Question:

A spherical iron ball 10cm 10 \, \text{cm} in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50cm3/min 50 \, \text{cm}^3/\text{min} . When the thickness of ice is 15cm 15 \, \text{cm} , then the rate at which the thickness of ice decreases is:

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For problems involving spherical volumes, note that the rate of change of volume depends on the square of the radius, especially when the shape is uniformly shrinking or melting.
Updated On: Jan 22, 2025
  • 56πcm/min \frac{5}{6\pi} \, \text{cm/min}
  • 154πcm/min \frac{1}{54\pi} \, \text{cm/min}
  • 118πcm/min \frac{1}{18\pi} \, \text{cm/min}
  • 136πcm/min \frac{1}{36\pi} \, \text{cm/min}
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The Correct Option is C

Solution and Explanation

Given: dVdt=50cm3/min, \text{Given: } \frac{dV}{dt} = 50 \, \text{cm}^3/\text{min},
ddt(43πr3)=50, \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 50,
3r2drdt=1504πdrdt=504πr2. 3r^2 \frac{dr}{dt} = \frac{150}{4\pi} \quad \Rightarrow \quad \frac{dr}{dt} = \frac{50}{4\pi r^2}.
\text{Substitute } r = 15: (drdt)r=15=504π×225=118πcm/min. \left( \frac{dr}{dt} \right)_{r = 15} = \frac{50}{4\pi \times 225} = \frac{1}{18\pi} \, \text{cm/min}. Final Answer: 118πcm/min \boxed{\frac{1}{18\pi} \, \text{cm/min}}
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