Between each pair of vertebrae in the spinal column is a cylindrical disc of cartilage. Typically, this disc has a radius of about `3.0xx10^(-2)m` and a thickness of about `7.0xx10^(-3)m`. The shear modulus of cartilage is `1.2xx10^(7)N//m^(2)`. Suppose a shearing force of magnitude 11 N is applied parallel to the top surface of the disc while the bottom surface remains fixed in place. How far does the top surface move relative to the bottom surface ?

Options

A. `2.0xx10^(-5)m`
B. `4.1xx10^(-5)m`
C. `2.3xx10^(-6)m`
D. `3.5xx10^(-6)m`

Answer

To find out how far the top surface of the cylindrical disc of cartilage moves relative to the bottom surface when a shearing force is applied, we can use the relationship defined by the shear modulus. Here’s a step-by-step solution:

### Step 1: Understand the Shear Modulus Formula
The shear modulus \( S \) is defined as:
\[ S = \frac{\text{Shear Stress}}{\text{Shear Strain}} \]
Where:
- Shear Stress \( \tau = \frac{F}{A} \)
- Shear Strain \( \gamma = \frac{\Delta x}{L} \)

### Step 2: Rearrange the Formula
We can rearrange the formula to find the displacement \( \Delta x \):
\[ S = \frac{F/A}{\Delta x/L} \implies \Delta x = \frac{F \cdot L}{S \cdot A} \]

### Step 3: Calculate the Area \( A \)
The area \( A \) of the circular disc is given by:
\[ A = \pi r^2 \]
Where \( r = 3.0 \times 10^{-2} \, m \). Therefore:
\[ A = \pi (3.0 \times 10^{-2})^2 = \pi \times 9.0 \times 10^{-4} \approx 2.827 \times 10^{-3} \, m^2 \]

### Step 4: Substitute the Known Values
Now we substitute the known values into the rearranged formula:
- \( F = 11 \, N \)
- \( L = 7.0 \times 10^{-3} \, m \)
- \( S = 1.2 \times 10^{7} \, N/m^2 \)

Substituting these values:
\[ \Delta x = \frac{11 \, N \cdot 7.0 \times 10^{-3} \, m}{1.2 \times 10^{7} \, N/m^2 \cdot 2.827 \times 10^{-3} \, m^2} \]

### Step 5: Calculate \( \Delta x \)
Calculating the numerator:
\[ 11 \cdot 7.0 \times 10^{-3} = 0.077 \, N \cdot m \]
Calculating the denominator:
\[ 1.2 \times 10^{7} \cdot 2.827 \times 10^{-3} \approx 33980.4 \, N \]
Now, substituting these into the equation:
\[ \Delta x = \frac{0.077}{33980.4} \approx 2.27 \times 10^{-6} \, m \]

### Final Answer
Thus, the top surface moves approximately:
\[ \Delta x \approx 2.27 \, \mu m \]

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