b
(b)Given \({\nu _1} = \frac{{20}}{{60}} = \frac{1}{3}\,{\sec ^{ - 1}}\,{\rm{and}}\;{\nu _2} = \frac{{15}}{{60}} = \frac{1}{4}\,{\sec ^{ - 1}}\)
Now \(\nu  = \frac{1}{{2\pi }}\sqrt {\frac{{M{B_H}}}{I}}  = \frac{1}{{2\pi }}\sqrt {\frac{{MB\cos \phi }}{I}} \;\;\left( {{B_H} = B\cos \phi } \right)\)
\(\therefore \;\frac{{{\nu _1}}}{{{\nu _2}}} = \sqrt {\frac{{{B_1}\cos {\phi _1}}}{{{B_2}\cos {\phi _2}}}} \) \(==>\) \(\frac{{{B_1}}}{{{B_2}}} = {\left( {\frac{{{\nu _1}}}{{{\nu _2}}}} \right)^2}\,{\left( {\frac{{\cos {\phi _2}}}{{\cos {\phi _1}}}} \right)^2}\)
\(==>\) \(\frac{{{B_1}}}{{{B_2}}} = {\left( {\frac{{1/3}}{{1/4}}} \right)^2}\frac{{\cos 60^\circ }}{{\cos 30^\circ }} = \frac{{16}}{9} \times \frac{{1/2}}{{\sqrt 3 /2}} = \frac{{16}}{{9\sqrt 3 }}\).