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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2003

  • question_answer
    The rate of change of the surface area of the sphere of radius r when the radius is increasing at the rate of 2 cm/sec is proportional to :

    A)  \[\frac{1}{{{r}^{2}}}\]                                   

    B)  \[\frac{1}{r}\]

    C)  \[{{r}^{2}}\]                                      

    D)  \[r\]

    Correct Answer: C

    Solution :

    Here we have                 \[\frac{dr}{dt}=2\,cm/\sec \] and surface area of sphere is given by                 \[A=\frac{4}{3}\pi {{r}^{3}}\] \[\therefore \]  \[\frac{dA}{dt}=3\times \frac{4}{3}\pi {{r}^{2}}\frac{dr}{dt}\]                 \[=4\pi {{r}^{2}}(2)\]                 \[=8\pi {{r}^{2}}\] \[\therefore \]  \[\frac{dA}{dt}\propto {{r}^{2}}\]


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