A) \[\sqrt{2}\,s\]
B) \[3\sqrt{3}\,s\]
C) 3 s
D) 6 s
Correct Answer: B
Solution :
When magnets are placed in same position \[{{T}_{1}}=2\pi \sqrt{\frac{I}{({{M}_{1}}+{{M}_{2}})H}}\] \[T_{1}^{2}=4{{\pi }^{2}}\frac{I}{({{M}_{1}}-{{M}_{2}})H}\] Similarly, \[T_{2}^{2}=4{{\pi }^{2}}\frac{I}{({{M}_{1}}-{{M}_{2}})H}\] \[\therefore \] \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{{{M}_{1}}-{{M}_{2}}}{{{M}_{1}}+{{M}_{2}}}\] Using sum and difference method \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{T_{2}^{2}+T_{1}^{2}}{T_{2}^{2}-T_{1}^{2}}\] \[\frac{2M}{M}=\frac{T_{2}^{2}+{{(3)}^{2}}}{T_{2}^{2}-{{(3)}^{2}}}\] \[2T_{2}^{2}-2\times 9=T_{2}^{2}+9\] \[T_{2}^{2}=27\] \[{{T}_{2}}=3\sqrt{3}s.\]
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