Let thickness =xcm
Total volume V=34π(10+x)3
dtdV=4π(10+x)2dtdx…..(i)
Given dtdV=50cm3/min
At x=5cm
50=4π(10+5)2dtdx
dtdx=18π1cm/min
JEE Main 2020 — Mathematics Calculus
A spherical iron ball of 10cm radius is coated with a layer of ice of uniform thickness that melts at a rate of 50cm3/min . When the thickness of ice is 5cm , then the rate (in cm/min .) at which of the thickness of ice decreases, is:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
6π5
54π1
36π1
18π1
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