A cubical block 'A' of mass `m_(0)(=a^(3)rho=3 kg)` of edge 'a' and density `rho` floats in a liquid of density `3rho`. The lower surface of the cube just touches the free end of massless spring of spring of spring constant `k(=a^(2)rho g)` fixed at the bottom of the vessel. Another block 'B' of mass 'm' is lept over block 'A', so that is completely immersed in liquid without wetting the block 'B'. Then find the value of m in kg.

Answer

To solve the problem step by step, we need to analyze the forces acting on the cubical block 'A' and the additional block 'B' that is placed on top of it.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of block 'A', \( m_0 = 3 \, \text{kg} \)
- Density of block 'A', \( \rho \)
- Edge length of block 'A', \( a \)
- Density of liquid, \( \text{Density} = 3\rho \)
- Spring constant, \( k = a^2 \rho g \)

2. **Calculate the Volume of Block 'A':**
\[ V_A = a^3 \]

3. **Calculate the Weight of Block 'A':**
\[ W_A = m_0 g = 3g \, \text{N} \]

4. **Calculate the Buoyant Force Acting on Block 'A':**
- The buoyant force \( F_B \) is given by Archimedes' principle:
\[ F_B = \text{Density of liquid} \times \text{Volume submerged} \times g \]
- Initially, block 'A' is floating, and let \( x \) be the submerged height. The volume submerged is \( A \cdot x = a^2 x \):
\[ F_B = 3\rho \cdot a^2 x \cdot g \]

5. **Equilibrium Condition for Block 'A':**
- At equilibrium, the weight of block 'A' equals the buoyant force:
\[ W_A = F_B \implies 3g = 3\rho a^2 x g \]
- Cancel \( g \) from both sides:
\[ 3 = 3\rho a^2 x \implies x = \frac{1}{\rho a^2} \]

6. **Determine the Height of Block 'A' Above the Liquid:**
- The height of block 'A' above the liquid is:
\[ h = a - x = a - \frac{1}{3}a = \frac{2}{3}a \]

7. **When Block 'B' is Placed on Block 'A':**
- When block 'B' of mass \( m \) is placed on block 'A', it becomes fully submerged. The total downward force now includes the weight of block 'B':
\[ W_{total} = W_A + mg = 3g + mg \]

8. **Calculate the New Buoyant Force:**
- The entire volume of block 'A' is submerged, so:
\[ F_B = 3\rho \cdot a^3 \cdot g \]

9. **Set Up the New Equilibrium Condition:**
- The new equilibrium condition gives:
\[ 3g + mg = 3\rho a^3 g \]

10. **Substituting Known Values:**
- We know \( m_0 = 3 \, \text{kg} \) corresponds to \( \rho a^3 = 3 \):
\[ 3 + m = 9 \implies m = 9 - 3 = 6 \, \text{kg} \]

11. **Final Result:**
- The value of mass \( m \) is:
\[ m = 6 \, \text{kg} \]