A) \[\sqrt{\frac{\text{7}}{\text{48}}}l\]
B) \[\frac{\text{1}}{\text{8}}l\]
C) \[\frac{1}{4}l\]
D) \[\sqrt{\frac{\text{3}}{\text{8}}}l\]
Correct Answer: A
Solution :
[a] \[\text{l=}\frac{M{{l}^{2}}}{12}+M\times \left( \frac{{{l}^{2}}}{16} \right)=\frac{7M{{l}^{2}}}{48}\] \[\therefore M{{K}^{2}}=\frac{7M{{l}^{2}}}{48}\] \[\Rightarrow K=\sqrt{\frac{7}{48}}l\]
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