The Young’s modulus of three materials are in the ratio 2:2:1. Three wires made of these materials have their cross-sectional areas in the ratio 1:2:3. For a given stretching force the elongation's in the three wires are in the ratio

1:2:3

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3:2:1

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5:4:3

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6:3:4

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Solution

The correct option is D 6:3:4
$l = \frac{F L}{A Y}$ and for a given stretching force $l \infty \frac{1}{A Y}$
Let three wires have young's modulus 2Y, 2Y and Y and their cross sectional areas are A, 2A and 3A respectively.
$l_{1} : l_{2} : l_{3} = \frac{1}{A_{1} Y_{1}} : \frac{1}{A_{2} Y_{2}} : \frac{1}{A_{3} Y_{3}} = \frac{1}{A \times 2 Y} : \frac{1}{2 A \times 2 Y} : \frac{1}{3 A \times Y} = \frac{1}{2} : \frac{1}{4} : \frac{1}{3} = 6 : 3 : 4.$

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The correct option is D 6:3:4l=FLAY and for a given stretching force l∞1AY Let three wires have young's modulus 2Y, 2Y and Y and their cross sectional areas are A, 2A and 3A respectively. l1:l2:l3=1A1Y1:1A2Y2:1A3Y3=1A×2Y:12A×2Y:13A×Y=12:14:13=6:3:4.