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Q.

An aluminum rod, Young's modulus 7.0x109 newton/metre2 has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in m2 in order to support a load of 104 newton is

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a

1x 10-2

b

1.0x10-3

c

1.4 x 10-2

d

7.1x10-4

answer is D.

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Detailed Solution

Y=FLAl  or  A=EY×LlGiven (l/L)=0⋅2%=0⋅2×10−2F=104Nand Y=7⋅0×109N/m2∴ A=1047⋅0×109×0⋅2×10−2≈7⋅1×10−4m2
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