JEE PYQ: Mechanical Properties Of Fluids Question 14
Question 14 - 2020 (04 Sep 2020 Shift 2)
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:
(1) $gdS(x_2^2 + x_1^2)$
(2) $gdS(x_2 + x_1)^2$
(3) $\frac{3}{4}gdS(x_2 - x_1)^2$
(4) $\frac{1}{4}gdS(x_2 - x_1)^2$
Show Answer
Answer: (4) $\frac{1}{4}gdS(x_2 - x_1)^2$
Solution
Initial potential energy: $U_i = (\rho S x_1)g \cdot \frac{x_1}{2} + (\rho S x_2)g \cdot \frac{x_2}{2}$ Final potential energy: $U_f = (\rho S x_f)g \cdot \frac{x_f}{2} \times 2$ By volume conservation: $Sx_1 + Sx_2 = S(2x_f)$, so $x_f = \frac{x_1 + x_2}{2}$ $\Delta U = U_i - U_f = \rho S g\left[\frac{x_1^2}{2} + \frac{x_2^2}{2} - x_f^2\right]$ $= \rho S g\left[\frac{x_1^2}{2} + \frac{x_2^2}{2} - \left(\frac{x_1 + x_2}{2}\right)^2\right] = \frac{\rho S g}{4}(x_1 - x_2)^2$
BETA
