JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    A spherical balloon is being inflated at the rate of 35cc/min. The rate of increase in the surface area (in\[{{\operatorname{cm}}^{2}}/\min .\]) of the balloon when its diameter is 14 cm, is     JEE Main Online Paper ( Held On 25  April 2013 )

    A)                 10          

    B)                 \[\sqrt{10}\]

    C)                 \[100\] 

    D)                                        \[10\sqrt{10}\]

    Correct Answer: A

    Solution :

                    Volume of sphere \[V=\frac{4}{3}\pi {{r}^{3}}\] \[\frac{dV}{dt}=\frac{4}{3}.\pi .3{{r}^{2}}.\frac{dr}{dt}\] \[35=4\pi {{r}^{2}}.\frac{dr}{dt}\]or\[\frac{dr}{dt}=\frac{35}{4\pi {{r}^{2}}}\] Surface area of sphere \[=S=4\pi {{r}^{2}}\] \[\frac{dS}{dt}=4\pi \times 2r\times \frac{dr}{dt}=8\pi r.\frac{dr}{dt}\] \[\frac{dS}{dt}=\frac{70}{r}\]                      (By using (1)) Now, diameter =14 cm, r = 7\[\therefore \]\[\frac{dS}{dt}=10\]


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