Question:medium

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.

Show Hint

Use volume conservation (incompressibility) to relate the change in height to the change in radius.
Updated On: Aug 7, 2026
  • \(2.4\)
  • \(2.2\)
  • \(1.6\)
  • \(1.9\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Use stretch ratios instead of computing volumes directly.
In continuum mechanics, incompressibility means the product of the three stretch ratios along the three perpendicular directions equals $1$:
\[ \lambda_r \cdot \lambda_\theta \cdot \lambda_z = 1 \]
where $\lambda_z$ is the stretch along the cylinder's axis and $\lambda_r$, $\lambda_\theta$ are the stretches along the radial and circumferential directions.

Step 2: Use symmetry to simplify.
Because the cylinder is stretched only along its axis, with no directional preference in the cross-section, the radial and circumferential stretches must be equal: $\lambda_r = \lambda_\theta$. Call this common value $\lambda_{lat}$.
Substituting into the incompressibility condition:
\[ \lambda_{lat}^2 \cdot \lambda_z = 1 \implies \lambda_{lat} = \frac{1}{\sqrt{\lambda_z}} \]

Step 3: Plug in the given axial stretch.
The tissue is stretched by 10%, so $\lambda_z = 1.10$.
\[ \lambda_{lat} = \frac{1}{\sqrt{1.10}} = \frac{1}{1.0488} = 0.9535 \]

Step 4: Apply this lateral stretch ratio directly to the diameter.
Since diameter scales the same way as radius, the new diameter is the old diameter times $\lambda_{lat}$:
\[ d_1 = d_0 \times \lambda_{lat} = 2 \times 0.9535 = 1.907 \text{ cm} \]

Step 5: Round and match to the options.
$1.907$ cm rounds to $1.9$ cm, which is option (D).

Final Answer:
The stretched diameter is about $1.9$ cm.
\[ \boxed{1.9 \text{ cm}} \]
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