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 \(x_{1}\) and in the other, \(x_{2}\). 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 \({gdS}\left(x_{2}+x_{1}\right)^{2}\)
B \(\frac{3}{4} g d S\left(x_{2}-x_{1}\right)^{2}\)
C \(\frac{1}{4} g d S\left(x_{2}-x_{1}\right)^{2}\)
D \({gdS}\left(x_{2}^{2}+x_{1}^{2}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{4} g d S\left(x_{2}-x_{1}\right)^{2}\)

Step-by-step Solution

Detailed explanation

\(U _{ i }=\left(\rho Sx _{1}\right) g \cdot \frac{ x _{1}}{2}+\left(\rho Sx _{2}\right) g \cdot \frac{ x _{2}}{2}\) \(U _{ f }=\left(\rho Sx _{ f }\right) g \cdot \frac{ x _{ f }}{2} \times 2\) By volume conservation \(Sx _{1}+ Sx _{2}= S \left(2 x _{ f }\right)\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app

S.No.	\(MS\;(cm)\)	\(VS\) divisions
\((1)\)	\(0.5\)	\(8\)
\((2)\)	\(0.5\)	\(4\)
\((3)\)	\(0.5\)	\(6\)