A) \[\sqrt{{}}600\]
B) \[\sqrt{{}}1200\]
C) \[\sqrt{{}}1800\]
D) \[\sqrt{{}}445\]
Correct Answer: C
Solution :
: \[T=2\pi \sqrt{\frac{I}{MH}}=2\pi \sqrt{\frac{I}{LmH}}\] where \[M=Lm\] \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}=\sqrt{\frac{{{m}_{1}}}{2{{m}_{1}}}}=\frac{1}{\sqrt{2}}\] \[\therefore \] \[{{T}_{2}}=\frac{{{T}_{1}}}{\sqrt{2}}\] We know that frequency \[f=\frac{1}{T}\] or \[{{f}_{2}}=\sqrt{2}{{f}_{1}}=\sqrt{2}\times 30=\sqrt{2\times 900}=\sqrt{1800}\]
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