Solutions (^205)
Water formed as a result of melting ice has a
volume corresponding to the hatched region in the
figure. Since a part of this volume is above the
surface of water (h 1 > h 0 ), the level of water rises
after melting of ice. On the other hand, since
h 1 < h, oil fills the formed "hole", i.e. the total
level of liquid in the vessel falls.
1.110. Let x be the length of the part of the rod in
the tumbler, and y be the length of its outer part.
Then the length of the rod is x y, and the centre
of mass of the rod is at a distance (x+y)/2 from
its ends and at a distance (y — x)/2 from the outer
end. The condition of equilibrium is the equality
to zero of the sum of the moments of force about
the brim of the tumbler.
(Fb—FA) x
— Mg (2
-x)
where Fb = mbg = PalgV is the force of gravity
of the ball, and FA = pwgVI2 is the buoyant force,
where V = (4/3)nr^3 is the volume of the ball.
The required ratio is
y = 1 + 2 (Fb —FA)
1.5.
x (^) Mg
1.111. When the barometer falls freely, the force of
atmospheric pressure is no longer compensated by
the weight of the mercury column irrespective of
its height. As a result, mercury fills the barometer
tube completely, i.e. to the division 1050 mm.
1.112. In a vessel with a liquid moving horizontal-
ly with an acceleration a, the surface of the liquid
becomes an inclined plane. Its slope cp is deter-
mined from the condition that the sum of the force of
pressure F and the force of gravity nig acting on
an area element of the surface is equal to ma, and
the force of pressure is normal to the surface. Hence
a
tan cp =—
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