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A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness than melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is ?

sudhanshu , 11 Years ago
Grade 12
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
Let the thickness of ice to be ‘a’ at any instant ‘t’.
Volume ‘V’:
V = \frac{4}{3}\pi (10+a)^{3}
\frac{\partial V}{\partial t} = 4\pi (10+a)^{2}.\frac{\partial a}{\partial t} = 50 \frac{cm^{3}}{min}
4\pi (10+5)^{2}.\frac{\partial a}{\partial t} = 50 \frac{cm^{3}}{min}
\frac{\partial a}{\partial t} = \frac{1}{18\pi } \frac{cm}{min}
Thanks & Regards
Jitender Singh
IIT Delhi
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