Two rods of the same material and volume having circular cross-section are subjected to tension \( T \). Within the elastic limit, the same force is applied to both the rods. Diameter of the first rod is half of the second rod, then the extensions of the first rod to second rod will be in the ratio
Show Hint
Remember, the extension in a rod is inversely proportional to its cross-sectional area.
Step 1: Relation for extension.
The extension \( \Delta L \) in a rod under tension is given by:
\[
\Delta L = \frac{F L}{A Y}
\]
where \( F \) is the applied force, \( L \) is the length of the rod, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus.
Step 2: Area and length relation.
Since the diameters of the rods are in the ratio 1:2, the areas \( A_1 \) and \( A_2 \) are in the ratio \( (1:2)^2 = 1:4 \). Thus, the extensions are in the inverse ratio of the areas, i.e., 4:1.
Step 3: Conclusion.
Thus, the correct answer is (A) 4:1.