A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is :

Application of Derivatives PYQ — JEE Main 2019 · 12 Apr · Shift 1 · Q36 with Solution

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Correct answer: D

Let:

Since the ladder length is constant and equal to $2$ m,

$x^2 + y^2 = 2^2 = 4$

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

So,

$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$

$\frac{dx}{dt} = -\frac{y}{x}\frac{dy}{dt}$

The top slides down, so

$\frac{dy}{dt} = -25\text{ cm/s}$

We are asked for $\frac{dx}{dt}$ when the top is $1$ m above the ground, i.e. $y=1$ m.

From

$x^2 + y^2 = 4$

$x^2 + 1^2 = 4$

$x^2 = 3$

$x = \sqrt{3}\text{ m}$

$\frac{dx}{dt} = -\frac{1}{\sqrt{3}}(-25)$

$\frac{dx}{dt} = \frac{25}{\sqrt{3}}\text{ cm/s}$

This corresponds to:

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

So the correct option is D.