NEET and AIPMT NEET Physics Magnetism and Matter MCQ Questions Solved  Difficulty Level:

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$\mathrm{B}=\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\mathrm{r}}^{3}}\phantom{\rule{0ex}{0ex}}\mathrm{B}\text{'}=\frac{{\mathrm{\mu }}_{0}}{4\mathrm{\pi }}\frac{\mathrm{M}}{8{\mathrm{r}}^{3}}\phantom{\rule{0ex}{0ex}}\mathrm{B}\text{'}=\frac{\mathrm{B}}{16}$

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The magnetic field at a point x on the axis of a small bar magnet is equal to the field at a point y on the equator of the same magnet. The ratio of the distances of x and y from the centre of the magnet is
(a) ${2}^{-3}$

(b) ${2}^{-1}{3}}$

(c) ${2}^{3}$

(d) ${2}^{1}{3}}$ Difficulty Level:

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