A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 2 5cm / s , then the rate (in cm / s ) 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

Given a ladder of length l = 2m leans against a vertical wall. Now, the top of ladder begins to slide down the wall at the rate 2 5cm / s . 

Let the rate at which bottom of the ladder slides away from the wall on the 

horizontal ground is $\frac{\mathrm{dx}}{\mathrm{dt}}\mathrm{cm}/\mathrm{s}$

$\because {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{l}}^2\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=4[\because \mathrm{l}=2\mathrm{m}]\dots \left(\mathrm{i}\right) \mathrm{On} \mathrm{differentiating} \mathrm{both} \mathrm{sides} \mathrm{of} \mathrm{eq}.\left(\mathrm{i}\right) \mathrm{w}.\mathrm{r}.\mathrm{t}.$

' t ' , we get $2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dt}}=0\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\frac{\mathrm{dy}}{\mathrm{dt}}$

From eq (i), when $\mathrm{y}=1\mathrm{m}$ then ${\mathrm{x}}^2+1^2=4\Rightarrow {\mathrm{x}}^2=3\Rightarrow \mathrm{x}=\sqrt{3}\mathrm{m}[\because \mathrm{x}>0]$

On substituting  $\mathrm{x}=\sqrt{3}\mathrm{m}$ and $\mathrm{y}=1\mathrm{m}$  in $\mathrm{wq}.\left( \right.\mathrm{ii})$), We get $\frac{\mathrm{dx}}{\mathrm{dt}}=-\frac{1}{\sqrt{3}}\left(-\frac{25}{100}\right)\mathrm{m}/\mathrm{s}$

$\left[given \frac{dy}{dt}=-25\mathrm{cm}/\sec \right] =\frac{25}{\sqrt{3}}\mathrm{cm}/\mathrm{s}$

Hence, option (b) is correct.