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A U TUBE OF CONSTANT AREA OF CROSS SECTION A CONTAINS LIQUID OF DENSITY d . INITIALLY, THE LIQUID IN THE TWO ARMS ARE HELD WITH A LEVEL DIFFERENCE h . AFTER BEING RELEASED THE LEVEL EQUALIZE AFTER SOMETIME. THE WORK DONE BY GRAVITATIONAL FORCE ON THE LIQUID DURING THIS PROCESS IS WHAT ?

RAGINI SEIWAL , 7 Years ago
Grade 12
anser 1 Answers
Eshan
Dear student,

After the liquid in the two tubes reach equilibrium, the liquid initially at higher level decreases by height h/2, and liquid initially at lower level rises by height h/2.
Let the height of liquid in tube at lower level beh_1.

Initailly the total gravitational potential energy of the sytem=P_1=(\rho A h_1)(g)(\dfrac{h_1}{2})+(\rho A (h_1+h)(g)(\dfrac{h_1+h}{2}))

Finally the total gravitational potential energy of system =P_2=
2\times (\rho A (h_1+\dfrac{h}{2}))(g)(\dfrac{h_1+\dfrac{h}{2}}{2})=2\rho Ag\dfrac{(h_1+\dfrac{h}{2})^2}{2}
Hence work done by gravitational force=Decrease in potential energy=\rho Ag\dfrac{h^2}{4}
Last Activity: 7 Years ago
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