A water tank has the shape of a right circular cone with axis...

Plotting the diagram we have,

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

Now we know that volume of cone is given by $V = \dfrac{1}{3}\pi{r}^{2}h = \dfrac{1}{3}\pi{h}^{3}\tan^{2}\theta = \dfrac{9\pi}{48}{h}^{3} = \dfrac{3\pi}{16}{h}^{3}$ \left\{\text{by using}tan\theta = \dfrac{3}{4} = \dfrac{r}{h}\right\}$</p><p>Now differentiating above equation we get,</p><p>$⇒ \dfrac{dV}{dt} = \dfrac{3\pi}{16} \cdot 3{h}^{2} \cdot \dfrac{dh}{dt} = 6$</p><p>$⇒ \dfrac{dh}{dt} = \dfrac{2}{3\pi}\text{m/hr}$</p><p>Now, we know that curved surface area of cone is given by,</p><p>$S = \pirl = \dfrac{15}{16}\pi{h}^{2}$\left\{\text{as}sin\theta = \dfrac{3}{5} = \dfrac{r}{l} ⇒ l = \dfrac{5}{3}r\text{and}r = \dfrac{3}{4}h\right\}$

Now differentiating both side we get,

$⇒ \dfrac{dS}{dt} = \dfrac{15\pi}{16} \cdot 2h\dfrac{dh}{dt}$

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