Question

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $\mathrm{d}$. The area of the base of both vessels is S but the height of liquid in one vessel is ${\mathrm{x}}_1$ and in the other ${\mathrm{x}}_2$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

• A. $\mathrm{gdS}{\left({\mathrm{x}}_2+{\mathrm{x}}_1\right)}^2$
• B. $\mathrm{gdS}{\left({{\mathrm{x}}^2}_2+{\mathrm{x}}_1\right)}^2$
• C. $\frac{1}{4}\mathrm{gdS}{\left[{\mathrm{x}}_2-{\mathrm{x}}_1\right]}^2$
• D. $\frac{3}{4}\mathrm{gdS}{\left[{\mathrm{x}}_2-{\mathrm{x}}_1\right]}^2$

Answer 1

The correct option is C

$\frac{1}{4}\mathrm{gdS}{\left[{\mathrm{x}}_2-{\mathrm{x}}_1\right]}^2$

Step 1. Given data

It is given that, two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $\mathrm{d}$. The area of the base of both vessels is S but the height of liquid in one vessel is ${\mathrm{x}}_1$ and in the other ${\mathrm{x}}_2$.

We have to determine the value where energy of the system decreases.

Step 2. Concept to be used

The conservation of volume states that the amount of the conserved quantity at a point or within a volume can only change by the amount of the quantity which flows in or out of the volume.

Step 3. Determine conservation of volume.

By conservation of volume,

${\left({\mathrm{V}}_{\mathrm{system}}\right)}_{\mathrm{initial}}={\left({\mathrm{V}}_{\mathrm{system}}\right)}_{\mathrm{final}}$

${\mathrm{Sx}}_1+{\mathrm{Sx}}_2={\mathrm{Sx}}_{\mathrm{f}}+{\mathrm{Sx}}_{\mathrm{f}}$

$x_1+{\mathrm{x}}_2={\mathrm{x}}_{\mathrm{f}}+{\mathrm{x}}_{\mathrm{f}}$

$x_1+{\mathrm{x}}_2=2{\mathrm{x}}_{\mathrm{f}}$

${\mathrm{x}}_{\mathrm{f}}=\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2}$ --------- $\left(1\right)$

Step 4. Calculate initial energy of system.

Now, initial energy of system is,

${\left({\mathrm{U}}_{\mathrm{system}}\right)}_{\mathrm{initial}}={\mathrm{M}}_1{\mathrm{gh}}_1+{\mathrm{M}}_2{\mathrm{gh}}_2$

$={\mathrm{dV}}_1{\mathrm{gh}}_1+{\mathrm{dV}}_2{\mathrm{gh}}_2$

$={\mathrm{dSx}}_1\mathrm{g}\left(\frac{{\mathrm{x}}_1}{2}\right)+{\mathrm{dSx}}_2\mathrm{g}\left(\frac{{\mathrm{x}}_2}{2}\right)$

$=\frac{1}{2}{\mathrm{dSgx}}_1^2+\frac{1}{2}{\mathrm{dSgx}}_{{}_2}^2\mathrm{g}$

$=\frac{1}{2}\mathrm{dSg}\left({\mathrm{x}}_1^2+{\mathrm{x}}_2^2\right)$ ----------- $\left(2\right)$

Step 5. Calculate final energy of system.

Final energy of the system is,

${\left({\mathrm{U}}_{\mathrm{system}}\right)}_{\mathrm{Final}}=\mathrm{M}\text{'}\mathrm{gh}\text{'}+\mathrm{M}\text{'}\mathrm{gh}\text{'}$

$=\mathrm{dV}\text{'}\mathrm{gh}\text{'}+\mathrm{dV}\text{'}\mathrm{gh}\text{'}$

$={\mathrm{dSx}}_{\mathrm{f}}\mathrm{g}\left(\frac{{\mathrm{x}}_{\mathrm{f}}}{2}\right)+{\mathrm{dSx}}_{\mathrm{f}}\mathrm{g}\left(\frac{{\mathrm{x}}_{\mathrm{f}}}{2}\right)$

$=\frac{1}{2}{\mathrm{dSgx}}_{\mathrm{f}}^2+\frac{1}{2}{\mathrm{dSgx}}_{{}_{\mathrm{f}}}^2\mathrm{g}$

$={\mathrm{dSgx}}_{\mathrm{f}}^2$

$=\mathrm{dSg}{\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2}\right)}^2$ ---------- $\left(3\right)$

Step 6. Determine the energy of system.

So, the change in energy of system is,

$∆{\mathrm{U}}_{\mathrm{system}}={\left({\mathrm{U}}_{\mathrm{system}}\right)}_{\mathrm{final}}-{\left({\mathrm{U}}_{\mathrm{system}}\right)}_{\mathrm{initial}}$

$=\mathrm{dSg}{\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2}\right)}^2-\frac{1}{2}\mathrm{dSg}\left({\mathrm{x}}_1^2+{\mathrm{x}}_2^2\right)$

$=\mathrm{dSg}\left(\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+2{\mathrm{x}}_1{\mathrm{x}}_2}{4}\right)-\mathrm{dSg}\left(\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2}{2}\right)$

$=\mathrm{dSg}\left[\left(\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+2{\mathrm{x}}_1{\mathrm{x}}_2}{4}\right)-\left(\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2}{2}\right)\right]$

$=\mathrm{dSg}\left[\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+2{\mathrm{x}}_1{\mathrm{x}}_2-2\left({\mathrm{x}}_1^2+{\mathrm{x}}_2^2\right)}{4}\right]$

$=\frac{\mathrm{dSg}}{4}\left[{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+2{\mathrm{x}}_1{\mathrm{x}}_2-2{\mathrm{x}}_1^2-2{\mathrm{x}}_2^2\right]$

$=\frac{\mathrm{dSg}}{4}\left[-{\mathrm{x}}_1^2-{\mathrm{x}}_2^2+2{\mathrm{x}}_1{\mathrm{x}}_2\right]$

$=-\frac{\mathrm{dSg}}{4}\left[{\mathrm{x}}_1^2-{\mathrm{x}}_2^2+2{\mathrm{x}}_1{\mathrm{x}}_2\right]$

$=-\frac{1}{4}\mathrm{gdS}{\left[{\mathrm{x}}_2-{\mathrm{x}}_1\right]}^2$

So, energy of the system will decrease by $\frac{1}{4}\mathrm{gdS}{\left[{\mathrm{x}}_2-{\mathrm{x}}_1\right]}^2$

Hence, option $\mathrm{C}$ is correct answer.