A wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the;
(a) stress on the wire;
(b) strain in, the wire. [\(\pi = 3.142\)]
(a) stress on the wire;
(b) strain in, the wire. [\(\pi = 3.142\)]
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Correct Answer: Option n
Explanation:
(a) Stress = \(\frac{Force}{Area} = \frac{50}{3.142(1 \times 10^{-3})^2}\)
(b) Strain = \(\frac{extension}{length} = \frac{2.5 \times 10^{-5}}{5}\)
= 5 x 10\(^{-5}\)
(a) Stress = \(\frac{Force}{Area} = \frac{50}{3.142(1 \times 10^{-3})^2}\)
(b) Strain = \(\frac{extension}{length} = \frac{2.5 \times 10^{-5}}{5}\)
= 5 x 10\(^{-5}\)