Question

A piece of wood is taken deep inside a long column of water and released. It will move up
(a) with a constant upward acceleration
(b) with a decreasing upward acceleration
(c) with a deceleration
(d) with a uniform velocity

Answer 1

A piece of wood is taken deep inside a long column of water and released. It will move up with  a constant upward acceleration, option a is correct.

Explanation:

1. When object is submerged in liquid, an upward force acts on object by liquid. This force is known as buoyant force.

2. Now come to question, here piece of wood in water. There are two force act on object. One is gravitational force act in downward direction while second force is buoyant force act in upward direction.

3. Where

   Gravitational force(W) = mg = $\rho _{o}\times V\times g$   ...1)

  $\rho _{o}$ = density of wood.

  V = volume of wood

4. Buoyant force $F_{b}= \rho _{w}\times V\times g$      ...2)

  $\rho _{w}$ = density of water

  V = volume of wood

5. Net force act in upward direction density of water is greater than density of water.

6. so net force $F_{NET}= F_{b}-W$ = m×a

   From we see acceleration direction is also in direction of net force with constant force.

Answer 2

Conclusion:

With decreasing upward acceleration

Explanation:

When we release the wooden block inside a long column of water there are two force will act on block, one is body weight acts downward and other Buoyant force acts upward then resultant will be  $\textrm{Buoyant force} -\textrm{body weight}$.

 And initial acceleration will be $\frac{\textrm{Buoyant force} - \textrm{body weight}}{\textrm{mass of the wooden block}}$

When wooden block starts move to upward the viscous force will act on it downward, velocity increases then viscous force also increases and the net force on block is $\textrm{Buoyant force} -\textrm{body weight} -\textrm{Viscous force}$.

 The variable force is $\textrm{Viscous force}$ and it will increase that's why wooden block With decreasing upward acceleration.