The rate of increase of length of the shadow of a man 2 metres height, due to a lamp at 10 metres height, when he is moving away from it at the rate of 2 m/sec, is

$\frac{1}{2}$ m/sec

Let CE = y be the length of the shadow of the man at time t when he is x metres away from the lamp-post.

It is given that $\frac{d x}{d t}$ = 2 m/sec

Clearly, ΔABE ~ ΔCDE

$\Rightarrow \frac{A B}{C D}=\frac{A E}{C E}$

$\Rightarrow \frac{10}{2}=\frac{x+y}{y}$

$\Rightarrow 4 y=x$

$\Rightarrow 4 \frac{d y}{d t}=\frac{d x}{d t}$

$\Rightarrow \frac{d y}{d t}=\frac{2}{4}=\frac{1}{2}$ m/sec