An empty plastic box of mass 5 kg is observed to accelerate up at the rate of `g//6` when placed deep inside water. What mass of sand should be put inside the box so that it may accelerate dow at the rate of `g//6` ?

Options

A. 1 kg
B. 1.5 kg
C. 2 kg
D. 2.5 kg

Answer

To solve the problem step by step, we will analyze the forces acting on the plastic box in two scenarios: when it accelerates upwards and when we want it to accelerate downwards.

### Step 1: Analyze the first scenario (Box accelerating upwards)
1. **Identify the forces**:
- Weight of the box (downward force): \( W = mg = 5g \) (where \( m = 5 \, \text{kg} \))
- Buoyant force (upward force): \( F_b \)

2. **Given acceleration**: The box accelerates upwards with \( a = \frac{g}{6} \).

3. **Apply Newton's second law**:
\[ F_{\text{net}} = ma \]
Here, the net force is the buoyant force minus the weight of the box:
\[ F_b - mg = ma \]
Substituting the values:
\[ F_b - 5g = 5 \cdot \frac{g}{6} \]
Simplifying this gives:
\[ F_b - 5g = \frac{5g}{6} \]
Rearranging to find the buoyant force:
\[ F_b = 5g + \frac{5g}{6} = \frac{30g}{6} + \frac{5g}{6} = \frac{35g}{6} \]

### Step 2: Analyze the second scenario (Box accelerating downwards)
1. **Add mass of sand**: Let the mass of sand added be \( M \).

2. **Identify the forces**:
- Weight of the box with sand (downward force): \( W = (5 + M)g \)
- Buoyant force (upward force): \( F_b = \frac{35g}{6} \) (from the previous calculation)

3. **Given acceleration**: The box should accelerate downwards with \( a = \frac{g}{6} \).

4. **Apply Newton's second law**:
\[ F_{\text{net}} = ma \]
The net force is the weight of the box plus sand minus the buoyant force:
\[ (5 + M)g - F_b = (5 + M) \cdot \frac{g}{6} \]
Substituting \( F_b = \frac{35g}{6} \):
\[ (5 + M)g - \frac{35g}{6} = (5 + M) \cdot \frac{g}{6} \]

5. **Simplifying the equation**:
Multiply through by 6 to eliminate the fraction:
\[ 6(5 + M)g - 35g = (5 + M)g \]
This simplifies to:
\[ 30g + 6Mg - 35g = 5g + Mg \]
Rearranging gives:
\[ 30g - 35g = 5g + Mg - 6Mg \]
\[ -5g = 5g - 5Mg \]
\[ -10g = -5Mg \]
Dividing both sides by -5g:
\[ 2 = M \]

### Conclusion
The mass of sand that should be put inside the box so that it may accelerate down at the rate of \( \frac{g}{6} \) is \( \boxed{2 \, \text{kg}} \).