JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 2)
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 the ice
decreases, is :
$${5 \over {6\pi }}$$
$${1 \over {9\pi }}$$
$${1 \over {36\pi }}$$
$${1 \over {18\pi }}$$
Explanation
_10th_April_Evening_Slot_en_2_1.png)
$$V = {4 \over 3}\pi \left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$$
$$ \Rightarrow {{dV} \over {dt}} = 4\pi {(10 + h)^2}{{dh} \over {dt}}$$
$$ \Rightarrow - 50 = 4\pi {\left( {10 + 5} \right)^2}{{dh} \over {dt}}$$
$$ \Rightarrow {{dh} \over {dt}} = - {1 \over {18\pi }}$$
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