The magnetic field on the axis of the magnet at a distance $d_{1}=14\, cm ,$ can be written as:

$B_{1}=\frac{\mu_{0} 2 M}{4 \pi\left(d_{1}\right)^{3}}=H\ldots .(1)$

Where, $M=$ Magnetic moment $\mu_{0}=$ Permeability of free space

$H=$ Horizontal component of the magnetic field at $d_{1}$

If the bar magnet is turned through $180^{\circ},$ then the neutral point will lie on the equatorial line. Hence, the magnetic field at a distance $d_{2},$ on the equatorial line of the magnet can be written

as:

$B_{2}=\frac{\mu_{0} M}{4 \pi\left(d_{2}\right)^{3}}=H\dots(ii)$

Equating equations $(i)$ and $(ii)$, we get:

$\frac{2}{\left(d_{1}\right)^{3}}=\frac{1}{\left(d_{2}\right)^{3}}$

$\left(\frac{d_{2}}{d_{1}}\right)^{3}=\frac{1}{2}$

$\therefore d_{2}=d_{1} \times\left(\frac{1}{2}\right)^{1 / 3}$

$=14 \times 0.794=11.1 \,cm$

The new null points will be located $11.1 \;cm$ on the normal bisector.