A) \[0.2%\]increases
B) \[0.2%\]decrease
C) \[0.1%\]increase
D) \[0.1%\] decrease
Correct Answer: A
Solution :
For a given wire, \[R=\frac{PL}{S}\] with LX S = volume = V = constant So, that \[R=P\frac{{{L}^{2}}}{V}\] \[\Rightarrow \] \[\frac{\Delta R}{R}=2\frac{\Delta L}{L}\] \[=2(0.1%)=02%\](increase)You need to login to perform this action.
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