A) \[3\sqrt{3}\,\,:\,\,8\]
B) \[16\,\,:\,\,9\sqrt{3}\]
C) \[4\,\,:\,\,9\]
D) \[2\sqrt{3}\,\,:\,\,9\]
Correct Answer: B
Solution :
\[T\,\,\propto \,\,\frac{1}{\sqrt{B}}\] \[\frac{3}{4}\,\,=\,\,\frac{{{B}_{{{H}_{2}}}}}{{{B}_{{{H}_{1}}}}}\,\,=\,\,\frac{B{{e}_{2}}\,\cos \,\,60{}^\circ }{B{{e}_{1}}\,\cos \,\,30{}^\circ }\,\,\] \[\Rightarrow \,\,\,\,\,\frac{B{{e}_{1}}}{B{{e}_{2}}}\,=\,\,\frac{16}{9\sqrt{3}}\]You need to login to perform this action.
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