A block of material of specific gravity 0.4 is held submerged at a depth of 1m in a vessel filled with water. The vessel is accelerated upwards with acceleration of `a_(o) = g//5`. If the block is released at t = 0, neglecting viscous effects, it will reach the water surface at t equal to `(g = 10 m/s^(2))` :

Options

A. 0.6 sec
B. 0.33 sec
C. 1.2 sec
D. 1 sec

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and calculate the time taken for the block to reach the water surface after being released.

### Step 1: Understand the Given Information
- Specific gravity of the block (SG) = 0.4
- Depth of the block in water (h) = 1 m
- Acceleration of the vessel (a₀) = g/5, where g = 10 m/s²
- Therefore, a₀ = 10/5 = 2 m/s²

### Step 2: Calculate the Effective Acceleration
When the vessel is accelerated upwards, the effective gravitational acceleration (g_net) acting on the block is the sum of the gravitational acceleration (g) and the pseudo force due to the upward acceleration of the vessel.

\[ g_{net} = g + a₀ = g + \frac{g}{5} = g + 2 = 10 + 2 = 12 \text{ m/s²} \]

### Step 3: Calculate the Density of the Block
The specific gravity (SG) is defined as the ratio of the density of the block (ρ_b) to the density of water (ρ_w).

\[ SG = \frac{\rho_b}{\rho_w} \]
Given that SG = 0.4, we can express the density of the block as:
\[ \rho_b = 0.4 \cdot \rho_w \]

### Step 4: Calculate the Buoyant Force and Weight
The buoyant force (F_b) acting on the block is given by Archimedes' principle:

\[ F_b = \rho_w \cdot V \cdot g_{net} \]

The weight (W) of the block is:

\[ W = \rho_b \cdot V \cdot g \]

### Step 5: Set Up the Equation of Motion
The net force (F_net) acting on the block when it is released is the difference between the buoyant force and the weight of the block:

\[ F_{net} = F_b - W = \rho_w \cdot V \cdot g_{net} - \rho_b \cdot V \cdot g \]

Substituting the expression for ρ_b:

\[ F_{net} = \rho_w \cdot V \cdot g_{net} - (0.4 \cdot \rho_w) \cdot V \cdot g \]

Factoring out common terms:

\[ F_{net} = V \cdot \rho_w \left(g_{net} - 0.4g\right) \]

### Step 6: Calculate the Acceleration of the Block
The mass of the block can be expressed as:

\[ m = \rho_b \cdot V = (0.4 \cdot \rho_w) \cdot V \]

Using Newton's second law (F = ma), we can find the acceleration (a) of the block:

\[ a = \frac{F_{net}}{m} = \frac{V \cdot \rho_w \left(g_{net} - 0.4g\right)}{(0.4 \cdot \rho_w) \cdot V} \]

The volume (V) cancels out:

\[ a = \frac{g_{net} - 0.4g}{0.4} \]

Substituting the values of g and g_net:

\[ a = \frac{12 - 0.4 \cdot 10}{0.4} = \frac{12 - 4}{0.4} = \frac{8}{0.4} = 20 \text{ m/s²} \]

### Step 7: Use the Equation of Motion to Find Time
Using the equation of motion \( h = ut + \frac{1}{2} a t^2 \), where initial velocity (u) = 0:

\[ 1 = 0 + \frac{1}{2} \cdot 20 \cdot t^2 \]

Solving for t:

\[ 1 = 10 t^2 \implies t^2 = \frac{1}{10} \implies t = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}} \approx 0.316 \text{ seconds} \]

### Final Answer
The time taken for the block to reach the water surface is approximately **0.316 seconds**.

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