Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions (^199)
equal to the atmospheric pressure at the altitude
of the manometer. Thus, the reading of the ma-
nometer corresponds to the zero level since there is
no pressure difference.
1.102. We choose the zero level of potential energy
at the bottom of the outer tube. Then the potential
energy of mercury at the initial instant of time is
Wi = 2S/pmerg 2 —

/

pnwrgS/2•

The potential energy of mercury at the final instant

.5

Fig. 191

(the moment of separation of the inner tube,

Fig. 191) is (by hypothesis, 1 > h)

x h

Wf = 2Sxpmerg ) npmerg (x - 2 -) ,

where x is the level of mercury in the outer tube at
the moment of separation. This level can be found
from the condition of the constancy of the mercury
volume:

25x+ Sh 251, x =1--

2 '

The difference in the potential energies is equal

to the sum of the requited work A done by external

forces and the work done by the force of atmospher-