Solutions (^203)
When the lever is immersed in water, the force
of gravity is supplemented by the buoyan of
water, which is proportional to the volume of the
12
r2
Fig. 194
body. By hypothesis, F (^1) =-- F 2 , and therefore
(mig — F1)^11 (ms g — F^2 ) 12.
Thus, the equilibrium of the lever will be violated.
1.107. The volume of the submerged part of each
box changes by the same amount eV = m/p„-,
where m is the mass of the body, and pw is the den-
sity of water. Since the change in the level of water
in each vessel is determined only by AV and the
vessels are identical, the le%els of water in them
will change by the same amount.
1.108. Let the volume of the steel ball be V, and
let the volume of its part immersed in mercury be
Vo before water is poured and V 1 after water covers
the ball completely. The value of Vo can be found
from the condition
Pst V = Pmer Vo,
where pst and pmer are the densities of steel and
mercury. Since the pressure of water is transmitted
through mercury to the lower part of the ball, the
buoyant force exerted on it by water is
pw (V — V] ) g, where p, is the density of water,
while the buoyancy of mercury is pmerVig. The
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