A cubical block of wood of edge a and density `rho` floats in water of density `2rho`. The lower surface of the cube just touches the free end of a mass less spring of force constant K fixed at the bottom of the vessel. The weight W put over the block so that it is completely immersed in water without wetting the weight is

Options

A. `a (a^(2) rho g + k)`
B. `a (a rho g + 2k)`
C. `a ((a rho g)/(2) + 2k)`
D. `a (a^(2) lamda g + (k)/(2))`

Answer

To solve the problem step by step, we will analyze the forces acting on the cubical block of wood when it is completely immersed in water.

### Step 1: Understand the Problem
We have a cubical block of wood with edge length \( a \) and density \( \rho \) floating in water with density \( 2\rho \). The block is completely immersed in water, and we need to find the weight \( W \) that is placed on the block to achieve this condition.

### Step 2: Calculate the Volume and Weight of the Block
The volume \( V \) of the cubical block can be calculated as:
\[ V = a^3 \]
The weight of the block \( W_{\text{block}} \) can be calculated using its density:
\[ W_{\text{block}} = \text{density} \times \text{volume} \times g = \rho \cdot a^3 \cdot g \]

### Step 3: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the block when it is fully immersed in water can be calculated using Archimedes' principle:
\[ F_b = \text{density of water} \times \text{volume of displaced water} \times g \]
Since the block is fully immersed, the volume of displaced water is equal to the volume of the block:
\[ F_b = 2\rho \cdot a^3 \cdot g \]

### Step 4: Set Up the Force Balance Equation
When the block is fully immersed, the upward buoyant force must balance the downward forces, which include the weight of the block and the additional weight \( W \):
\[ F_b = W_{\text{block}} + W \]
Substituting the expressions we found:
\[ 2\rho \cdot a^3 \cdot g = \rho \cdot a^3 \cdot g + W \]

### Step 5: Solve for the Weight \( W \)
Rearranging the equation to solve for \( W \):
\[ W = 2\rho \cdot a^3 \cdot g - \rho \cdot a^3 \cdot g \]
\[ W = \rho \cdot a^3 \cdot g \]

### Step 6: Include the Spring Force
If we consider the spring force \( F_s \) when the block is pressed down by the weight \( W \), we can express the spring force as:
\[ F_s = K \cdot x \]
where \( x \) is the compression of the spring.

### Step 7: Final Expression for Weight \( W \)
The total downward force when the spring is compressed will be:
\[ W + W_{\text{block}} = F_b + F_s \]
Substituting the values we have:
\[ W + \rho \cdot a^3 \cdot g = 2\rho \cdot a^3 \cdot g + K \cdot x \]
From this, we can derive the final expression for \( W \):
\[ W = 2\rho \cdot a^3 \cdot g - \rho \cdot a^3 \cdot g - K \cdot x \]
This leads to the final expression for \( W \) in terms of \( a \), \( \rho \), \( g \), and \( K \).

### Conclusion
The weight \( W \) that must be placed on the block to keep it fully immersed in water without wetting the weight is given by:
\[ W = \rho \cdot a^3 \cdot g + K \cdot x \]