Explanation

(a) Let at a time $tdV$ be the decrease in volume of water in vessel in time$dt$. Therefore rate of decreases of water in vessel = rate of water flowing out of narrow tube
So $–\frac{dV}{dt} - \frac{\pi\left( p_{1} - p_{2} \right)r^{4}}{8\eta l}$
But $p_{1} = p_{2} = h\ \rho\ g$
$\therefore\ - \frac{dV}{dt} = \frac{\pi(h\ \rho\ g)r^{4}}{8\eta l} = \frac{(\pi\ \rho\ g\ r^{4})}{8\eta l \times A} \times (h \times A)$
Where $h \times A =$ Volume of water in vessel at a time $t = V$
$\therefore dV = - \left( \frac{\pi\ \rho\ g\ r^{4}}{8\eta l\ A} \right) \times V\ dt = - \lambda V\ dt$
Or $\frac{dV}{V} = - \lambda\ dt$
Where $\frac{\pi\ \rho\ g\ r^{4}}{8\eta l\ A} = \lambda =$ constant
Integrating it within the limits as time changes 0 to $t$, volume changes $V_{0}$ to $V$
Or $\log_{e}\frac{V}{V_{0}} = - \lambda\ t$ or $V = V_{0}e^{- \lambda t}$
Where $V_{0} =$ initial volume of water in vessel
$= A\ h_{0}$
Therefore, $h \times A = h_{0}Ae^{- \lambda t}$ or $h = h_{0}e^{- \lambda t}$
Thus, the variation of $h$ and $t$ will be represented by exponential curve as given by (a)