Question

Two liquids A and B have density 1.5 g/cm³ and 0.9 g/cm³ respectively. In which case the wooden piece of density 0.68 g/cm³ will have more portion inside the liquid while floating and why?​

Answer 1

Answer:

According to Archimedes' principle.

Mass of volume displaced is by object is buoyancy force.

Let the volume of wood is V.

then the volume submerged in liquid is V/4

let the density of wood be ρ.

let the density of liquid be ρ

l

=0.8cm

−3

.

weight of wood ρVg.

buoyancy force =

4

ρ

l

V

g

now weight of block is equal to buoyancy force.

ρVg=

4

ρ

l

V

g

So ρ=

4

0.8

=.2gcm

−3

then the volume submerged in oil is 60% of is 3V/5

weight of wood ρVg

buoyancy force =

5

ρ

oil

3V

g

now weight of block is equal to buoyancy force.

ρVg=

5

ρ

oil

3V

g

So ρ

oil

=

3

ρ×5

=

3

0.2×5

=0.33gcm

−3

Answer 2

Answer:

According to Archimedes' principle.

Mass of volume displaced is by object is buoyancy force.

Let the volume of wood is V.

then the volume submerged in liquid is V/4

let the density of wood be ρ.

let the density of liquid be ρ

l

=0.8cm

−3

.

weight of wood ρVg.

buoyancy force =

4

ρ

l

V

g

now weight of block is equal to buoyancy force.

ρVg=

4

ρ

l

V

g

So ρ=

4

0.8

=.2gcm

−3

then the volume submerged in oil is 60% of is 3V/5

weight of wood ρVg

buoyancy force =

5

ρ

oil

3V

g

now weight of block is equal to buoyancy force.

ρVg=

5

ρ

oil

3V

g

So ρ

oil

=

3

ρ×5

=

3

0.2×5

=0.33gcm

−3

Step-by-step explanation:

hope it will be helpful to you dear friend