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8. 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.

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Given =  \frac{dV}{dt} = 900 \ cm^{3}/s
To find =   \frac{dr}{dt}   at r = 15 cm
Solution:-

Volume of sphere(V) =  \frac{4}{3}\pi r^{3}
\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})} {dr}.\frac{dr}{dt} = \frac{4}{3}\pi\times 3r^{2} \times \frac{dr}{dt}
                               
\frac{dV}{dt}= 4 \pi r^{2} \times \frac{dr}{dt}    
\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times(15)^{2}} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/s
Hence,  the rate at which the radius of the balloon increases when the radius is 15 cm is  \frac{1}{\pi} \ cm/s
 

Posted by

Gautam harsolia

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