Krotov

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 (hl > hg), the level of water rises after melting of ice. On the other hand, since hl < h, oil hlls 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 sr: —l— y, and the centre of mass of ,_the rod is at a distance (x-|—y)/2 from its ends and at a distance (y — :::)/2 from the outer end. Tl1e condition of equilibrium is the equality to zero of the sum of the moments. of force about the brim of the tumbler. , M 1 —x <m.——n.> ¤:=—¢’i=—g—-—i. where Fb = mbg = palgV is the force of gravity of the ball, and F A = pwgV/2 is the buoyant force, where V = (4/3);rtr°* is the volume of the ball.

The required ratio is 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 iills the barometer tube completely, i.e. to the division 1050 mm.

1.112. In a vessel with a liquid moving horizontally with an accele1·ation a, the surface of the liquid becomes an inclined plane. Its slope cp is determined from the condition that the sum of the force of pressure F and the force of gravity mg acting on an area element of the surface is equal to ma, and the force of pressure is no1·mal to the surface. Hence tan = il- , (P 8