Choose the correct answer from the given four options: A ladder, … - Vidyadip

MCQ

Choose the correct answer from the given four options:
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 10cm/ 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
$\frac{1}{10}\text{radian/}\sec$
✓
$\frac{1}{20}\text{radian/}\sec$
C
$20\text{ radian/}\sec$
D
$10\text{ radian/}\sec$

✓

Answer

Correct option: B.

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

Let the angle between floor and the ladder be $\theta.$
Let at any time 't' AB = x cm and BC = y cm

$\therefore\ \sin\theta=\frac{\text{x}}{500}$ and $\cos\theta=\frac{\text{y}}{500}$
$\Rightarrow\ \text{x}=500\sin\theta$ and $\text{y}=500\cos\theta$
Also it is given that $\frac{\text{dx}}{\text{dt}}=10\text{cm/s}$
$\Rightarrow\ 500\cdot\cos\theta\cdot\frac{\text{d}\theta}{\text{dt}}=10\text{cm}/ \text{s}$
$\Rightarrow\ \frac{\text{d}\theta}{\text{dt}}=\frac{10}{500\cos\theta}=\frac{1}{50\cos\theta}$
For $\text{y}=2\text{m}=20\text{cm},$
$\frac{\text{d}\theta}{\text{dt}}=\frac{1}{50\cdot\frac{\text{y}}{500}}=\frac{10}{\text{y}}$ $=\frac{10}{200}=\frac{1}{20}\text{radian/}\sec$

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