A constant immersion hydrometer floats vertically in water when the weight of its float is 30 g. When it is made to float in a liquid (miscible in water), the weight of its float is increased by 10 g. Find the weight of the hydrometer when it is floated in a mixture containing equal masses of water and the liquid.

Explanation

In constant immersion Hydrometer, will the level of the immersion of the hydrometer remain same (constant)?
Find the weight of the displaced water using law of floatation Is the volume of the displaced water = Volume of the immersed part of hydrometer = 30 `cm^3` What is the weight of the hydrometer when it Ooats in the liquid? By using law of floatation, find the density of the liquid. Since, density of the mixture contains equal m= of liquid and water,
SO `p_("mixture")`= `(M_w+M_L)/(V_W+V_L)`
But `V_1=M_L/D_L` and `V_W = (Mw)/(D_w)`
By applying, find the value of `P_("mixture")` Substitute the values in
W=U=V_(p_("mixture"))`
`W=U=V_(p_("mixture"))`
This gives the weight of the hydrometer
34.28 g