Question

A ladder, $5$ meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of $10 \,cm/sec$, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is $2$ metres from the wall is

• A. $\frac{1}{10}$ radian/sec
• B. $\frac{1}{20}$ radian/sec
• C. $20$ radian/sec
• D. $10$ radian/sec

Answer 1

Answer: (B)

$\frac{1}{20}$ radian/sec

Answer 2

Given $\frac{dy}{dt} =10\, cm/s$, $x = 2m = 200\, cm$ and Length of ladder $= 5\, m = 500 \,cm$ $\Rightarrow$ Let $?ACB = \theta$ Now $sin \,\theta = \frac{y}{500}$ Diff w.r.t. $t$, we get $cos\,\theta \frac{d\theta}{dt} = \frac{1}{500} \frac{dy}{dt}$ $\Rightarrow \frac{200}{500} \times \frac{d\theta }{dt}$ $= \frac{1}{500} \times 10$ $\Rightarrow \frac{d\theta }{dt} = \frac{10}{200}$ $\Rightarrow \frac{d\theta }{dt} = \frac{1}{20}$ rad/s.