A hydrometer stem has a length 30 cm. If the hydrometer is immersed in water , its floatation bulb just sinks. If the same hydrometer is immersed in a liquid having density of 500 kg m^(-3), two-third of the stem is immersed. Find the least specific gravity of a liquid that can be measured using the hydrometer.

A hydrometer stem has a length 30 cm. If the hydrometer is immersed in

1.	A hydrometer stem has a length 30 cm. If the hydrometer is immersed in water , its floatation bulb just sinks. If the same hydrometer is immersed in a liquid having density of 500 kg m^(-3), two-third of the stem is immersed. Find the least specific gravity of a liquid that can be measured using the hydrometer.
Answer» Solution :Let 'V' be volume of the hydrometer upto the beginning of its setm (i.e., volume of gravity bulb+floatation bulb) Let 'a' be the area of cross SECTION of the stem. When the hydrometer is immersed in water, `V xx 1`= mass of hydrometer. DENSITY of the given liquid`=500 kg m^(-3)=0.5 g cm^(-3)` Length of the stem immersed in the liquid `=(2)/(3) ("total length")=(2)/(3)xx30=20cm` `therefore` volume of the hydrometer in the liquid =V+20a `therefore ` mass of liquid DISPLACED =(V+20a)0.5=mass of hydrometer `therefore(V+20a)0.5=V IMPLIES V+20a=2V` or `a(V)/(20)` let `RHO` be the density of the liquid on which the hydrometer floats upto its full length of stem. Then mass of hydrometer `=(V+30a)rho=V` `implies rho= (V)/(V+30a)=(V)/(V+30((V)/(20)))` `rho =0.4` `therefore ` least specific gravity of the liquid that can be measured using the hydfrometer is 0.4.