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

(i) In constant immersion Hydrometer, will the level of the immersion of the hydrometer ramain 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 floats in the liquid ?
By using law of floatation, find the density of the liquid.
Since, density of the mixture contains equal masses of liquid and water,
so, `rho_("mixture")=(M_(w)+M_(L))/(V_(w)+V_(L))=`?
But, `V_(L)=(M_(L))/(D_(L)) " and " V_(w)=(M_(w))/(D_(w))`
By applying, find the value of `rho_("mixture")`.
Substitute the values in,
`W=U=Vrho_("mixture"g)`
This gives the weight of the hydrometer.
(ii)34.28 g