A) 0.2 mm
B) 0.5 mm
C) 2 mm
D) 1 mm
Correct Answer: B
Solution :
Youngs modulus is given as \[Y=\frac{F}{A}\times \frac{L}{\Delta L}=\frac{mgL}{\pi {{r}^{2}}\times \Delta L}\] ?(i) Given: Length of wire L = 4 m Diameter of wire \[=2\,mm=2\times {{10}^{-3}}m\] So, radius of wire \[r=1\times {{10}^{-3}}m\] Youngs modulus \[\gamma =2\times {{10}^{11}}N/{{m}^{2}}\] Weight loaded on the wire m = 8 kg Putting the given value in Eq. (1), we get \[2\times {{10}^{11}}=\frac{8\times 9.8\times 4}{3.14\times {{(1\times {{10}^{-3}})}^{2}}\times \Delta L}\] \[\Rightarrow \] \[\Delta L=\frac{8\times 9.8\times 4}{3.14\times {{(1\times {{10}^{-3}})}^{2}}\times 2\times {{10}^{11}}}\] \[=44.9\times {{10}^{-5}}=50\times {{10}^{-5}}=0.5\times {{10}^{-3}}m\] \[=0.5\,mm\]
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