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 $t d V$ be the decrease in volume of water in vessel in time $d t$.
Therefore rate of decrease of water in vessel $=$ rate of water flowing out of narrow tube
So, $\frac{d V}{d t}=\frac{\pi\left(p_{1}-p_{2}\right) r^{4}}{8 \eta l}$
But, $p_{1}=p_{2}=h \rho g$
$\therefore -\frac{d V}{d t}=\frac{\pi(h \rho g) r^{4}}{8 \eta l}=\frac{\left(\pi \rho g r^{4}\right)}{8 \eta l \times A} \times(h \times A)$
where $h \times A=$ volume of water in vessel at a time $t=V$
$\therefore d V=-\left(\frac{\pi \rho g r^{4}}{8 \eta l A}\right) \times V d t=-\lambda V d t$
or $\frac{d V}{V}=-\lambda d t$
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 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).