question_answer

A man of 2 m height walks at a uniform speed of 6 km/h away from a lamp post of 6 m height. The rate at which the length of his shadow increases is

A)  2 km/h     

B)  1 km/h

C)  3 km/h     

D)  6 km/h

E)  3/2 km/h

Correct Answer: C

Solution :

In\[\Delta ADC,\]                 \[\tan \theta =\frac{6}{x+y}\] and in\[\Delta BCE,\] \[\therefore \]  \[\frac{2}{x}=\frac{6}{x+y}\Rightarrow x+y=3x\] \[\Rightarrow \]               \[y=2x\] On differentiating w.r.t. t, we get                 \[\frac{dy}{dt}=2\frac{dx}{dt}\] \[\Rightarrow \]               \[6=2\frac{dx}{dt}\]        \[\left( \because \frac{dy}{dt}=6given \right)\] \[\Rightarrow \]               \[\frac{dx}{dt}=3\,km/h\]