Question

Two identical cylenderical vessels with their bases at same level contains liquid of density p to different height $h_1$ and $h_2$. The work done by gravity in equalising the levels, when two vessels are connected by a small tube is

• A. $A(h_1\sim h_2)gp$
• B. $(h_1\sim h_2)pg$
• C. $\frac{1}{4}(h_1\sim h_2{)}^{2}gAp$
• D. $\frac{1}{4}(h_1\sim h_2)gAp$

Answer 1

Answer: (C)

$\frac{1}{4}(h_1\sim h_2{)}^{2}gAp$

given that, density $=P$

$\text{let,}{h}_2<h_1$

let at equilibrium the height of liquid = h

since, volume of water is constant

$So,A{h}_1+A{h}_2+V_0=V_0+2Ah$

$V_0$=volume of water in pipe

$h=\frac{h_1+h_2}{2}$

potential energy of the system, initially

$\begin{array}{cc}{U}_i& =m_{1}g\frac{h_1}{2}+m_{2}g\frac{h_2}{2}\\ & =A\cdot h_1\cdot P\cdot \frac{h_1}{2}g+A\cdot h_2\cdot P\cdot \frac{h_2}{2}\cdot g\\ & =\frac{AP{h}_1^{2}g}{2}+\frac{AP{h}_2^2}{2}g\\ & =\frac{APg}{2}(h_1^2+h_2^2)\end{array}$

potential energy at equilibrium

$\begin{array}{cc}{U}_f& =\mathrm{mg}\frac{h}{2}+mg\frac{h}{2}\\ & =mgh\\ & =A\rho g{h}^2\end{array}$

$=\frac{APg{h}^2}{4}(h_1+h_2{)}^2$

$\therefore$ work done $w=v_i-v_f$

or $w=\frac{\mathrm{APg}}{2}(h_1^2+h_2^2-\frac{h_1^2+h_2^2+2h_1{h}_2}{2})$

or $w=\frac{APg{(h_1-h_2)}^2}{4}$

Correct option is $\left(C\right)$.