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Q.
A man $1.6\, m$ high walks at the rate of $30 \,m / \min$ away from a lamp which is $4\, m$ above ground. How fast is the man's shadow lengthening?
ManipalManipal 2013
Solution:
Let $P Q=4\, m$ be the height of pole and, $A B=1.6 \,m$ be height of man.
Let the end of shadow and it is at a distance of $l$ from $A$ when the man is at a distance $x$ from $P Q$ at some instant.
Since, $\Delta P Q R$ and $\Delta A B R$ are similar,
we have, $\frac{P Q}{A B}=\frac{P R}{A R}$
$\Rightarrow \frac{4}{1.6}=\frac{x+l}{l}$
$\Rightarrow 2 x=3 l$
$\Rightarrow \frac{2 d x}{d t}=3 \cdot \frac{d l}{t d}\left\{\right.$ given $\left.\frac{d x}{d t}=30\, m / \min \right\}$
$\Rightarrow \frac{d l}{d t}=\frac{2}{3} \times 30\, m / \min =20\, m / \min$
