Q. A man of 2m 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 increase is
A) 2km/h
B) 1km/h
C) 3km/h
D) 6km/h
Explanation
(c)
$In\ \mathrm{\Delta}ADC\ ,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tan{\theta = \frac{6}{x + y}}$
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And in $\mathrm{\Delta}BCE$, $\tan{\theta = \frac{2}{x}}$
$\therefore\ \frac{2}{x} = \frac{6}{x + y\ } \Rightarrow y = 2x$
On differentiating w. r. t. $t,$we get
$\frac{dy}{dt} = 2\frac{dx}{dt} \Rightarrow \frac{dx}{dt} = 3km/h$ $\left\lbrack \because\frac{dy}{dt} = 6,\ given \right\rbrack\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.