Question

A hydrometer stem has a length 0.3 m. If the hydrometer is immersed in water, its floating bulb just sinks. If the same hydrometer is immersed in a liquid of density 0.5 g cm^–3, two third of the stem is immersed. Determine the least specific gravity of a liquid that can be measured using the hydrometer.

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Answer 1

The correct option is D

0.4

Let the conditions of hydrometer be displayed below ;the hydrometer's bulb is immersed in water of density

$$\begin{gathered} 1\mathrm{g}/{\mathrm{cm}}^3\mathrm{with}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{stem}=30\mathrm{cm} \\ \mathrm{let}\mathrm{area}\mathrm{of}\mathrm{cross}\mathrm{section}\mathrm{of}\mathrm{stem}=\mathrm{a} \\ \mathrm{Volume}=\mathrm{V} \end{gathered}$$

Now when the density of the liquid is $0.5\mathrm{g}/{\mathrm{cm}}^3$ in which the hydrometer is immersed ,two thirds of the stem is immersed which is shown below

Also least specific gravity exists for hydrometer which immerses completely or sinks completely inside the liquid
When we look at initial stage m the hydrometer upon balancing force ;
$$\begin{gathered} \mathrm{upward}\mathrm{force}=\mathrm{downward}\mathrm{force} \\ {\mathrm{\rho }}_{\mathrm{l}}{\mathrm{V}}_{\mathrm{l}}\mathrm{g}=\mathrm{mg} \\ \Rightarrow 0.1\times {\mathrm{V}}_{\mathrm{l}}=\mathrm{m} \\ \mathrm{m}=\mathrm{V}\mathrm{grams} \\ \mathrm{Now}\mathrm{when}\mathrm{looking}\mathrm{at}\mathrm{second}\mathrm{case}\mathrm{the}\mathrm{upward}\mathrm{force}=\mathrm{downward}\mathrm{force} \\ {\mathrm{\rho }}_{\mathrm{l}}{\mathrm{V}}_{\mathrm{l}}\mathrm{g}=\mathrm{mg} \\ \Rightarrow 0.5[\mathrm{V}+20\mathrm{a}]=\mathrm{V} \\ \mathrm{V}+20\mathrm{a}=2\mathrm{V} \end{gathered}$$

Now when looking at third case; where the hydrometer is completely immersed in liquid

$$\begin{gathered} \mathrm{here}\mathrm{too}\mathrm{upward}\mathrm{force}=\mathrm{downward}\mathrm{force} \\ {\mathrm{\rho }}_{\mathrm{l}}{\mathrm{V}}_{\mathrm{l}}\mathrm{g}=\mathrm{mg} \\ \mathrm{\rho }[\mathrm{V}+30\mathrm{a}]=\mathrm{V}(\mathrm{as}\mathrm{mass}\mathrm{m}=\mathrm{V}) \\ \mathrm{by}\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\text{'}\mathrm{a}\text{'}\mathrm{in}\mathrm{above}\mathrm{equation}\mathrm{we}\mathrm{get} \\ \Rightarrow \mathrm{\rho }=\frac{20\mathrm{V}}{50\mathrm{V}} \\ \mathrm{\rho }=0.4\mathrm{g}/{\mathrm{cm}}^3 \\ \mathrm{Now}\mathrm{the}\mathrm{relative}\mathrm{density}\mathrm{or}\mathrm{specific}\mathrm{gravity}\mathrm{of}\mathrm{liquid}=\frac{\mathrm{density}\mathrm{of}\mathrm{liquid}}{\mathrm{density}\mathrm{of}\mathrm{water}\mathrm{at}{1}^{\circ }\mathrm{C}} \\ =\frac{0.4}{1}=0.4\mathrm{g}/{\mathrm{cm}}^3 \end{gathered}$$

Hence option (c) is the correct answer.