JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    A spherical balloon is filled with \[4500\pi \] cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of \[72\pi \] cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is:   AIEEE  Solved  Paper-2012

    A) \[\frac{9}{7}\]                                      

    B) \[\frac{7}{9}\]

    C) \[\frac{2}{9}\]                                      

    D) \[\frac{9}{2}\]

    Correct Answer: C

    Solution :

                 \[V=\frac{4}{3}\pi {{r}^{3}}\]                                          \[4500\pi =\frac{4\pi {{r}^{3}}}{3}\]              \[\frac{dV}{dt}=4\pi {{r}^{2}}\left( \frac{dr}{dt} \right)\]                   \[45\times 45\times 3={{r}^{3}}\]                                                                 \[r=15\,\,m\] after \[49\,\min =(4500-49.72)\pi =972\,\pi \,{{m}^{3}}\] \[972\pi =\frac{4}{3}\pi {{r}^{3}}\] \[{{r}^{3}}=3\times 243=3\times {{3}^{5}}\] \[r=9\] \[72\pi =4\pi \times 9\times 9\left( \frac{dr}{dt} \right)\] \[\frac{dr}{dt}=\left( \frac{2}{9} \right)\]


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