A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the…

1. $25$
2. $25\sqrt{3}$
3. $\frac{25}{3}$
4. $\frac{25}{\sqrt{3}}$

Solution

$\frac{dy}{dt}=-25cm/\mathrm{s}\mathrm{e}\mathrm{c}$ , $\frac{dx}{dt}=?$
Now, $x^2+y^2=2^2=4$
differentiating w.r.t. $t$ both side
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ $x^2+y^2=4$ when $y=1$ $\Rightarrow x^2=3\Rightarrow x=\sqrt{3}$
$\Rightarrow \frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$
$\Rightarrow {\frac{dx}{d{t}_{ }}}_{\left(\sqrt{3},1\right)}=-\frac{1}{\sqrt{3}}\times \left(-25\right)=\frac{25}{\sqrt{3}} cm/\mathrm{s}\mathrm{e}\mathrm{c}$

Asked in: JEE Main 2019 (12 Apr Shift 1)

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