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

Column $I$	Column $II$
$A$. Stoke's law	$I$. Pressure and energy
$B$. Turbulence	$II$. Hydraulic lift
$C$. Bernoulli's Principle	$III$. Viscous drag
$D$. Pascal's law	$IV$. Reynold's number

Similar Questions

$A$ fluid container containing a liquid of density $\rho$ is accelerating upward with acceleration $a$ along an inclined plane of inclination $\alpha$ as shown. Then the angle of inclination $\theta$ of the free surface is:

Match the columns $I$ and $II$:

Column $I$ Column $II$

$A$. Stoke's law $I$. Pressure and energy

$B$. Turbulence $II$. Hydraulic lift

$C$. Bernoulli's Principle $III$. Viscous drag

$D$. Pascal's law $IV$. Reynold's number

The correct match is:

$A$ motor is used to deliver water at a certain rate through a given horizontal pipe. To deliver $n$-times the water through the same pipe in the same time, the power of the motor must be increased as follows:

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