A ladder, 5 meter long, standing oh 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

The correct option is 1/20 radian/sec.

Let us set up the coordinate system to analyze the motion of the ladder. Let the vertical wall be along the $y$-axis and the horizontal floor be along the $x$-axis.
Let $x$ represent the distance of the lower end of the ladder from the wall (along the floor), and let $y$ represent the height of the top of the ladder from the floor (along the wall) at any time $t$.
Let $\theta$ be the angle between the floor and the ladder.

The ladder has a constant length of $L=5\text{meters}$.
By using right-triangle trigonometry, we can relate $y$, $L$, and $\theta$:
$y=L\sin \left(\theta \right)=5\sin \left(\theta \right)$

We are given that the top of the ladder slides downwards at a rate of $10\text{cm/sec}$.
Converting this rate to meters per second to match the unit of length of the ladder:

The negative sign represents that the height $y$ is decreasing over time.

We want to find the rate of change of the angle, $\frac{d\theta }{dt}$, when the lower end of the ladder is $x=2\text{meters}$ from the wall.

First, let's find the value of $\cos \left(\theta \right)$ at this specific instant. From the geometry of the right triangle:
$\cos \left(\theta \right)=\frac{x}{L}=\frac{2}{5}$

Now, we differentiate the equation $y=5\sin \left(\theta \right)$ with respect to time $t$ using the chain rule:
$\frac{dy}{dt}=5\cos \left(\theta \right)·\frac{d\theta }{dt}$

Substitute the known values $\frac{dy}{dt}=-\frac{1}{10}$ and $\cos \left(\theta \right)=\frac{2}{5}$ into the differentiated equation:
$-\frac{1}{10}=5·\left(\frac{2}{5}\right)·\frac{d\theta }{dt}$

Simplifying the right side of the equation:
$-\frac{1}{10}=2·\frac{d\theta }{dt}$

Solving for $\frac{d\theta }{dt}$:
$\frac{d\theta }{dt}=-\frac{1}{20}\text{radian/sec}$

The negative sign confirms that the angle $\theta$ is decreasing. Therefore, the rate at which the angle is decreasing is $\frac{1}{20}\text{radian/sec}$.