$\mathrm{d}=\frac{\mathrm{1}}{2}(\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta) \mathrm{t}_{2}$

$t_{1}=\sqrt{\frac{2 d}{g \sin \theta}}$

$\mathrm{t}_{2}=\sqrt{\frac{2 \mathrm{d}}{\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta}}$

According to question, $\mathrm{t}_{2}=\mathrm{nt}_{1}$ $\sqrt[n]{\frac{2 d}{g \sin \theta}}=\sqrt{\frac{2 d}{g \sin \theta-\mu g \cos \theta}}$

$\mu,$ applicable here, is kinetic friction as the block moves over the inclined plane. $\mathrm{n}=\frac{1}{\sqrt{1-\mu_{\mathrm{k}}}}\left(\because \cos 45^{\circ}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}\right)$

$\mathrm{n}^{2}=\frac{1}{1-\mu_{\mathrm{k}}} \quad$ or $1-\mu_{\mathrm{k}}=\frac{1}{n^{2}}$

Or $\mu_{\mathrm{k}}=1-\frac{1}{\mathrm{n}^{2}}$

$\geqslant \mu\left(2 \sin ^{2} \frac{\theta}{2}\right)$

$\therefore \cot \frac{\theta}{2} \geqslant \mu$