Question

A man of height $2 \,m$ walks at a uniform speed of $7 \,m / min$ away from a lamp post of height $9 \,m$. 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 $A B$ be the lamp-post and $P Q$ the man, $C P$
be his shadow at time $t$.
Let $A P=x, P C=y$, Also $A B=9 \,m , P Q=2 \,m$
Now, $\Delta C A B$ and $\Delta C P Q$ are equiangular and hence similar.

$\therefore \frac{P C}{A C}=\frac{P Q}{A B}$
$\Rightarrow \frac{y}{x+y}=\frac{2}{9}$
$\Rightarrow 9\, y=2 x+2 y$
$\Rightarrow 7 \,y=2 x$
$\Rightarrow x=\frac{7}{2} y$
$\Rightarrow \frac{d x}{d t}=\frac{7}{2} \frac{d y}{d t}($ differentiating w.r.t. $t)$
But $\frac{d x}{d t}=7 \,m / min$
$\therefore 7=\frac{7}{2} \frac{d y}{d t}$
$\Rightarrow \frac{d y}{d t}=2 \,m / min$
$\therefore $ Length of shadow is increases at $2 \,m / min$.