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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    If the volume of a spherical ball is increasing at the rate of \[4\pi \,cc/\sec \]., then the rate of increase of its radius (in cm. / sec.), when the volume is 288\[\pi \]cc, is

    A)  \[\frac{1}{6}\]                                     

    B)  \[\frac{1}{9}\]

    C)  \[\frac{1}{24}\]                        

    D)  \[\frac{1}{36}\]  

    Correct Answer: D

    Solution :

    \[V=\frac{4}{3}\,\pi {{r}^{3}}\] Now, \[\frac{dV}{dt}=4\pi {{r}^{2}}\,\frac{dr}{dt}\] \[\Rightarrow \,\,4\pi =4\pi {{r}^{2}}\,\frac{dr}{rt}\] \[\therefore \,\,\frac{dr}{dt}\,=\frac{1}{{{r}^{2}}}\,=\frac{1}{36}\] As \[V=288\,\pi \Rightarrow \,\,r=6\]


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