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

Answer 1

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 ${}^{\prime }{t}^{\prime }$ $AB=xcm$ and $BC=ycm$

So,

$\sin \theta =\frac{x}{500}$ and ,

$\cos \theta =\frac{y}{500}$

or, $x=500\sin \theta$

or, $y=500\cos \theta$

Also given that

$\frac{dx}{dt}=10cm/s$

implies that,

$500.\cos \theta .\frac{d\theta }{dt}=10$

implies that,

$\frac{d\theta }{dt}=\frac{1}{50\cos \theta }$

$\frac{d\theta }{dt}=\frac{1}{50.\frac{y}{500}}$

$=\frac{10}{y}$

$=\frac{10}{200}$

Hence the requires angle is

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