A uniform shear force of magnitude 0.01 Newton (N) is applied to the top surface of a cubical tissue sample with side of length 1 cm (marked with dashed lines in the figure below). In the deformed configuration (marked with solid lines), the angle \(\theta = 5\) degrees. The shear modulus of the tissue is kilopascals (kPa). Assume the tissue to be a linear, isotropic and homogenous elastic solid.
Step 1: Identify the shear stress.
Shear stress is the applied force divided by the area over which it acts, parallel to that area. Here the force \(F = 0.01\) N acts on the top face of the cube, whose area is
\[ A = (1\ \text{cm})\times(1\ \text{cm}) = 1\ \text{cm}^2 = 1\times10^{-4}\ \text{m}^2 \]
So the shear stress is
\[ \tau = \frac{F}{A} = \frac{0.01}{1\times10^{-4}} = 100\ \text{Pa} \]
Step 2: Identify the shear strain.
For a cube sheared through an angle \(\theta\) as shown, the shear strain \(\gamma\) is defined as the tangent of the angle of distortion between the originally vertical face and its deformed position.
\[ \gamma = \tan(\theta) = \tan(5^{\circ}) \]
Using \(\tan(5^{\circ}) \approx 0.0875\):
\[ \gamma \approx 0.0875 \]
Step 3: Apply Hooke's law for shear.
For a linear, isotropic, homogeneous elastic solid, the shear modulus \(G\) relates shear stress and shear strain by
\[ G = \frac{\tau}{\gamma} \]
Substituting the values found above:
\[ G = \frac{100}{0.0875} \approx 1143\ \text{Pa} \]
Step 4: Convert to kilopascals.
\[ G \approx 1143\ \text{Pa} = 1.143\ \text{kPa} \approx 1.14\ \text{kPa} \]
Final Answer:
The shear modulus is about 1.14 kPa, matching option (A). The other options (10.40, 5.14, 3.14) come from errors such as using the wrong power of ten in the area conversion or mixing up the strain definition.
\[ \boxed{G \approx 1.14\ \text{kPa}} \]
