#### 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

Let 5m/500cm be the length of the ladder, which is the hypotenuse of the right triangle formed in the above figure.

Now let the angle between the ladder and the floor be β, so

$\\\sin \beta=\frac{\text { opposite side }}{\text { hypotenuse }} \\\\\Rightarrow \sin \beta=\frac{\mathrm{x}}{500}$
Differentiate both sides with respect to time t and get
$\\\Rightarrow \frac{\mathrm{d}(\sin \beta)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\frac{\mathrm{x}}{500}\right)}{\mathrm{dt}}$
Apply derivatives and get
$\\\Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{500} \frac{\mathrm{dx}}{\mathrm{dt}}$

Now, the top of the ladder slides downwards at the rate of
$10 \mathrm{~cm} / sec$

$\frac{d x}{d t}=10 \mathrm{~cm} / \mathrm{sec}$
So the equation is
$\\ \Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{500} \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{500} \times 10 \\\Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{50} \ldots \ldots (ii)$
Now,
$\cos \beta=\frac{\text { adacent side }}{\text { hypotenuse }}$

\begin{aligned} &\text { Substitute (iii) in (ii) and get }\\ &\Rightarrow \cos \beta \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{50}\\ &\Rightarrow \frac{y}{500} \times \frac{d \beta}{d t}=\frac{1}{50}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{1}{50} \times \frac{500}{\mathrm{y}}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}=\frac{10}{\mathrm{y}}\\ &\begin{array}{l} \text { Now when } y=200 \mathrm{~cm} \text { , we get } \end{array}\\ &\Rightarrow \frac{\mathrm {d \beta}}{\mathrm{dt}}_{\mathrm{y}=200}=\frac{10}{\mathrm{y}}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}_{\mathrm{y}=200}=\frac{10}{200}\\ &\Rightarrow \frac{\mathrm{d} \beta}{\mathrm{dt}}_{y=200}=\frac{1}{20} \end{aligned}

Hence the rate at which the angle between the floor and the ladder is decreasing is $\frac{1}{20}$ radian/sec

So the correct answer is option B.