Question

A $2m$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25cm/sec.,$ then the rate $(\text{in}cm/sec.)$ 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 :

• A. $\frac{25}{3}$
• B. $\frac{25}{\sqrt{3}}$
• C. $25$
• D. $25\sqrt{3}$

Answer 1

The correct option is B $\frac{25}{\sqrt{3}}$


$x^2+y^2=4$
$\Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}=0\Rightarrow x\frac{dx}{dt}+y(-25)=0$
At $y=1,x=\sqrt{4-1}=\sqrt{3}$
$\therefore \frac{dx}{dt}=\frac{25}{\sqrt{3}}$