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Question

A ladder, 5 meter long, standing on a horizontal floor, leans against 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
110radian/sec
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B
120radian/sec
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C
20radian/sec
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D
10radian/sec
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Solution

The correct option is B $$\frac { 1 }{ 20 } radian/sec $$
The required problem be,
Let, angle between floor and the ladder be $$\theta$$ 
Let at any time $$'t'$$ $$AB = x \,cm$$ and $$BC = y \,cm$$
So,
$$\sin \theta  = \dfrac{x}{{500}}$$ and ,
$$\cos \theta  = \dfrac{y}{{500}}$$
or, $$ x = 500 \sin \theta $$
or, $$ y = 500 \cos \theta $$
Also given that
$$\dfrac{{dx}}{{dt}} = 10\,cm/s$$
implies that,
$$500.\cos \theta .\dfrac{{d\theta }}{{dt}} = 10$$
implies that,
$$\dfrac{{d\theta }}{{dt}} = \dfrac{1}{{50\cos \theta }}$$
$$\dfrac{{d\theta }}{{dt}} = \dfrac{1}{{50.\dfrac{y}{{500}}}}$$

$$ = \dfrac{{10}}{y}$$

$$ = \dfrac{{10}}{{200}}$$

Hence the requires angle is 

$$ = \dfrac{1}{{20}}rad/s$$


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