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Question on hydrostatics?
A cylindrical beaker of height = 30 cm and diameter = 20 cm contains two immiscible liquids mercury and water of R.D. 13.6 and 1 respectively. A solid sphere of diameter = 14 cm floats in such a way that half the volume is dipped in mercury and the other half in water. The top of the sphere touches the uppersurface of water.
a) Find the density of the solid.
b) What happens if we add some more water to the beaker without disturbing the equilibrium. (will the body rise\dip more\no change)
c) If we add some other liquid of density less than that of mercury and more than that of water what will happen.(As compared to (a))
1 Answer
- MadhukarLv 710 years agoFavorite Answer
a)
Let d = density of the solid, g/c.c.
Volume of sphere = V
Weight of the sphere = buoyancy due to water + buoyancy due to mercury
=> V * d * g = (V/2) * 1 * g + (V/2) * 13.6 * g
=> d = (1/2) (1 + 13.6) = 7.3 g/c.c.
b)
Sphere will remain floating half-dipped in water and half-dipped in mercury as in a).
c)
If the other liquid added is immiscible with water and mercury, it will form a layer between mercury and water and exert more upward force of buoyancy than that of water which will make the sphere come out of mercury to some extent depending upon the density of the added liquid.
