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  • A compass needle of oscillation magnetometer oscillates 20 times per minute at a place <img alt="\mathrm{P \: of \: dip \: 30^{\circ}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathr

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place \mathrm{P \: of \: dip \: 30^{\circ}}. The number of oscillations per minute become 10 at another place \mathrm{Q \: of\: 60^{\circ} \: dip}. The ratio of the total magnetic field at the two places \mathrm{\left(B_{Q}: B_{P}\right)} is :
 

Option: 1

\sqrt{3}: 4


Option: 2

4: \sqrt{3}


Option: 3

\sqrt{3}: 2


Option: 4

2: \sqrt{3}


Answers (1)

best_answer

\mathrm{\delta _{1}=30\degree\: \: \left ( At\: pt.P \right )}

\mathrm{\delta _{2}=60\degree\: \: \left ( At\: pt.Q \right )}

\mathrm{T=2\pi \sqrt{\frac{I}{MB}}}

\mathrm{f=\frac{1}{2\pi }\sqrt{\frac{MB}{I}}}

\mathrm{\frac{f_{1}}{f_{2}}=\frac{20}{10}=\sqrt{\frac{\left ( BH \right )_{1}}{\left (BH \right )_{2}}}=\sqrt{\frac{B_{1}\cos 30\degree}{B_{2}\cos60\degree }}}

\mathrm{\frac{2}{1}=\sqrt{\frac{B_{1}}{B_{2}}}\times \frac{\sqrt{3}}{2}\times \frac{2}{1}}

\mathrm{\frac{B_{P}}{B_{Q}}=\frac{B_{1}}{B_{2}}=\frac{4}{\sqrt{3}}}

Hence 1 is correct option.

Posted by

Sayak

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