A) elongated to 50.1414 cm
B) contracted by 2.0 mm
C) stretched by 0.707 mm
D) of length changed to 49.293 cm
E) of length changed to 50.2 cm
Correct Answer: C
Solution :
The work done by wire is stored as potential energy in the wire \[U=\frac{1}{2}\,\times \] youngs modulus \[\times \,{{(strain)}^{2}}\] Given, \[y=2\times {{10}^{10}}N{{m}^{-2}},\] Strain, \[\frac{l}{L}=\frac{l}{50\times {{10}^{-2}}},\] \[U=2\times {{10}^{-2}}J\] \[\therefore \] \[2\times {{10}^{-2}}=\frac{1}{2}\times 2\times {{10}^{10}}\times {{\left( \frac{1}{50\times {{10}^{-2}}} \right)}^{2}}\] \[\Rightarrow \] \[l\approx 0.707\,mm\] (stretched)
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