A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semivertical angle is tan^-1 3 4. Water is poured in…

Plotting the diagram we have,

Now given, $\mathrm{tan}\theta =\frac{3}{4}=\frac{r}{h}$tanθ=43​=hr​ and let water is poured at constant rate $\frac{dV}{dt}=6$dtdV​=6 cubic meter per hour,

Now we know that volume of cone is given by $V=\frac{1}{3}\pi {r}^{2}h=\frac{1}{3}\pi {h}^{3}{\mathrm{tan}}^{2}\theta =\frac{9\pi }{48}{h}^{3}=\frac{3\pi }{16}{h}^{3}$V=31​πr2h=31​πh3tan2θ=489π​h3=163π​h3 ${\text{by using }\mathrm{tan}\theta =\frac{3}{4}=\frac{r}{h}}$by using tanθ=43​=hr​

Now differentiating above equation we get,

$$\Rightarrow \frac{dV}{dt}=\frac{3\pi }{16}\cdot 3{h}^{2}\cdot \frac{dh}{dt}=6$$⇒dtdV​=163π​⋅3h2⋅dtdh​=6

$$\Rightarrow \frac{dh}{dt}=\frac{2}{3\pi }\text{m/hr}$$⇒dtdh​=3π2​m/hr

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

$S=\pi rl=\frac{15}{16}\pi {h}^{2}$S=πrl=1615​πh2 ${\text{as}\mathrm{sin}\theta =\frac{3}{5}=\frac{r}{l}\Rightarrow l=\frac{5}{3}r\text{and }r=\frac{3}{4}h}$assinθ=53​=lr​⇒l=35​rand r=43​h

Now differentiating both side we get,

$$\Rightarrow \frac{dS}{dt}=\frac{15\pi }{16}\cdot 2h\frac{dh}{dt}$$⇒dtdS​=1615π​⋅2hdtdh​

$\Rightarrow \frac{dS}{dt}=5{m}^{2}/\mathrm{hr}$⇒dtdS​=5m2/hr ${\text{as }\frac{dh}{dt}=\frac{2}{3\pi }\text{m/hr}}$as dtdh​=3π2​m/hr