Save The Fish

Let the rate of the liquid flowing out of the hole be : $\frac{\mathrm dV}{\mathrm dt}$ = -(rate of volume of the liquid lost as it flows out) . The minus sign is for indicating the decrease in volume . Let : $A$ and : $a$ be the area of cross-section of the base of the container and the hole respectively . Let : $v$ be the velocity in m/s with which the liquid flows out of the hole. Thus , the volume of liquid which flows out per second from the hole = : $av$ . Therefore ,

: $\frac{\mathrm dV}{\mathrm dt} = -av$ : $(Eq. 1)$

Let mass: $m$ of the liquid be at height : $h$ above the base of the container . It has potential energy =: $mgh$ . When this liquid flows out of the hole, it will have a kinetic energy = : $\frac{1}{2}mv^2$ . Hence, using conservation of energy and assuming no loss of energy due to viscosity of liquid or forces of cohesion , we can say that : $mgh=\frac{1}{2}mv^2$

: $\Rightarrow v=\sqrt{2gh}$ : $(Eq.2)$

From : $Eq.1$ and : $Eq.2$ , we conclude : $\frac{\mathrm dV}{\mathrm dt}=-a\sqrt{2gh}$ . Writing the volume of liquid in the container as a function of time , we have : $V(t)=Ah(t)$ where : $h$ is the height of the liquid in the container as a function of time . Thus, we can write : $\frac{\mathrm d\big (V(t)\big )}{dt}=A\frac{\mathrm d\big (h(t)\big )}{dt}$ . Therefore, : $A\frac{\mathrm d\big (h(t)\big )}{dt}=-a\sqrt{2gh}\Rightarrow \frac{\mathrm d\big (h(t)\big )}{dt}=-\frac{a}{A}\sqrt{2gh}$ .

Simply put, we have : $\frac{\mathrm dh}{\mathrm dt}=-k\sqrt{h}$ where : $k=\frac{a}{A}\sqrt{2g}$ .

After variable separation, we have : $\frac{\mathrm dh}{\sqrt{h}}=-k$ : $\mathrm dt$ . Let the height of the container be : $H$ . As the height of the liquid decreases from : $H$ to zero , time increases from zero to : $t$ . Integrating the above obtained differential equation after applying proper limits, we have

: $\int_H^0 \frac{\mathrm dh}{\sqrt{h}}=\int_0^t k \, \mathrm{dt} \Rightarrow -2\sqrt{H}=-kt \Rightarrow t=\frac{2\sqrt{H}}{k}$ .

Substituting for : $k$ , we have : $t=\frac{A}{a}\sqrt\frac{2H}{g}$ .

Therefore, observe that by doubling the height, we have to multiple $t$ by $\sqrt{2}$ . Hence, the answer is $\sqrt{2}\times 10 =14.142135...$ .

Using toricelli's law -dh/dt= c(h)^0.5

Integrate this to get that timeis proportional to square root of height.

So for double height time will be root 2 times

as simple as that