Target Exam

CUET

Subject

Section B1

Chapter

Applications of Derivatives

Question:

A man of height 2 metres walks at a uniform speed of 5km/h away form a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases.

Options:

1.5 km/h

2.0 km/h

2.5 km/h

5.0 km/h

Correct Answer:

2.5 km/h

Explanation:

The correct answer is Option (3) → 2.5 km/h ##

In Fig, Let $AB$ be the lamp-post, the lamp being at the position $B$ and let $MN$ be the man at a particular time $t$ and let $AM = l$ metres. Then, $MS$ is the shadow of the man. Let $MS = s$ metres.

Note that $\quad \Delta MSN \sim \Delta ASB$

or $\quad \frac{MS}{AS} = \frac{MN}{AB}$

or $\quad AS = 3s$ (as $MN = 2$ and $AB = 6$ (given))

Thus $\quad AM = 3s - s = 2s$. But $AM = l$

So $\quad l = 2s$

Therefore $\quad \frac{dl}{dt} = 2 \frac{ds}{dt}$

Since $\frac{dl}{dt} = 5$ km/h. Hence, the length of the shadow increases at the rate $\frac{5}{2}$ km/h.