A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water?
[Take, density of water `= 10^(3) kg//m^(3)`]

Options

A. 87.5 kg
B. 30.1 kg
C. 46.3 kg
D. 65.4 kg

Explanation

To solve the problem, we need to determine the maximum weight that can be placed on a cubical block of side 0.5 m that floats on water with 30% of its volume submerged. We will use the principles of buoyancy and Archimedes' principle to find the solution.

### Step-by-Step Solution:

1. **Calculate the Volume of the Cubical Block:**
The volume \( V \) of a cube is given by the formula:
\[ V = \text{side}^3 \]
For a side length of 0.5 m:
\[ V = (0.5 \, \text{m})^3 = 0.125 \, \text{m}^3 \]

2. **Determine the Volume of Water Displaced:**
Since 30% of the block is submerged, the volume of water displaced \( V_d \) is:
\[ V_d = 0.3 \times V = 0.3 \times 0.125 \, \text{m}^3 = 0.0375 \, \text{m}^3 \]

3. **Calculate the Weight of the Water Displaced:**
The weight of the water displaced \( W_d \) can be calculated using the density of water (\( \rho = 1000 \, \text{kg/m}^3 \)) and the gravitational acceleration (\( g \approx 9.81 \, \text{m/s}^2 \)):
\[ W_d = \rho \cdot V_d \cdot g = 1000 \, \text{kg/m}^3 \cdot 0.0375 \, \text{m}^3 \cdot 9.81 \, \text{m/s}^2 \]
\[ W_d = 1000 \cdot 0.0375 \cdot 9.81 = 368.25 \, \text{N} \]

4. **Determine the Weight of the Block:**
The weight of the block \( W_b \) when it is floating is equal to the weight of the water displaced:
\[ W_b = W_d = 368.25 \, \text{N} \]

5. **Calculate the Maximum Additional Weight:**
Let \( W \) be the maximum weight that can be added to the block without fully submerging it. When the maximum weight is added, the total weight of the block plus the additional weight should equal the weight of the water displaced when the block is fully submerged (100% submerged):
\[ W + W_b = \rho \cdot V \cdot g \]
The weight of the water displaced when fully submerged:
\[ W_{full} = 1000 \cdot 0.125 \cdot 9.81 = 1226.25 \, \text{N} \]

6. **Set Up the Equation:**
\[ W + 368.25 = 1226.25 \]

7. **Solve for \( W \):**
\[ W = 1226.25 - 368.25 = 858 \, \text{N} \]

8. **Convert Weight to Mass:**
To find the mass \( m \) that corresponds to the weight \( W \):
\[ W = m \cdot g \implies m = \frac{W}{g} = \frac{858}{9.81} \approx 87.5 \, \text{kg} \]

Thus, the maximum weight that can be placed on the block without fully submerging it under water is approximately **87.5 kg**.