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Question

# 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

A
1:2:3
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B
3:2:1
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C
5:4:3
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D
6:3:4
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Solution

## 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.

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