A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its…

Plotting the diagram we have,

Now given, $\tan \theta =\frac{3}{4}=\frac{r}{h}$ and  let water is poured at constant rate $\frac{dV}{dt}=6$ cubic meter per hour,

Now we know that volume of cone is given by $V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi h^3{\tan }^2\theta =\frac{9\pi }{48} h^3=\frac{3\pi }{16} h^3$ $\left\{\text{by using }\tan \theta =\frac{3}{4}=\frac{r}{h}\right\}$

Now differentiating above equation we get,

$\Rightarrow \frac{dV}{dt}=\frac{3\pi }{16}·3h^2·\frac{dh}{dt}=6$

$\Rightarrow \frac{dh}{dt}=\frac{2}{3\mathrm{\pi }}\text{m/hr}$

Now, we know that curved surface area of cone is given by,

 $S=\pi \mathrm{r}l=\frac{15}{16}\pi h^2$ $\left\{\text{as} \sin \theta =\frac{3}{5}=\frac{r}{l}\Rightarrow l=\frac{5}{3}r \text{and }r=\frac{3}{4}h\right\}$

Now differentiating both side we get,

$\Rightarrow \frac{dS}{dt}=\frac{15\pi }{16}·2\mathrm{h}\frac{dh}{dt}$

$\Rightarrow \frac{dS}{dt}=5 {\mathrm{m}}^2/\mathrm{hr}$ $\left\{\text{as }\frac{dh}{dt}=\frac{2}{3\mathrm{\pi }}\text{m/hr}\right\}$