A) 1 : 5
B) 1 : 8
C) 5 : 8
D) 8 : 5
Correct Answer: C
Solution :
Let M1 and M2 be the magnetic moments of magnets and H the horizontal component of earth?s field. We have \[\tau =MH\sin \theta \]. If f is the twist of wire, then \[\tau =C\varphi ,\]C being restoring couple per unit twist of wire \[\Rightarrow \ C\varphi =MH\sin \theta \] Here \[{{\varphi }_{1}}=({{180}^{o}}-{{30}^{o}})={{150}^{o}}\]\[=150\times \frac{\pi }{180}\]rad \[{{\varphi }_{2}}=({{270}^{o}}-{{30}^{o}})={{240}^{o}}\]\[=240\times \frac{\pi }{180}\]rad So,\[C{{\varphi }_{1}}={{M}_{1}}H\sin \theta \](For deflection \[\theta ={{30}^{o}}\]of I magnet) \[C{{\varphi }_{2}}={{M}_{2}}H\sin \theta \] (For deflection \[\theta ={{30}^{o}}\]of II magnet) Dividing \[\frac{{{\varphi }_{1}}}{{{\varphi }_{2}}}=\frac{{{M}_{1}}}{{{M}_{2}}}\] \[\Rightarrow \frac{{{M}_{1}}}{{{M}_{2}}}=\frac{{{\varphi }_{1}}}{{{\varphi }_{2}}}=\frac{150\times \left( \frac{\pi }{180} \right)}{240\times \left( \frac{\pi }{180} \right)}=\frac{15}{24}=\frac{5}{8}\] \[\Rightarrow {{M}_{1}}:{{M}_{2}}=5:8\]
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