A) 4 sec
B) 2 sec
C) \[\frac{1}{2}\sec \]
D) \[\sqrt{2}\sec \]
Correct Answer: D
Solution :
\[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}};\ \therefore \frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{({{B}_{H}})}_{2}}}{{{({{B}_{H}})}_{1}}}}\Rightarrow {{T}_{2}}={{T}_{1}}\sqrt{\frac{{{({{B}_{H}})}_{1}}}{{{({{B}_{H}})}_{2}}}}\] Here n1=30 oscillation /min \[=\frac{1}{2}\]oscillation/sec \[\therefore \ \frac{\mu }{4\pi }.\frac{{{m}^{2}}}{{{r}^{2}}}=50gm-wt\] \[\therefore {{T}_{2}}=2\sqrt{\frac{{{B}_{H}}}{2{{B}_{H}}}}=2\times \frac{1}{\sqrt{2}}=\sqrt{2}\sec \]
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