CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The radius of a circle is increasing at the rate of 0.1 cm/s. When the radius of the circle is 5 cm, the rate of change of its area, is:

    A)  \[-\pi \,c{{m}^{2}}/s\]  

    B)         \[10\pi \,c{{m}^{2}}/s\]

    C)  \[0.1\pi \,c{{m}^{2}}/s\]              

    D)         \[5\pi \,c{{m}^{2}}/s\]

    E)  \[\pi \,c{{m}^{2}}/s\]

    Correct Answer: E

    Solution :

    Given that,\[\frac{dr}{dt}=0.1\text{ }cm/s\] \[\therefore \]Area,       \[A=\pi {{r}^{2}}\] On differentiating w.r.t. r, we get \[\Rightarrow \]               \[\frac{dA}{dt}=2\pi r\frac{dr}{dt}\] \[\therefore \] \[{{\left. \frac{dA}{dt} \right|}_{r=5}}=10\pi \times 0.1=\pi c{{m}^{2}}/s\]


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