Question

Consider an incompressible cylindrical tissue with a diameter of 2 cm and a height of 3 cm. If this tissue is stretched by 10% in the axial direction, its diameter in the stretched configuration is cm.
Assume homogeneous deformation of the tissue.

• A. \(2.4\)
• B. \(2.2\)
• C. \(1.6\)
• D. \(1.9\)

Hint:

Use volume conservation (incompressibility) to relate the change in height to the change in radius.

Answer 1

The Correct Option is D

Step 1: Set up the incompressibility condition.
For an incompressible material, the total volume stays constant during deformation. Modeling the tissue as a cylinder, its volume is
\[ V = \pi r^2 h \]
where \(r\) is the radius and \(h\) is the height.

Step 2: Write down the initial dimensions.
Initial diameter \(d_0 = 2\) cm, so initial radius \(r_0 = 1\) cm. Initial height \(h_0 = 3\) cm.
Initial volume:
\[ V_0 = \pi (1)^2 (3) = 3\pi \text{ cm}^3 \]

Step 3: Find the new height after axial stretching.
The tissue is stretched by 10% along its axis (the height direction), so the axial stretch ratio is \(\lambda_z = 1.10\).
\[ h_1 = h_0 \times 1.10 = 3 \times 1.10 = 3.3 \text{ cm} \]

Step 4: Apply the incompressibility condition to find the new radius.
Since volume is conserved and the deformation is homogeneous,
\[ \pi r_1^2 h_1 = \pi r_0^2 h_0 \]
\[ r_1^2 = r_0^2 \times \frac{h_0}{h_1} = (1)^2 \times \frac{3}{3.3} = \frac{1}{1.1} = 0.9091 \]
\[ r_1 = \sqrt{0.9091} = 0.9535 \text{ cm} \]

Step 5: Convert radius back to diameter.
\[ d_1 = 2 r_1 = 2 \times 0.9535 = 1.907 \text{ cm} \]
Rounding to one decimal place gives about \(1.9\) cm, matching option (D).

Step 6: Check why the other options are wrong.
Options (A) 2.4 and (B) 2.2 would both increase the diameter, but stretching a fixed volume of tissue along its length must shrink its cross-section, not expand it. Option (C) 1.6 is too small; it would need a much larger axial stretch than 10%.

Final Answer:
\[ \boxed{d_1 \approx 1.9 \text{ cm}} \]