A) \[\frac{\sqrt{3}}{4}\]
B) \[\frac{\sqrt{3}}{2}\]
C) \[\frac{2}{\sqrt{3}}\]
D) \[\frac{4}{\sqrt{3}}\]
Correct Answer: A
Solution :
\[{{B}_{{{H}_{1}}}}={{B}_{1}}\cos 60{}^\circ and\,{{B}_{{{H}_{2}}}}={{B}_{2}}\cos 30{}^\circ \] |
Using \[\frac{{{T}^{2}}}{T_{2}^{2}}=\frac{{{B}_{{{H}_{2}}}}}{{{B}_{{{H}_{1}}}}}\] |
Or \[\frac{{{B}_{{{H}_{1}}}}}{{{B}_{{{H}_{2}}}}}=\frac{T_{2}^{2}}{T_{1}^{2}}={{\left( \frac{3}{6} \right)}^{2}}=\frac{1}{4}\] |
Now \[\frac{{{B}_{1}}\cos 60{}^\circ }{{{B}_{2}}\cos 30{}^\circ }=\frac{1}{4}\therefore \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\sqrt{3}}{4}\] |
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