A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of ice decreases is
(A) 1/36π
(B) 5/6π
(C) 1/9π
(D) 1/18π
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A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm³/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of ice decreases is
(A) 1/36π
(B) 5/6π
(C) 1/9π
(D) 1/18πq✓✓
Answered by
0
Answer:1/33π
Step-by-step explanation:
Volume of iron ball=4/3 πr³
=4/3 π1000
=4000π/3
Volume of ball including ice =4/3 πr³
=4/3 π3375
=13500π/3
Volume of ice=Total volume(including the ice ones)-Volume of iron ball
=13500π/3-4000π/3
=9500π/3
Decrease in thickness at cm/sec.=50*3/9500π
=3/190π
=1/33π
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