Question

.A U-tube of cross section A and 2A asshown contains liquid of density d.Initially, the liquid in the two arms areheld with a level difference h. Afterbeing released the levels equalize aftersome time. The work done by gravityforces on the liquid in the process is

Answer 1

Explanation:

Given .A U-tube of cross section A and 2A as shown contains liquid of density d. Initially, the liquid in the two arms are held with a level difference h. After being released the levels equalize after some time. The work done by gravity forces on the liquid in the process is

• We need to find the work done by gravity forces on the liquid will be
• Now Area will be A and let the difference be x1 and area will be 2A and also difference be x2.  
• So A x1 = 2 Ax2
• Or x1 / x2 = 2A / A
•                 = 2/1
•                 = 2 : 1
• Now x1 + x2 should be equal
• So x1 + x2 = h
• Or x2 = h/3
• Now the level from x2 be of zero potential
• So we have initial potential energy
• So U initial = h x A x ρ x g
•               Now multiply by h/2, we get
•             So U initial = h^2 A ρ g / 2
• Now the final potential energy  
• So U final = 2 ρ x2 A
• Multiply by g and x/2 we get
• U final = 2 ρ x^2 A g / 2  
•             = ρ x^2 A g , so there is an increase in potential energy.
• Now the height of x1 was downward,  
• So if we have h – x1 = h – h/3 we get 2h / 3
• So h - 2h / 3 = h/3 and centre of mass = h/6
• So U final = ρ x2^2 A g + ρh^2 A g / 18
• Now squaring x2 we get (x2)^2 = h^2 / 9
• Now U final = 3 ρ h^2 A g / 18
• Therefore U final – U initial = ρh^2 A g / h – h^2 A ρ g / 2
•                                             = - ρ h^2 A g / 3 which is change in potential energy. The negative sign is the work done by the gravitational force.
• So we get ρh^2 Ag/3

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Answer 2

Answer:

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