A force of 40 N is applied at the end of a wire 4m long and produces an extension of 0.24mm. If the diameter of the wire is 2.00mm, calculate the;
(i) stress on the wire; (ii) strain in the wire.
(i) stress on the wire; (ii) strain in the wire.
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Correct Answer: Option n
Explanation:
(i) Stress = \(\frac{force}{\text{cross sectiona area}}\)
= \(\frac{40}{3.142 (1 \times 10^{-3})^2}\) = 12.74 x 10\(^{6}\)Nm\(^{-2}\)
(ii) Strain = \(\frac{e}{I}\) = \(\frac{24 \times 10^{-5}}{4}\)
= 6.0 x 10\(^{-5}\)
(i) Stress = \(\frac{force}{\text{cross sectiona area}}\)
= \(\frac{40}{3.142 (1 \times 10^{-3})^2}\) = 12.74 x 10\(^{6}\)Nm\(^{-2}\)
(ii) Strain = \(\frac{e}{I}\) = \(\frac{24 \times 10^{-5}}{4}\)
= 6.0 x 10\(^{-5}\)