KVPY Sample Paper KVPY Stream-SX Model Paper-11

A bar magnet suspended at a place P where Dip angle is \[60{}^\circ \] gives 10 oscillations per minute. The same bar magnet suspended at another place Q where dip angle is \[30{}^\circ \] gives 20 osciilations per minute. The ratio of magnetic Field at \[P\] and \[Q\],\[\frac{{{B}_{P}}}{{{B}_{Q}}}\] is

A) \[\frac{\sqrt{3}}{4}\]

B) \[\frac{\sqrt{3}}{2}\]

C) \[\frac{2}{\sqrt{3}}\]

D) \[\frac{4}{\sqrt{3}}\]

Correct Answer: A

Solution :

\[{{B}_{{{H}_{1}}}}={{B}_{1}}\cos 60{}^\circ and\,{{B}_{{{H}_{2}}}}={{B}_{2}}\cos 30{}^\circ \]

Using \[\frac{{{T}^{2}}}{T_{2}^{2}}=\frac{{{B}_{{{H}_{2}}}}}{{{B}_{{{H}_{1}}}}}\]

Or \[\frac{{{B}_{{{H}_{1}}}}}{{{B}_{{{H}_{2}}}}}=\frac{T_{2}^{2}}{T_{1}^{2}}={{\left( \frac{3}{6} \right)}^{2}}=\frac{1}{4}\]

Now \[\frac{{{B}_{1}}\cos 60{}^\circ }{{{B}_{2}}\cos 30{}^\circ }=\frac{1}{4}\therefore \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\sqrt{3}}{4}\]