a man 2 m high walks at a uniform speed 5m/hr away from a lamp from 6 m high .The rate at which the length of his shadow increases is

AB = height of lamp

CE = height of man

BC = y = distance b/w lamp n man

DC = x = length of shadow

Θ = angle ADB

now

tanΘ = 2/x (in smaller triangle)

tanΘ = 6/(x+y) .......[in larger triangle]

since both are same

so , 2/x = 6/(x+y)

=> 2x = y

x = y/2

now differentiating both sides w.r.to time ,t

dx/dt = (1/2)*dy/dt

dy/dt = 5m/hr.(given)

so, dx/dt = 2.5 m/hr.