When a hydrometer having a stem diameter of .3 in is place in water, the stem protrudes 3.15 in above the water surface. If the water is replaced with a liquid having a specific gravity of 1.1, how much of the stem would protrude above the liquid surface? The hydrometer weighs .042 lb.
3 Answers

The result depends on the mass (0.042 lb).
Let v(w) be the volume of water displaced, and v(l) be the volume of another liquid displaced.
By Archimedes' law, the volume of water displaced by the hydrometer can be calculated in the following way
In water,
v(w) = mass/density
v(w) = (0.042 lb)(1 ft^3/62.42 lb)(12 in/1 ft)^3 = 1.1627 in^3
In another liquid,
v(l) = v(w)/1.1 = 1.0570 in^3
The change in reduced volume equals to the change of the volume above the water.
v(w)v(l) = pi r^2 ∆h
Solve for ∆h,
∆h = (1.16271.0570)/(.15^2 pi) = 1.50 in
h = 3.15+1.50 = 4.65 in, the height of the stem above the liquid surface.

Let l = 3.15 in above the water, then in the other fluid L = (Rho/rho)l = (1.1rho/rho)l = 1.1(3.15) = 3.46 in above the surface of the other liquid. Rho = 1.1 rho, which is the specific gravity of the other liquid and rho is the water density.
The above follows from W = mg = rho g V = Rho g v = B; where W is the hydrometer weight, rho is water density, V = aD is displaced volume of water, Rho is other liquid density, v = ad is displaced volume of other liquid, and B is buoyancy. The a is the cross sectional area of the hydrometer and the D and d are the respective depths the gauge sinks into the liquids.
Then rho/Rho = v/V = ad/aD = d/D so that D = (rho/Rho)d; where D is the depth the hydrometer sinks in the other liquid, d is the depth in water. In words, this means the hydrometer will sink less in the other more dense liquid than in the water.
Naturally, if it sinks less in the denser liquid more of the hydrometer will show above the surface. As the depth the hydrometer sinks is just the complement of how much shows above the surface, we can show that where rho/Rho = d/D for the depth, Rho/rho = L/l for the length above the surface. That is, proportionally more length above shows when less depth is reached. And that is why the length in the denser fluid is L = (Rho/rho)l = 1.1 X 3.15 in.

The analyzing won't exchange. yet, i assume if the fish is splashing around the dimensions will bounce. there is an previous humorous tale that if a fish is swimming interior the water, it would not upload any weight. this is no longer actual. including a fish is particularly like including greater water. they have on the subject of an identical density. And, confident it somewhat is buoyant rigidity that keeps the fish off the backside.