Aspire Faculty ID #15866 · Topic: JEE Main 2019 (12 April Morning Shift) · Just now
JEE Main 2019 (12 April Morning Shift)

A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate of 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 1m above the ground is:

Solution

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: $\boxed{\dfrac{25}{\sqrt{3}}\ \text{cm/sec}}$

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