Q97 / 180 · KCET 2015 · Mathematics
A balloon which always remains spherical is being inflated by pumping in \( 10 \) cubic centimeters
of gas per second. Find the rate at which the radius of the balloon is increasing when the
radius is \( 15 \mathrm{cms} \).
of gas per second. Find the rate at which the radius of the balloon is increasing when the
radius is \( 15 \mathrm{cms} \).
- Correct answerA\( \frac{1}{90 \Pi} \mathrm{cm} s^{-1} \)
- B\( \frac{1}{9 \Pi} \mathrm{cm} s^{-1} \)
- C\( \frac{1}{30 \Pi} \mathrm{cm} s^{-1} \)
- D\( \frac{1}{\Pi} \mathrm{cm} s^{-1} \)
Solution
As we know that, the volume of sphere is given by
\[ V=\frac{4}{3} \Pi r^{3} \]
\[ \begin{array}{l} \text { So, } \frac{d V}{d t}=4 \Pi r^{2} \frac{d r}{d t} \\ \text { Here } \frac{d V}{d t}=10 \mathrm{~cm}^{3} \mathrm{~s}^{-1} \text { and } r=15 \mathrm{~cm} \text { we get } \\ 10=4 \Pi(15)^{2} \frac{d r}{d t} \\ \Rightarrow \frac{d r}{d t}=\frac{10}{4 \Pi\left(15^{2}\right)}=\frac{1}{90 \Pi} \end{array} \]
\[ V=\frac{4}{3} \Pi r^{3} \]
\[ \begin{array}{l} \text { So, } \frac{d V}{d t}=4 \Pi r^{2} \frac{d r}{d t} \\ \text { Here } \frac{d V}{d t}=10 \mathrm{~cm}^{3} \mathrm{~s}^{-1} \text { and } r=15 \mathrm{~cm} \text { we get } \\ 10=4 \Pi(15)^{2} \frac{d r}{d t} \\ \Rightarrow \frac{d r}{d t}=\frac{10}{4 \Pi\left(15^{2}\right)}=\frac{1}{90 \Pi} \end{array} \]
Hence, the correct answer is option: 'A'.