The specific gravity of a liquid is to be measured by a hydrometer. A relation between specific gravity and the vertical distance from the reference level is to be obtained, and the amount of lead that needs to be added into the tube for a certain hydrometer is to be determined.
Assumptions 1 The weight of the glass tube is negligible relative to the weight of the lead added. 2 The curvature of the tube bottom is disregarded.
Properties We take the density of pure water to be 1000 kg/m³.
Analysis (a) Noting that the hydrometer is in static equilibrium, the buoyant force F_{\text {B }} exerted by the liquid must always be equal to the weight W of the hydrometer. In pure water (subscript w), we let the vertical distance between the bottom of the hydrometer and the free surface of water be z_{0} . Setting F_{B, w}=W in this case gives
W_{\text {hydro }}=F_{B, w}=\rho_{w} g V_{\text {sub }}=\rho_{w} g A z_{0} (1)
where A is the cross-sectional area of the tube, and \rho_{w} is the density of pure water.
In a fluid lighter than water \left(\rho_{f}<\rho_{w}\right), the hydrometer will sink deeper, and the liquid level will be a distance of \Delta z \text { above } z_{0} \text {. Again setting } F_{B}=W \text { gives }
W_{\text {hydro }}=F_{B, f}=\rho_{f} g V_{\text {sub }}=\rho_{f} g A\left(z_{0}+\Delta z\right) (2)
This relation is also valid for fluids heavier than water by taking Δz to be a negative quantity. Setting Eqs. (1) and (2) here equal to each other since the weight of the hydrometer is constant and rearranging gives
\rho_{w} g A z_{0}=\rho_{f} g A\left(z_{0}+\Delta z\right) \rightarrow \quad SG _{f}=\frac{\rho_{f}}{\rho_{w}}=\frac{z_{0}}{z_{0}+\Delta z}
which is the relation between the specific gravity of the fluid and Δz. Note that z_{0} is constant for a given hydrometer and Δz is negative for fluids heavier than pure water.
(b) Disregarding the weight of the glass tube, the amount of lead that needs to be added to the tube is determined from the requirement that the weight of the lead be equal to the buoyant force. When the hydrometer is floating with half of it submerged in water, the buoyant force acting on it is
F_{B}=\rho_{w} g V_{\text {sub }}
Equating F_{B} to the weight of lead gives
W=m g=\rho_{w} g V_{\text {sub }}
Solving for m and substituting, the mass of lead is determined to be
m=\rho_{w} V_{\text {sub }}=\rho_{w}\left(\pi R^{2} h_{\text {sub }}\right)=\left(1000 kg / m ^{3}\right)\left[\pi(0.005 m )^{2}(0.1 m )\right]= 0 . 0 0 7 8 5 kg
Discussion Note that if the hydrometer were required to sink only 5 cm in water, the required mass of lead would be one-half of this amount. Also, the assumption that the weight of the glass tube is negligible is questionable since the mass of lead is only 7.85 g.
