AIIMS 2018 Physics Young's Modulus MCQ Question
Two wires of equal lengths are made of the same material wire. A has a diameter that is twice as that of wire B. If identical weights are suspended from the ends of these wires the increase on length is
Four times for wire A as for wire B.
Twice for wire A as for wire B.
half for wire A as for wire B.
one-fourth for wire A as for wire B.
Correct Answer
Detailed Explanation
The increase in length of a wire under tension is determined by Young's modulus (Y), which is a material property, and the formula for elongation (ΔL) is given by ΔL = (F * L) / (A * Y), where F is the force applied, L is the original length, and A is the cross-sectional area. Since wire A has a diameter twice that of wire B, its cross-sectional area (A = π(d/2)²) is four times larger, which compensates for the increased diameter when calculating elongation, resulting in the same increase in length for both wires when identical weights are applied. Therefore, options A, B, and C are incorrect as they suggest differing elongations based on diameter, neglecting the relationship between area and elongation.
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