A street lamp is fixed on a lamp-post at a height of 3.3 m from the ground. A boy 110 cm tall walks away from the base of this lamp post at a speed of 0.8 m/s. The length of the shadow of the boy after 4 seconds is:

1. 1.1 m

2. 1.6 m

3. 2.1 m

4. 2.6 m

Let AB = 3.3 m be the lamp post and QP = 1.1 m be the position of boy after 4 seconds.

AP is the distance moved in 4s at 0.8m/s = 4(0.8) = 3.2 m

PM is the length of the shadow of boy.

In triangle AMB and PMQ,

∠MAB = ∠MPQ = 90°

∠AMB = ∠PMQ [Common angle]

∴ ΔAMB ∼ ΔPMQ [By AA similarity].

Then ratios of corresponding sides of similar triangles are equal.

$\Rightarrow \dfrac{AM}{PM} = \dfrac{AB}{PQ} \\[1em] \Rightarrow \dfrac{AP + PM}{PM} = \dfrac{AB}{PQ} \\[1em] \Rightarrow \dfrac{3.2 + x}{x} = \dfrac{3.3}{1.1} \\[1em] \Rightarrow 3.2 + x = 3x \\[1em] \Rightarrow 3.2 = 2x \\[1em] \Rightarrow x = \dfrac{3.2}{2} \\[1em] \Rightarrow x = 1.6 \text{ m}.$

Hence, option 2 is the correct option.