A 2 m ladder leans against a vertical wall. | vedclass

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(D) Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the height of the top of the ladder from the ground. The length of th

Vedclass · Accepted Answer

$\frac{25}{\sqrt{3}}$

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(D) Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the height of the top of the ladder from the ground. The length of the ladder is $L = 2 \ m = 200 \ cm$.By Pythagoras theorem,we have $x^2 + y^2 = 200^2$.Differentiating with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.Given that the top slides down at $25 \ cm/sec$,we have $\frac{dy}{dt} = -25 \ cm/sec$.When $y = 1 \ m = 100 \ cm$,we find $x$ using $x^2 + 100^2 = 200^2$,so $x^2 = 40000 - 10000 = 30000$,which gives $x = \sqrt{30000} = 100\sqrt{3} \ cm$.Substituting these values into the differentiated equation: $(100\sqrt{3}) \frac{dx}{dt} + (100)(-25) = 0$.$(100\sqrt{3}) \frac{dx}{dt} = 2500$.$\frac{dx}{dt} = \frac{2500}{100\sqrt{3}} = \frac{25}{\sqrt{3}} \ cm/sec$.

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