A) \[3\sqrt{3}:8\]
B) \[16:9\sqrt{3}\]
C) \[4:9\]
D) \[2\sqrt{3}:9\]
Correct Answer: B
Solution :
Given \[{{\nu }_{1}}=\frac{20}{60}=\frac{1}{3}{{\sec }^{-1}}\text{and}\ {{\nu }_{2}}=\frac{15}{60}=\frac{1}{4}{{\sec }^{-1}}\] Now\[\nu =\frac{1}{2\pi }\sqrt{\frac{M{{B}_{H}}}{I}}=\frac{1}{2\pi }\sqrt{\frac{MB\cos \varphi }{I}}\ \ \left( \because {{B}_{H}}=B\cos \varphi \right)\] \[\therefore \ \frac{{{\nu }_{1}}}{{{\nu }_{2}}}=\sqrt{\frac{{{B}_{1}}\cos {{\varphi }_{1}}}{{{B}_{2}}\cos {{\varphi }_{2}}}}\] Þ \[\frac{{{B}_{1}}}{{{B}_{2}}}={{\left( \frac{{{\nu }_{1}}}{{{\nu }_{2}}} \right)}^{2}}\,{{\left( \frac{\cos {{\varphi }_{2}}}{\cos {{\varphi }_{1}}} \right)}^{2}}\] Þ \[\frac{{{B}_{1}}}{{{B}_{2}}}={{\left( \frac{1/3}{1/4} \right)}^{2}}\frac{\cos 60{}^\circ }{\cos 30{}^\circ }=\frac{16}{9}\times \frac{1/2}{\sqrt{3}/2}=\frac{16}{9\sqrt{3}}\].
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