A marble column with a cross-sectional area of 27.2 cm2 supports a load of 7.0 × 104 N. The marble has a Young’s modulus of 60 GPa and a compressive strength of 200 MPa.

If the column is 2.00 m high, how much is its length changed by supporting the load?

m

To determine how much the length of the marble column changes under the given load, we’ll use the formula for deformation under compression:

\[
\Delta L = \frac{F \cdot L}{A \cdot E}
\]

where:
- \( F \) is the applied force (load),
- \( L \) is the original length of the column,
- \( A \) is the cross-sectional area,
- \( E \) is the Young’s modulus of the material.

Let's go through each step:

### 1. Convert Units
We have:
- \( A = 27.2 \, \text{cm}^2 = 27.2 \times 10^{-4} \, \text{m}^2 = 0.00272 \, \text{m}^2 \),
- \( E = 60 \, \text{GPa} = 60 \times 10^9 \, \text{Pa} \),
- \( F = 7.0 \times 10^4 \, \text{N} \),
- \( L = 2.00 \, \text{m} \).

### 2. Calculate the Deformation
Using the formula:

\[
\Delta L = \frac{(7.0 \times 10^4) \times 2.00}{0.00272 \times 60 \times 10^9}
\]

#### Step-by-Step Calculation
1. Compute \( A \cdot E \):
   \[
   0.00272 \times 60 \times 10^9 = 1.632 \times 10^8 \, \text{N/m}^2
   \]
   
2. Compute \( F \cdot L \):
   \[
   (7.0 \times 10^4) \times 2.00 = 1.4 \times 10^5 \, \text{N} \cdot \text{m}
   \]

3. Calculate \( \Delta L \):
   \[
   \Delta L = \frac{1.4 \times 10^5}{1.632 \times 10^8} \approx 8.58 \times 10^{-4} \, \text{m} = 0.858 \, \text{mm}
   \]

### Answer
The length of the column changes by approximately **0.858 mm** under the load.

Would you like more details on the calculation process? 

#### Related Questions
1. How would the deformation change if the cross-sectional area were doubled?
2. What would happen to the deformation if the Young’s modulus of marble was 120 GPa?
3. How does Young's modulus relate to the material’s compressive strength?
4. What is the significance of the compressive strength in this context?
5. What would be the deformation if the column height was halved?

#### Tip:
For materials under stress, remember that Young’s modulus quantifies stiffness—higher values mean less deformation under a given load.

A marble column with a cross-sectional area of 27.2 cm² supports a load of 7.0 × 10⁴ N. The marble has a Young’s modulus of 60 GPa and a compressive strength of 200 MPa. If the column is 2.00 m high, how much is its length changed by supporting the load?

Find out how to calculate the change in length of a marble column under a 7.0 × 10⁴ N load with a cross-sectional area of 27.2 cm². This problem explores key physics concepts like stress, strain, and Young’s modulus with a step-by-step solution. Ideal for Grades 11-12.

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