A) \[{{L}_{A}}=2{{L}_{B}}\,and\,{{M}_{A}}={{M}_{B}}/2\]
B) \[{{L}_{A}}=4{{L}_{B}}\] regardless of masses
C) \[{{L}_{A}}={{L}_{B}}/4\] regardless of masses
D) \[{{L}_{A}}=2{{L}_{B}}\,\,\,and\,{{M}_{A}}=2{{M}_{B}}\]
Correct Answer: C
Solution :
[c] \[{{f}_{A}}=\frac{1}{2\pi }\sqrt{\frac{g}{{{L}_{A}}}}\] and \[{{f}_{B}}=\frac{{{f}_{A}}}{2}=\frac{1}{2\pi }\sqrt{\frac{g}{{{L}_{B}}}}\] \[\therefore \,\,\frac{{{f}_{A}}}{{{f}_{A}}/2}=\frac{1}{2\pi }\sqrt{\frac{g}{{{L}_{A}}}}\times 2\pi \sqrt{\frac{{{L}_{B}}}{g}}\Rightarrow 2=\sqrt{\frac{{{L}_{B}}}{{{L}_{A}}}}\] \[\Rightarrow \,4=\frac{{{L}_{B}}}{{{L}_{A}}},\] regardless of mass.
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