Question

the length of the stem of a hydrometer is 10cm and the volume of the stem is 10 cm3.while floating in pure water,it sinks to the 5cm mark.what will be the density of the liquid in which the hydrometer

• sinks to the top of the stem
• sinks to the bottom of the stem

Answer 1

Here,

Length = 10 cm

Let mass of the hydrometer = M

Total volume of the hydrometer = 10 cm³

Now since the hydrometer sinks to 5cm mark that is half of the volume of the stem is under water, the volume of hydrometer under water is 5cm³

Total volume of hydrometer undermeter under water = 5 cm³

So water displaced = 5cm³

Mass of water displaced = 1× 5 gram

Now weight of water displaced = buoyant force

Mg = 5g

=> M = 5 ---1.

1. Let the density of the liquid where hyrometer sinks to the top be ρ

Volume of liquid displaced = 10cm³

Mg = ρg(10)

=> M = ρ(10)

=> 5 = ρ(10)

=> ρ = 0.5 g/c.c.

1. Let the density of the liquid where hyrometer sinks to the top be ρ

Volume of liquid displaced = 10cm³

Mg = ρg(10)

=> M = ρ(10)

=> 5 = ρ(10)

=> ρ = 0.5 g/c.c.

2. The 0 cm mark of the scale corresponds to a density of 0.5 g/c.c

5^th mark of the scale = 1 g/c.c (because the hydrometer floats in water sinking 5cm mark)

Thus setting up the scale division, per unit division the corresponding density becomes

(1-0.5)/5 = 0.1 g/c.c.

Now the scale corresponding to length of the hydrometer stem is given by

ρ = 0.1x + 0.5

This scale fits our observation, as because when x = 0, ρ = 0.5 then if x = 5, ρ = 1

Now if bottom of the scale is above the liquid density of the liquid will be corresponding to x = 10.

ρ = 0.1×10 + 0.5 = 1.5 g/c.c

Thus density of the liquid = 1.5 g/cc