Question

A man of height $2m$ walks at a uniform speed of $7m/min$ away from a lamp post of height $9m$. The rate $(m/min)$ at which the length of his shadow increases is

• A. $2$
• B. $\frac{5}{2}$
• C. $3$
• D. $\frac{7}{2}$

Answer 1

Answer: (A)

Let $AB$ be the lamp-post and $PQ$ the man, $CP$
be his shadow at time $t$.
Let $AP=x,PC=y$, Also $AB=9m,PQ=2m$
Now, $\Delta CAB$ and $\Delta CPQ$ are equiangular and hence similar.

$\therefore \frac{PC}{AC}=\frac{PQ}{AB}$
$\Rightarrow \frac{y}{x+y}=\frac{2}{9}$
$\Rightarrow 9y=2x+2y$
$\Rightarrow 7y=2x$
$\Rightarrow x=\frac{7}{2}y$
$\Rightarrow \frac{dx}{dt}=\frac{7}{2}\frac{dy}{dt}($ differentiating w.r.t. $t)$
But $\frac{dx}{dt}=7m/min$
$\therefore 7=\frac{7}{2}\frac{dy}{dt}$
$\Rightarrow \frac{dy}{dt}=2m/min$
$\therefore$ Length of shadow is increases at $2m/min$.