A cube of iron whose sides are of length l l, is put into mercury.The weight of iron cube is w w. The density of iron is rho 1 rho1, that of mercury is rho m rhom. The depth to which the cube sinks is given by the expression-
Answers
The height of the block above the mercury level is h = 4.71 cm
Explanation:
Correct statement:
A cubical iron block of side 10 cm is floating on mercury in a vessel. What is the height of the block above the mercury level?
Given that the density of iron is 7.2 gm/c.c and density of the mercury =13.6 gm/c.c
Give data:
Density of iron is 7.2 gm/cm^3
Density of the mercury =13.6 gm/cm^3
Length of side of iron block = 10 cm
Let height of the lock = x
Volume of lock inside mercury = 10 x 10 x(10-x)
Weight of this volume of mercury = vm x 13.6 x 981 = upthrust on block
Total volume of block = 10 x 10 x 10 = 1000 cm^3
Weight "W" = Vρig = V x 7.2 x 981
Weight of mercury displaced = Weight of iron block
100 x (10-x) x 13.6 x 981 = 1000 x 7.2 x 981
100 x (10-x) x 13.6 = 1000 x 7.2
(10-x) x 1360 = 7200
10-x = 7200 / 1360
10-x = 5.29
10 - 5.29 = x
h = 4.71 cm
The height of the block above the mercury level is h = 4.71 cm
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