🎥 Video solution / Text Solution of this question is given below:
Let
$x$ = distance of the bottom of the ladder from the wall (cm)
$y$ = height of the top of the ladder above the ground (cm)
Length of ladder = $2,\text{m} = 200,\text{cm}$
So the relation is
$x^2 + y^2 = 200^2$
Differentiate w.r.t. time $t$:
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
$\Rightarrow \dfrac{dx}{dt} = -\dfrac{y}{x}\dfrac{dy}{dt}$
Given:
$\dfrac{dy}{dt} = -25\ \text{cm/s}$ (negative since top slides down)
At the instant when
$y = 1,\text{m} = 100,\text{cm}$
Find $x$:
$x = \sqrt{200^2 - 100^2} = \sqrt{40000 - 10000} = \sqrt{30000} = 100\sqrt{3}$
Now substitute:
$\dfrac{dx}{dt} = -\dfrac{100}{100\sqrt{3}}(-25)$
$\dfrac{dx}{dt} = \dfrac{25}{\sqrt{3}}\ \text{cm/s}$
Final Answer:
$\boxed{\dfrac{25}{\sqrt{3}}\ \text{cm/sec}}$