Water is dropped at the rate of 2 m³/s into a cone of semi-vertical angle $45°$. If the rate at which periphery of water surface changes when the height of the water in the cone is 2 m is d, then the value of 5d is _______

We have

$\begin{array}{l} \frac{\mathrm{dV}}{\mathrm{dt}}=2 \mathrm{or} \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{3}{\mathrm{\pi r}}^3\right)=2 \left[\mathrm{Here}, \mathrm{r}=\mathrm{h} \mathrm{as} \mathrm{\theta }=45°\right]\\ \mathrm{or} {\mathrm{\pi r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{or} \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{2}{{\mathrm{\pi r}}^2} \left(1\right)\\ \mathrm{Now}, \mathrm{perimeter} =2\mathrm{\pi r}=\mathrm{p} \left(\mathrm{let}\right)\\ \therefore \frac{\mathrm{d}}{\mathrm{dt}}\left(2\mathrm{\pi r}\right)=2\mathrm{\pi }\left(\frac{2}{{\mathrm{\pi r}}^2}\right)=\frac{4}{{\mathrm{r}}^2} \left[\mathrm{Using} \mathrm{equation} \left(1\right)\right] \left(2\right)\\ \mathrm{When} \mathrm{h}=2\mathrm{m}, \mathrm{r}=2\mathrm{m}\\ \mathrm{Hence}, \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{4}{4}=1 \mathrm{m}/\mathrm{s}.\end{array}$