A) 6 mm
B) 9 mm
C) 12 mm
D) 18 mm
Correct Answer: C
Solution :
We have, \[l={{l}_{0}}{{e}^{-\mu x}}\] \[\therefore \] \[\frac{1}{8}={{e}^{-4.36}}\] \[\Rightarrow \] \[{{\left( \frac{1}{2} \right)}^{3}}={{e}^{-4.36}}\] \[={{({{e}^{-\mu x}})}^{3}}\] or \[-3\mu x=-36\mu \] \[\Rightarrow \] \[x=12\,mm\]You need to login to perform this action.
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