Question

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?

• A. $22$ m/min
• B. $20$ m/min
• C. $15$ m/min
• D. $25$ m/min

Answer 1

Answer: (B)

$20$ m/min

Let $PQ=4$ m be the height of pole and $AB=1.6$ m be height of man.

Let the end of shadow is $R$ and it is at a distance of $l$ from $A$ when the man is at a distance $x$ from $PQ$ at some instant.

Since, $\Delta PQR$ and $\Delta ABR$ are similar.
We have, $\frac{PQ}{AB}=\frac{PR}{AR}$
$\Rightarrow \frac{4}{1.6}=\frac{x+l}{l}$
$\Rightarrow 2x=3l$
$\Rightarrow \frac{2dx}{dt}=3\cdot \frac{dl}{dt}$ ....{given $\frac{dx}{dt}=30$ m/min}
$\Rightarrow \frac{dl}{dt}=\frac{2}{3}\times 30=20$ m/min