A man 1.6 m high walks at the rate of $30 \mathrm{~m} / \mathrm{min}$ away from a lamp which is 4 m above ground. How fast is the man's shadow lengthening?

• A

  $22 \mathrm{~m} / \mathrm{min}$
• Correct answer
  B

  $20 \mathrm{~m} / \mathrm{min}$
• C

  $15 \mathrm{~m} / \mathrm{min}$
• D

  $25 \mathrm{~m} / \mathrm{min}$

Solution

Let $P Q=4 m$ be the height of pole and, $A B=1.6 \mathrm{~m} \quad$ 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 $P Q$ at some instant. Since, $\triangle P Q R$ and $\triangle A B R$ are similar,
$\begin{array}{ll}\text { we have, } \frac{P Q}{A B}=\frac{P R}{A R} \\ \Rightarrow & \frac{4}{1.6}=\frac{x+l}{l} \\ \Rightarrow & 2 x=3 l\end{array}$ $\Rightarrow \quad \frac{2 d x}{d t}=3 \cdot \frac{d l}{d t} \quad\left\{\right.$ given $\left.\frac{d x}{d t}=30 \mathrm{~m} / \mathrm{min}\right\}$ $\Rightarrow \quad \frac{d l}{d t}=\frac{2}{3} \times 30 \mathrm{~m} / \mathrm{min}=20 \mathrm{~m} / \mathrm{min}$.

Hence, the correct answer is option: 'B'.