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Question: An aluminum rod (Young's modulus \(\left. = 7 \times 10 ^ { 9 } \mathrm {~N} / \mathrm { m } ^ { 2 }...

An aluminum rod (Young's modulus =7×109 N/m2)\left. = 7 \times 10 ^ { 9 } \mathrm {~N} / \mathrm { m } ^ { 2 } \right) has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in order to support a load of 10410 ^ { 4 }Newton's is

A
B

1.4×10−3 m21.4 \times 10 ^ { - 3 } \mathrm {~m} ^ { 2 }

C

3.5×10−3 m23.5 \times 10 ^ { - 3 } \mathrm {~m} ^ { 2 }

D

7.1×10−4 m27.1 \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }

Answer

7.1×10−4 m27.1 \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }

Explanation

Solution

Y=F/A strain ⇒A=FY× strain Y = \frac { F / A } { \text { strain } } \Rightarrow A = \frac { F } { Y \times \text { strain } }

= 1047×109×0.002\frac { 10 ^ { 4 } } { 7 \times 10 ^ { 9 } \times 0.002 }= 114×10−2\frac { 1 } { 14 } \times 10 ^ { - 2 } =7.1×10−4 m2= 7.1 \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }