The cross-sectional area of vessel A is 50 cm² and it contains water to a height 30 cm. The vessel B has an area of cross-section of 25 cm². The two vessels are connected with a thin tube as shown in the figure, When the tap is slowly opened, and the water attained an equilibrium in both vessels. The reduction in the potential energy of the water is (Density of water is 1000 kgm-³)

1) 7.5 J
2) 22.5 J
3) 0.75 J
4) 8.5 J
5) 75 J

Please show the working along with a brief explanation.​

The reduction in the potential energy of the water is approximately 7.5 J.

option 1

We can use the principle of conservation of energy to determine the reduction in potential energy of the water.

Initially, the water in vessel A has a certain amount of potential energy due to its height above the bottom of the vessel. When the water flows through the tube and reaches vessel B, its height above the bottom of vessel B is lower than that of vessel A, which means that its potential energy has decreased.

The potential energy of the water in vessel A is given by:

PE_A = mgh_A

The mass of the water in vessel A is given by:

m = density x volume

volume = A x h_A

Substituting for m and simplifying, we get:

PE_A = density x A x h_A x g

Similarly, the potential energy of the water in vessel B is:

PE_B = density x A_B x h_B x g

At equilibrium, the height of the water in the two vessels will be the same, so we can set h_A = h_B = h.

Also, since the water is in equilibrium, the pressure at the bottom of both vessels must be the same. This means that the pressure difference between the top and bottom of the water column in vessel A (due to the weight of the water) must be balanced by the pressure difference between the top and bottom of the water column in vessel B.

The pressure difference in vessel A is:

P_A = density x g x h_A

and the pressure difference in vessel B is:

P_B = density x g x h_B

Since the pressure difference must be balanced, we have:

P_A - P_B = density x g x h_A - density x g x h_B = 0

which simplifies to:

h_A = h_B x A_B / A

Substituting for h_A and h_B in the expressions for PE_A and PE_B, we get:

PE_A = density x A x h x g

PE_B = density x A_B x h x g x A / A_B

The reduction in potential energy of the water is:

ΔPE = PE_A - PE_B = density x g x h x (A - A_B x A / A_B)

which simplifies to:

ΔPE = density x g x h x (A - A_B)

Substituting the given values, we get:

ΔPE = 1000 kg/m³ x 9.8 m/s² x 0.3 m x (50 cm² - 25 cm²)

Converting the area units to m², we get:

ΔPE = 1000 kg/m³ x 9.8 m/s² x 0.3 m x (0.005 m² - 0.0025 m²)

Simplifying, we get:

ΔPE = 7.4 J

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