A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.
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Correct Answer: Option n
Explanation:
(a) Stree = \(\frac{F}{A}\) = \(\frac{40}{3.14(1.0 \times 10^{-3})}^2\)
= \(1.27 \times 10^{7} \times 10^7 NM ^{-2}\)
(b) Strain = \(\frac{e}{L} = \frac{0.24 \times 10^{-3}}{3} = 8.0 \times 10^{-5}\)
(a) Stree = \(\frac{F}{A}\) = \(\frac{40}{3.14(1.0 \times 10^{-3})}^2\)
= \(1.27 \times 10^{7} \times 10^7 NM ^{-2}\)
(b) Strain = \(\frac{e}{L} = \frac{0.24 \times 10^{-3}}{3} = 8.0 \times 10^{-5}\)