Verified Solution • 2025–26 Session moderate

QJEEM20Ph513a: JEE Main 2020 Physics - Class 11 JEEM20Ph513a Solution: Two identical cylindrical vessels are kept on the ground and…

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other, x2 . 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:

Multiple Choice Options

  • A gdS(x2 + x1)2
  • B gdS (x22+x12)\left( {x_2^2 + x_1^2} \right)
  • C 14gdS(x2−x1)2{1 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}
  • D 34gdS(x2−x1)2{3 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}

Step-by-Step Verified Solution

ui=[dSx1.x12+dSx2.x22]g{dSx1→m, x12→h(C.O.M)}{u_i} = \left[ {dS{x_1}.{{{x_1}} \over 2} + dS{x_2}.{{{x_2}} \over 2}} \right]g\left\{ {dS{x_1} \to m,\,{{{x_1}} \over 2} \to h(C.O.M)} \right\}

Here total volume remains same.

∴\therefore Vi = Vf

⇒\Rightarrow S(x1 + x2) = S(h + h)

⇒\Rightarrow h = x1+x22{{{x_1} + {x_2}} \over 2}

uf = (dSh)g h2×2{h \over 2} \times 2

⇒\Rightarrow uf=[dS(x1+x22)×(x1+x24)×2]g{u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} \right) \times \left( {{{{x_1} + {x_2}} \over 4}} \right) \times 2} \right]g

∴\therefore ui−uf=dsg[x122+x222−(x1+x2)24]{u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} \right)}^2}} \over 4}} \right]

=dsg(x1−x2)24= dsg{{{{\left( {{x_1} - {x_2}} \right)}^2}} \over 4}

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