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A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases is
- 1/(18π)
- 1/(54π)
- 1/(36π)
- 5/(6π)
Correct answer: 1/(18π)
Solution
The volume of the ice layer can be expressed in terms of its thickness, and as it melts, the rate of change of volume relates to the rate of decrease in thickness. By applying the formula for the volume of a sphere and differentiating with respect to time, we find that the rate at which the thickness decreases when the ice is 5 cm thick is indeed 1/(18π) cm/min.
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