A small block of wood of specific gravity 0.5 is subnerged at a depth of 1.2 m in a vessel filled with water. The vessel is accelerated upwards with an acceleration `a_(0) = (g)/(2)`. Time taken by the block to reach the surface, if it is released with zero initial velocity is `(g = 10 m//s^(2))`

Options

A. 0.6 s
B. 0.4 s
C. 1.2 s
D. 1 s

Answer

To solve the problem, we need to determine the time taken by a small block of wood with a specific gravity of 0.5 to reach the surface of a water-filled vessel that is accelerating upwards with an acceleration of \( a_0 = \frac{g}{2} \). The block is submerged at a depth of 1.2 m and is released from rest.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Specific gravity of the block, \( SG = 0.5 \)
- Depth of the block, \( h = 1.2 \, m \)
- Acceleration due to gravity, \( g = 10 \, m/s^2 \)
- Upward acceleration of the vessel, \( a_0 = \frac{g}{2} = 5 \, m/s^2 \)

2. **Calculate the Weight and Buoyant Force:**
- The weight of the block \( W = m \cdot g \)
- The volume of the block \( V = \frac{m}{\rho_{wood}} \) where \( \rho_{wood} = SG \cdot \rho_{water} = 0.5 \cdot 1000 \, kg/m^3 = 500 \, kg/m^3 \)
- The buoyant force \( F_b = \rho_{water} \cdot V \cdot g = 1000 \cdot \frac{m}{500} \cdot g = 2m \cdot g \)

3. **Determine the Net Acceleration of the Block:**
- The net force acting on the block when the vessel is accelerated upwards can be calculated using:
\[ F_{net} = F_b - W = 2mg - mg = mg \]
- The effective acceleration of the block relative to the water is:
\[ a_{block} = \frac{F_{net}}{m} = g \]
- However, since the vessel is accelerating upwards, we need to consider the effective acceleration of the block relative to the upward acceleration of the vessel:
\[ a_{relative} = g - a_0 = g - \frac{g}{2} = \frac{g}{2} = 5 \, m/s^2 \]

4. **Use Kinematic Equation to Find Time:**
- The block starts from rest and moves a distance \( h = 1.2 \, m \) under the effective acceleration \( a_{relative} = 5 \, m/s^2 \).
- Using the kinematic equation:
\[ h = \frac{1}{2} a_{relative} t^2 \]
- Plugging in the values:
\[ 1.2 = \frac{1}{2} \cdot 5 \cdot t^2 \]
- Rearranging gives:
\[ t^2 = \frac{2 \cdot 1.2}{5} = \frac{2.4}{5} = 0.48 \]
- Therefore, taking the square root:
\[ t = \sqrt{0.48} \approx 0.693 \, seconds \]

5. **Final Calculation:**
- The time taken by the block to reach the surface is approximately \( 0.693 \, seconds \).

### Summary of the Solution:
The time taken by the block to reach the surface is approximately \( 0.693 \, seconds \).