Question

Question #0ab87

Answer 1

`5/4rho.pi.g.r^2.h^2 " "J`

Explanation:

Let the origin of the coordinate axes be located in the plane of surface in tube A.
Suppose the liquid level in tube rise a height `z`. Now consider an infinitesimal disk of liquid of width `dz` at this height. The mass of this disk
`dm="density" times "Volume"=rho xx(pir^2times dz)` .....(1)
We know that gravitational potential energy is given by the expression:

`PE=mgh`, for the disk under consideration and with reference to origin, the increase in potential energy:

`dPE=dmxxgxxz`, insert value of `dm` from (1)

`dPE=rho.pi.g.r^2.z.dz`
Total potential energy increase is integral of the expression for height limits from `0 " to " h`
`PE=int_0^hdPE=int_0^hrho.pi.g.r^2.z.dz`
`PE=rho.pi.g.r^2.z^2/2]_0^h`, ignoring constant of integration for a definite integral
`PE=1/4rho.pi.g.r^2.h^2` .....(2)

Assuming that problem is about finding total increase in potential energy of the system.

We observe that due to atmospheric pressure, there will be equal rise of liquid in the tube B.
Therefore, increase of potential energy due to tube B can be found by inserting radius of tube B in (2)
`PE_B=1/4rho.pi.g.(2r)^2.h^2`
`PE_B=rho.pi.g.r^2.h^2`
Total increase`=PE_A+PE_B=1/4rho.pi.g.r^2.h^2+rho.pi.g.r^2.h^2`
`=5/4rho.pi.g.r^2.h^2 " "J`