}{dt} =\frac{dV}{dr}.\frac{dr}{dt}\;\;\;[By\; Chain\; Rule]\end{align}
\begin{align} =\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right).\frac{dr}{dt}\end{align}
\begin{align} =4\pi r^2.\frac{dr}{dt}\end{align}
It is given that
\begin{align} \frac{dV}{dt}=900\; cm^3/s\end{align}
\begin{align} \therefore 900=4\pi r^2.\frac{dr}{dt}\end{align}
\begin{align} \Rightarrow \frac{dr}{dt}=\frac{900}{4\pi r^2}=\frac{225}{\pi r^2}\end{align}
Therefore, when radius = 15 cm,
\begin{align} \frac{dr}{dt}=\frac{225}{\pi (15)^2}=\frac{1}{\pi }\end{align}
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is
\begin{align} \frac{1}{\pi }\end{align}
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