Application of Derivatives 2 Question 1
####1. A spherical iron ball of radius $10 \mathrm{~cm}$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \mathrm{~cm}^{3} / \mathrm{min}$. When the thickness of the ice is $5 \mathrm{~cm}$, then the rate at which the thickness (in $\mathrm{cm} / \mathrm{min}$ ) of the ice decreases, is
(2019 Main, 10 April II)
(a) $\frac{1}{9 \pi}$
(b) $\frac{1}{18 \pi}$
(c) $\frac{1}{36 \pi}$
(d) $\frac{5}{6 \pi}$
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Answer:
(b)
Solution:
- Let the thickness of layer of ice is $x \mathrm{~cm}$, the volume of spherical ball (only ice layer) is
$ V=\frac{4}{3} \pi\left[(10+x)^{3}-10^{3}\right] $
On differentiating Eq. (i) w.r.t. ’ $t$ ‘, we get
$ \frac{d V}{d t}=\frac{4}{3} \pi\left(3(10+x)^{2}\right) \frac{d x}{d t}=-50 $
[-ve sign indicate that volume is decreasing as time passes].
$\Rightarrow \quad 4 \pi(10+x)^{2} \frac{d x}{d t}=-50$
At $x=5 \mathrm{~cm}$
$ \frac{d x}{d t}\left[4 \pi(10+5)^{2}\right]=-50 $
$\Rightarrow \quad \frac{d x}{d t}=-\frac{50}{225(4 \pi)}=-\frac{1}{9(2 \pi)}=-\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$
So, the thickness of the ice decreases at the rate of $\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$.
