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.

Explanation

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.