When a magnet is placed vertically on a horizontal board, number of neutral points obtained on the board is

$\mathrm{Consider} \mathrm{a} \mathrm{bar} \mathrm{magnet} \mathrm{placed} \mathrm{on} \mathrm{a} \mathrm{table} \mathrm{perpendicular} \mathrm{to} \mathrm{horizontal} \mathrm{component} \mathrm{of} \mathrm{magnetic} \mathrm{field}.$

P will be a neutral point if 

${\mathrm{B}}_{\mathrm{S}}={\mathrm{B}}_{\mathrm{H}}+{\mathrm{B}}_{\mathrm{n}}\cos \theta$

A will be a neutral point if 

${\mathrm{B}}_{\mathrm{n}}\mathrm{cos\alpha }={\mathrm{B}}_{\mathrm{S}}+{\mathrm{B}}_{\mathrm{H}}$ and this is not possible.

As the north pole is further away than the south pole. so ${\mathrm{B}}_{\mathrm{n}}<{\mathrm{B}}_{\mathrm{S}}$.

${\mathrm{B}}_{\mathrm{n}}\mathrm{cos\alpha }\text{ is even lesser than }{\mathrm{B}}_{\mathrm{n}}$.

So, ${\mathrm{B}}_{\mathrm{n}}\mathrm{cos\alpha }\text{ cannot be equal to }{\mathrm{B}}_{\mathrm{S}}+{\mathrm{B}}_{\mathrm{H}}$.

Hence only one such point possible.