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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    The volume of a spherical balloon is increasing at the rate of 40 cubic centrimetre per minute. The rate of change of the surface of the balloon at the instant when its radius is 8 centimetre, is     [Roorkee 1983]

    A) \[\frac{5}{2}\] sq cm/min

    B) 5 sq cm/min

    C) 10 sq cm/min

    D) 20 sq cm/min

    Correct Answer: C

    Solution :

    • Here \[f'(x)=2x\log x+x\] and \[S=4\pi {{r}^{2}}\]           
    • Þ \[\frac{dV}{dt}=4\pi {{r}^{2}}\frac{dr}{dt}\] Þ \[f''(1)=3+2{{\log }_{e}}1\]           
    • \[\therefore \frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi \times 8\times \frac{5}{32\pi }=10\].


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