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

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v/3

v/6

v/2

answer is A.

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Detailed Solution

$L_{1} L_{2} = street lamp$

$M_{1} M_{2} = man,$

$S = length of shadow of man from figure$

$\frac{3 h}{x + S} = \frac{h}{S}$

$3 S = x + S \Rightarrow 2 S = x$

$\Rightarrow 2 \frac{dS}{dt} = \frac{dx}{dt} \Rightarrow \frac{dS}{dt} = \frac{V_{Man}}{2} = \frac{v}{2}$

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