A) 0.25 mm
B) 0.65 mm
C) 1.65 mm
D) 0.35 mm
Correct Answer: B
Solution :
[b] \[\Delta L=\frac{LF}{AY},\,\,\]where \[L=3m,\] \[A=\pi {{(1.0\times {{10}^{-3}}m)}^{2}}=3.14\times {{10}^{-6}}{{m}^{2}}\] and since each wire supports one-quarter of the load, \[F=\frac{(50kg)(9.8m/{{s}^{2}})}{4}=123N\] \[\Delta L=\frac{(3m)(123N)}{(3.14\times {{10}^{-6}}{{m}^{2}})(1.8\times {{10}^{11}}N/{{m}^{2}})}\] \[=65\times {{10}^{-5}}m\,\,or\,\,0.65mm\]
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