Question

Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a smaller average density than box B. When the boxes are released, they accelerate upward to the surface. Which box has the greater acceleration?

Answer 1

Answer:

Box A

Explanation:

Let mass of each wooden box =m

Density of box A=$\rho_A$

Density of box B =$\rho_B$

$\rho_A$

Density,$\rho=\frac{m}{V}$

$V_A=\frac{m}{\rho_A}$

$V_B=\frac{m}{\rho_B}$

Density of inversely proportional to volume.

The volume of box with smaller density is larger than the volume of box with large density.

$V_A>V_B$

When the boxes are submerged under water.

Then, the buoyant force=$\rho_l Vg$

Where V=Volume of displace fluid.

$\rho=$Density of fluid

Buoyant force of box A=$\rho_lV_Ag$

Buoyant force of box  B=$\rho_lV_Bg$

Force=Buoyant force

$ma_A=\rho_lV_Ag$

$ma_B=\rho_lV_Bg$

Acceleration is directly proportional to volume.

Therefore,the box with large volume has greater acceleration.

Hence, the box A has greater acceleration.

Answer 2

Answer:

Explanation:

Let boxes A and B has equal mass of M and volume be Va and Vb

their density be da and db

given da < db

M / Va < M/Vb

Va> Vb

Buoyant force on box A in water = weight of displaced water

= volume of displaced water x density of water x g

= Va x d x g , d is density of water

Aa  = acceleration on box A

= buoyant force on A  / mass of A  

= Va x d x g / M

Similarly acceleration on box B

Ab = Va x d x g / M

Since Va > Vb

Aa > Ab

So box A will have greater acceleration.