A spherical balloon is being inflated so that its volume increase uniformly at the rate of 40 cm3 /minute. The rate of increase in its surface area when the radius is 8 cm, is 

A spherical balloon is being inflated so that its volume increase uniformly at the rate of 40 cm3 /minute. The rate of increase in its surface area when the radius is 8 cm, is 

  1. A

    10cm2 /minute 

  2. B

    20cm2 /minute 

  3. C

    40 cm2 /minute 

  4. D

    none of these

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    Solution:

    Let V be the volume, S be the surface area and r be the radius of the balloon. Then,  

    V=43πr3 and S=4πr2

     dVdt=4πr2drdt and  dSdt=8πrdrdt

    When r=8cm and dVdt=40cm3/ minute  

    dVdt=4π2drdt40=4π×82drdtdrdt=532π

    putting , r=8,drdt=532π in dSdt=8πrdrdt 

    dSdt=8π×8×532π=10cm2/ minute 

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