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Hydrometer use and stability of floating bodies

Short answer

A hydrometer floats to measure liquid density; stem scale reads specific gravity. For floating bodies, stability requires metacentre M above centre of gravity G (GM > 0), giving a restoring moment W·GM sinθ.

Explain the use of hydrometer and shortly explain the conditions of stability of floating bodies.

Hydrometer

A hydrometer is a floating instrument used to find the specific gravity (or density) of a liquid directly. It is a sealed glass tube with a weighted bulb at the bottom (so that it floats upright) and a graduated stem at the top.

Principle: it floats in equilibrium, so weight = buoyant force. Its weight WW is constant, so a denser liquid needs less displaced volume and the hydrometer floats higher; a lighter liquid makes it sink deeper.

W=ρgVsub⇒ρ=WgVsubW = \rho g V_{sub}\quad\Rightarrow\quad \rho = \frac{W}{g V_{sub}}

If the stem has area aa and sinks by an extra length xx, then Vsub=V0+axV_{sub} = V_0 + a x. The scale is calibrated in specific gravity, which is read at the liquid surface. The scale is not uniform: because the hydrometer floats higher in a denser liquid, the surface cuts the stem lower down, so the low values are at the top of the stem and the high values at the bottom, and the divisions get closer together towards the bottom (high-density end).

Uses: testing milk, acid, battery electrolyte, alcohol, sugar solutions and sea water.

      |  <- stem with scale (reads S)
 ~~~~~|~~~~~ liquid surface
      |
    (   )  <- bulb
    (___)  <- lead shots (ballast)

Stability of floating bodies

A body floats in equilibrium when weight = buoyant force and the centre of gravity GG and the centre of buoyancy BB lie on the same vertical line. For stability, a small tilt must produce a restoring couple. The metacentre M is the point where the line of action of buoyancy through the new centre of buoyancy cuts the original vertical axis.

ConditionPositionResult
Stable equilibriumM above G (GM>0GM > 0)Restoring couple, body returns
Unstable equilibriumM below G (GM<0GM < 0)Overturning couple, body capsizes
Neutral equilibriumM coincides with G (GM=0GM = 0)Body stays in the new position
GM=IV−BGGM = \frac{I}{V} - BG

where II is the second moment of the waterline area about the longitudinal axis, VV is the volume of liquid displaced and BGBG is the distance between BB and GG (GG above BB). The restoring moment is W⋅GMsin⁡θW\cdot GM\sin\theta. A body with a large metacentric height is more stable, but rolls quickly.

✓ Answer checked by Claude Opus 5.5 (high effort)

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