A ladder of length 13m has one end resting against a vertical wall and the other on the ground. If the lower end moves away from the wall at a speed 2m/min then the speed(in m/min) at which upper end falls when the bottom is 5m away from the wall is

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5/6

$1 / \sqrt{2}$

6/5

12/5

answer is C.

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Detailed Solution

$\begin{array}{l} x^{2} + y^{2} = (13)^{2} \\ \Rightarrow 2 x \cdot \frac{d x}{d t} + 2 y \cdot \frac{d y}{d t} = 0 \\ \Rightarrow 5 \times 2 + 12 \times \frac{d y}{d t} = 0 \\ [at x = 5 y, y = \sqrt{13^{2} - 5^{2}} = 12] \\ \Rightarrow \frac{d y}{d t} = \frac{- 5}{2} \end{array}$
The speed of by which upper end falls is $\frac{5}{6}$

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