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# A particle starts from rest from top of a smooth inclined plane of a given base. The time of fall is least when inclination of the plane to horizontal is

$(a)\;75 ^{\circ} \quad (b)\;45^{\circ} \quad (c)\;15^{\circ} \quad(d)\;60^ {\circ}$

Using $s=ut+\large\frac{1}{2}$$at^2 \large\frac{a}{\cos \theta}=\frac{1}{2}$$(g \sin \theta)t^2$
$\quad t=\sqrt {\large\frac{2a}{g \sin \theta \cos \theta}}$
$\qquad= \sqrt {\large\frac{4a}{g \sin 2 \theta}}$
t is minimum when $sin\; 2 \theta=1$
$\theta=45^{\circ}$
Hence b is the correct answer.
edited May 26, 2014