A cylindrical specimen is subjected to plastic deformation in tension up to a uniform elongation of 10%. The final cross-sectional area of the gage section is found to be 20 mm\(^2\). The initial cross-sectional area of the gage section is _________ mm\(^2\) (rounded off to an integer).
The elongation (or strain) is given as 10%, which means the final area is 90% of the original area (because 10% elongation corresponds to 10% reduction in cross-sectional area). Let \( A_0 \) be the initial area and \( A_f \) be the final area. We have: \[ A_f = A_0 \times (1 - {strain}) \] Substituting the values: \[ 20 = A_0 \times (1 - 0.10) \] \[ 20 = A_0 \times 0.90 \] Solving for \( A_0 \): \[ A_0 = \frac{20}{0.90} = 22.22 \, {mm}^2 \] Thus, the initial cross-sectional area of the gage section is 22 mm\(^2\).
Answer: 22 mm\(^2\).
Bird : Nest :: Bee : __________
Select the correct option to complete the analogy.