1.

At a place on earth, horizontal component of earth's magnetic field is \[{{B}_{1}}\] and vertical component of earth's magnetic field is \[{{B}_{2}}\]. If a magnetic needle is kept vertical, in a plane making angle \[\alpha \] with the horizontal component of magnetic field, then square of time period of oscillation of needle when slightly distributed is proportional to

A. \[\frac{1}{\sqrt{{{B}_{1}}\,\cos \,\alpha }}\]
B. \[\frac{1}{\sqrt{{{B}_{2}}}}\]
C. \[\frac{1}{\sqrt{{{({{B}_{1\,}}\cos \,\alpha )}^{2}}+B_{2}^{2}}}\]
D. infinite
Answer» D. infinite


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