(^200) Aptitude Test Problems in Physics
is pressure acting on the surface of mercury in
the outer tube and on the upper (sealed) end of the
inner tube. The displacement of mercury in the
outer tube is 1 x, the corresponding work of the
force of atmospheric pressure being poS (1 — x) =
PmergSh (1 — x).
The displacement of the sealed end of the in-
ner tube is 1 — (1 + x), the corresponding work
being —poSx = — pmergShx.
Therefore, the required work of external
forces is
A= Wf — WI — PmergSh (1— 2x)
3
=pmergSh (1 --- 4
h
).
1.103. The pressure at the bottom of the "vertical"
cylinder is p = Po pw gh, where Po is the atmo-
spheric pressure, pw is the density of water, and g
is the free-fall acceleration. According to Pascal's
Fig. 192
law, the same pressure is exerted on the lower part
of the piston in the "horizontal" cylinder. The
total pressure of water on the part of the piston
separated from the lower part by a distance x
along the vertical is p — pgx (Fig. 192).
Let us consider the parts of the piston in the
form of narrow (of width Ax) horizontal strips
separated by equal distances a from its centre. The
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