Two identical cylindrical vessels are kept on | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The initial potential energy of the system is given by the sum of the potential energies of the liquid in both vessels:$U_{i} = (dSx_{1})g \cdot \

Vedclass · Accepted Answer

$\frac{1}{4} gdS(x_{2}-x_{1})^{2}$

Vedclass · Answer

(C) The initial potential energy of the system is given by the sum of the potential energies of the liquid in both vessels:$U_{i} = (dSx_{1})g \cdot \frac{x_{1}}{2} + (dSx_{2})g \cdot \frac{x_{2}}{2} = \frac{dSg}{2}(x_{1}^{2} + x_{2}^{2})$By the principle of conservation of volume,the total volume of liquid remains constant:$Sx_{1} + Sx_{2} = S(x_{f} + x_{f}) = 2Sx_{f}$$x_{f} = \frac{x_{1} + x_{2}}{2}$The final potential energy of the system is:$U_{f} = 2 	imes (dSx_{f})g \cdot \frac{x_{f}}{2} = dSgx_{f}^{2} = dSg \left( \frac{x_{1} + x_{2}}{2} ight)^{2}$The change in potential energy is $\Delta U = U_{f} - U_{i}$:$\Delta U = dSg \left[ \left( \frac{x_{1} + x_{2}}{2} ight)^{2} - \frac{x_{1}^{2} + x_{2}^{2}}{2} ight]$$\Delta U = dSg \left[ \frac{x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2}}{4} - \frac{2x_{1}^{2} + 2x_{2}^{2}}{4} ight]$$\Delta U = dSg \left[ \frac{2x_{1}x_{2} - x_{1}^{2} - x_{2}^{2}}{4} ight] = -\frac{dSg}{4}(x_{1} - x_{2})^{2}$The magnitude of the change in energy is $\frac{1}{4} gdS(x_{2} - x_{1})^{2}$.

Similar Questions

In the figure shown,the heavy cylinder (radius $R$) resting on a smooth surface separates two liquids of densities $2\rho$ and $3\rho$. The height '$h$' for the equilibrium of the cylinder must be

State whether the following statements are True or False:
$(i)$ In incompressible flow,the density of a fluid changes with time and position.
$(ii)$ Non-viscous flow can be a streamline flow.
$(iii)$ In a steady flow,the velocity of the fluid at each point remains identical with time.

Vedclass Test Series

Exam Paper Generator

Online Exam Module