A spherical iron ball of rdius 10cm is coated with a layer of ice of uniform 

thickness that melts at a rate of  50 \: cm^{3}/min. When the thickness of 

the ice is 5 cm , then the rate at which the thickness ( in cm / min ) of the ice

decreases, is :

  • Option 1)

    \frac{1}{18\pi}

  • Option 2)

    \frac{1}{36\pi}

  • Option 3)

    \frac{5}{6\pi}

  • Option 4)

    \frac{1}{9\pi}

 

Answers (1)

Volume of ice , V=\frac{4}{3}\pi(R+x)^{3}-\frac{4}{3}\pi R^{3}

                         \frac{dV}{dt}=4\pi (R+x)^{2}\frac{dx}{dt}=50

                          \frac{dx}{dt}=\frac{50}{4\pi (R+x)^{2}}

                   at x = 5,

                         \frac{dx}{dt}=\frac{50}{4\pi (10+5)^{2}}=\frac{1}{18\pi}cm/min

So, option (1) is correct. 

                      


Option 1)

\frac{1}{18\pi}

Option 2)

\frac{1}{36\pi}

Option 3)

\frac{5}{6\pi}

Option 4)

\frac{1}{9\pi}

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