A massive stone pillar `20 m` high and of uniform cross section rests on a rigid base and supports a vertical load of `5.0 xx 10^(5)N` at its upper end. If the compressive stress in the pillar is not exceed `1.6 xx10^(6)N//m^2`, what is the minimum cross-sectional area of the pillar? (Density of the stone `=2.5xx10^(3) kg//m^(3)` take `g= 10 N//kg)`

Options

A. `0.15m^(2)`
B. `0.25m^(2)`
C. `0.35m^(2)`
D. `0.45m^(2)`

Explanation

To solve the problem step by step, we need to determine the minimum cross-sectional area of the stone pillar that can support the given load without exceeding the specified compressive stress.

### Step 1: Understand the Given Data
- Height of the pillar, \( h = 20 \, m \)
- Load on the pillar, \( F = 5.0 \times 10^5 \, N \)
- Maximum compressive stress, \( \sigma_{max} = 1.6 \times 10^6 \, N/m^2 \)
- Density of the stone, \( \rho = 2.5 \times 10^3 \, kg/m^3 \)
- Acceleration due to gravity, \( g = 10 \, N/kg \)

### Step 2: Calculate the Weight of the Pillar
The weight \( W \) of the pillar can be calculated using the formula:
\[ W = m \cdot g \]
Where \( m \) is the mass of the pillar. The mass can be calculated from the density and volume:
\[ m = \rho \cdot V \]
The volume \( V \) of the pillar can be expressed as:
\[ V = A \cdot h \]
Where \( A \) is the cross-sectional area. Thus, we can write:
\[ m = \rho \cdot (A \cdot h) \]
Substituting this into the weight formula:
\[ W = \rho \cdot (A \cdot h) \cdot g \]

### Step 3: Calculate the Total Force Acting on the Pillar
The total force \( F_{total} \) acting on the pillar is the sum of the applied load and the weight of the pillar:
\[ F_{total} = F + W = F + \rho \cdot (A \cdot h) \cdot g \]

### Step 4: Write the Stress Formula
The stress \( \sigma \) in the pillar is given by:
\[ \sigma = \frac{F_{total}}{A} \]
Substituting for \( F_{total} \):
\[ \sigma = \frac{F + \rho \cdot (A \cdot h) \cdot g}{A} \]

### Step 5: Set Up the Equation
We know that the stress must not exceed the maximum compressive stress:
\[ 1.6 \times 10^6 = \frac{5.0 \times 10^5 + \rho \cdot (A \cdot h) \cdot g}{A} \]
Substituting \( \rho \), \( h \), and \( g \):
\[ 1.6 \times 10^6 = \frac{5.0 \times 10^5 + (2.5 \times 10^3) \cdot (A \cdot 20) \cdot 10}{A} \]

### Step 6: Simplify the Equation
This simplifies to:
\[ 1.6 \times 10^6 = \frac{5.0 \times 10^5 + 5.0 \times 10^4 \cdot A}{A} \]
Multiplying through by \( A \):
\[ 1.6 \times 10^6 A = 5.0 \times 10^5 + 5.0 \times 10^4 A \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 1.6 \times 10^6 A - 5.0 \times 10^4 A = 5.0 \times 10^5 \]
\[ (1.6 \times 10^6 - 5.0 \times 10^4) A = 5.0 \times 10^5 \]
\[ (1.55 \times 10^6) A = 5.0 \times 10^5 \]

### Step 8: Solve for Area \( A \)
Now, solving for \( A \):
\[ A = \frac{5.0 \times 10^5}{1.55 \times 10^6} \]
Calculating this gives:
\[ A \approx 0.3226 \, m^2 \]

### Step 9: Conclusion
Thus, the minimum cross-sectional area of the pillar is approximately \( 0.3226 \, m^2 \).