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Question: On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section <img src=...

On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section , the force required is [Y for steel=2.2×1011 N/m2]\left. = 2.2 \times 10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 } \right]]

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Answer
Explanation

Solution

F=YAlL=2.2×1011×2×10−6×5×10−42F = \frac { Y A l } { L } = \frac { 2.2 \times 10 ^ { 11 } \times 2 \times 10 ^ { - 6 } \times 5 \times 10 ^ { - 4 } } { 2 }