Physics - Fluid Question with Solution | TestHub

$$\text{ Let the rate of falling water level be }-\frac{dh}{dt}$$
Initially at $t=0;h=h$ $$t=t;h=0$$ Then, $A\left(-\frac{dh}{dt}\right)=\pi a^2\cdot v$ $$dt=-\frac{A}{\pi a^2\sqrt{2gh}}dh$$ $[\because$ velocity of efflux of liquid $v=\sqrt{2gh}]$ Integrating both sides $${\int }_0^{t}dt=-\frac{A}{\sqrt{2g}\pi a^2}{\int }_h^0{h}^{-1/2}dh$$ $$[t{]}_0^t=-\frac{A}{\sqrt{2g}\pi a^2}\cdot {\left[\frac{h^{1/2}}{1/2}\right]}_h^0$$ $$t=\frac{\sqrt{2}A}{\pi a^2}\sqrt{\frac{h}{g}}$$