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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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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

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

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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. 
answered Jul 3, 2013 by meena.p
edited May 26, 2014 by lmohan717

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