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 density of iron is 7.2 gm/cc and density of mercury is 13.6 gm/cc.
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4
answer is 4.71 cm
weight can be determined by multiplying density and volume of the cube.
for a floating body wt = upthrust
by the foumula of upthrust volume of cube submerged can be obtained grom which the height of cube submerged can be obtained.
10 - the height obtained gives the answer.
weight can be determined by multiplying density and volume of the cube.
for a floating body wt = upthrust
by the foumula of upthrust volume of cube submerged can be obtained grom which the height of cube submerged can be obtained.
10 - the height obtained gives the answer.
Answered by
6
(a) Let h be the height of the iron block above mercury.
In case of flotation,
Weight of the block = buoyant force
i.e., x3 ρ g = [(x - h) σ g]x2
Or h = x (1 - ρ/σ) = 10 (1 - 7.2/13.6)
= 4.70cm.
(b) Let y be the height of the mercury level.
For equilibrium of the block,
x3 ρg = [σwgy + σHg g(x - y)]x2
xσ = (x - y) σHg + y σw
or y = x (σHg-σw)/(σHg-σw)
= 10 (13.6-7.2)/(13.6-1) = 5.08cm
In case of flotation,
Weight of the block = buoyant force
i.e., x3 ρ g = [(x - h) σ g]x2
Or h = x (1 - ρ/σ) = 10 (1 - 7.2/13.6)
= 4.70cm.
(b) Let y be the height of the mercury level.
For equilibrium of the block,
x3 ρg = [σwgy + σHg g(x - y)]x2
xσ = (x - y) σHg + y σw
or y = x (σHg-σw)/(σHg-σw)
= 10 (13.6-7.2)/(13.6-1) = 5.08cm
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