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
B
C
D

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 $$= \dfrac{1}{{20}}rad/s$$Maths

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