Ts Eamcet
Application Of Derivatives
TG EAPCET 2025 (Online) 3rd May Evening Shift
MCQ+1 / -02025
A man of 5 feet height is walking away from a light fixed at a height of 15 feet at the rate of of $K$ miles/hour. If the rate of increase of his shadow is $\frac{11}{5}$ feet $/ \mathrm{sec}$, then $K=($ Take 1 mile $=5280$ feet $)$
