Question

A man of height ‘h’ is walking away from a street lamp with a constant speed ‘v’. The height of the street lamp is 3h. The rate at which the length of the man’s shadow is increasing when he is at a distance 10h from the base of the street lamp is:

• A. $\frac{v}{2}$
• B. $\frac{v}{3}$
• C. $2v$
• D. $\frac{v}{6}$

Answer 1

The correct option is A $\frac{v}{2}$



$tan\theta =\frac{3h}{10h+x}=\frac{h}{x}\Rightarrow 3hx=10h^2+hx2hx=10h^2\Rightarrow x=5h$
(length of the shadow at this instant)
Also, $\frac{k_1}{10h+x}=\frac{k_2}{x}{k}_{1}x=k_2(10h+x)$
Now $k_1\frac{dx}{dt}=k_2(d\frac{\left(10h\right)}{dt}+\frac{dx}{dt})\Rightarrow (k_1-x_2)\left(\frac{dx}{dt}\right)=k_{2}V\Rightarrow \frac{dx}{dt}=\frac{k_{2}V}{k_1-k_2}=\frac{hv}{3h-h}=\frac{hv}{2h}=\frac{V}{2}$
So, option (A)