Question

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

$fraction numerator 1 over denominator 36 space straight pi end fraction space cm divided by min$
$fraction numerator 1 over denominator 18 space straight pi end fraction space cm divided by min$
$fraction numerator 1 over denominator 54 space straight pi end fraction space cm divided by min$
$fraction numerator 5 over denominator 6 space straight pi end fraction space cm divided by min$

Solution

fraction numerator 1 over denominator 18 space straight pi end fraction space cm divided by min

dv over dt space equals space 50 4 πr squared space dr over dt space equals space 50 rightwards double arrow space dr over dt space equals space fraction numerator 50 over denominator 4 straight pi space left parenthesis 15 right parenthesis squared end fraction space where space straight r space equals space 15 space equals space fraction numerator 1 over denominator 16 space straight pi end fraction

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