Question

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

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

Answer 1

The correct option is A $\frac{1}{20}\text{radian/sec}$


Let the angle between floor and the ladder be $\theta$
Let at any time ‘𝑡’,
$\text{AB = x cm and BC = y cm}$.
$\therefore \text{sin}\theta =\frac{\text{x}}{500}$
$\Rightarrow \text{x}=500\text{sin}\theta$ …(1)
$\text{cos}\theta =\frac{\text{y}}{500}$
$\Rightarrow \text{y}=500\text{cos}\theta$ …(2)
Also, it is given that
$\frac{\text{dx}}{\text{dt}}=10\text{cm/s}$
Differentiating the eq. (1) with respect to 𝑡,
$\Rightarrow 500\text{cos}\theta \frac{\text{d}\theta }{dt}=\frac{\text{dx}}{\text{dt}}$
$\Rightarrow 500\text{cos}\theta \frac{\text{d}\theta }{\text{dt}}=10$
Given : $\frac{\text{dx}}{\text{dt}}=10\text{cm/s}$
$\Rightarrow \frac{\text{d}\theta }{\text{dt}}=\frac{1}{50\text{cos}\theta }$
For $\text{y = 2m = 200cm}$,
$\Rightarrow \frac{\text{d}\theta }{\text{dt}}=\frac{1}{50.\frac{\text{y}}{500}}$ [From eq. 2]
$\Rightarrow \frac{\text{d}\theta }{\text{dt}}=\frac{10}{\text{y}}$
$\Rightarrow \frac{\text{d}\theta }{\text{dt}}=\frac{10}{200}=\frac{1}{20}\text{radian/sec}$.
Hence, the rate at which the angle between the floor and the ladder is decreasing is $\frac{1}{20}\text{radian/sec}$.
Hence, option B is correct