Question

Water in a vessel of uniform cross-section escapes through a narrow tube at the base of the vessel. Which graph given below represents the variation of the height $h$ of the liquid with time $t?$

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Let at a time $tdV$ be the decrease in volume of water in vessel in time $dt$.
Therefore rate of decrease 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, $p_1=p_2=h\rho g$
$\therefore -\frac{dV}{dt}=\frac{\pi (h\rho g)r^4}{8\eta l}=\frac{\left(\pi \rho g{r}^4\right)}{8\eta l\times A}\times (h\times A)$
where $h\times A=$ volume of water in vessel at a time $t=V$
$\therefore dV=-\left(\frac{\pi \rho g{r}^4}{8\eta lA}\right)\times Vdt=-\lambda Vdt$
or $\frac{dV}{V}=-\lambda dt$
where, $\frac{\pi \rho g{r}^4}{8\eta lA}=\lambda =$ constant
Integrating it within the limits as time changes $0$ to $t$, volume changes from $V_0$ 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, $h\times A=h_{0}A{e}^{-\lambda t}$
or $h=h_0{e}^{-\lambda t}$
Thus, the variation of $h$ and $t$ will be represented by exponential curve as given by (a).