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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 an inclined plane of a given base. The time of fall is least when the inclination of the plane to the horizon is

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

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Let the base be of length 'a'
the length $AC=a/\cos \theta$
The body travels along AC. and the component of $g$ along $AC=g \sin \theta$.
using $s=ut+\large\frac{1}{2} $$at^2$
$\large\frac{a}{\cos \theta}=\frac{1}{2}$$ (g \sin \theta)t^2$
$t=\sqrt {\large\frac{2a}{g \sin \theta \cos \theta}}=\sqrt {\large\frac{4a}{g \sin 2 \theta}}$
t is minimum when $\sin 2 \theta=1$
$2 \theta=90^{\circ}\quad or \quad \theta=45^{\circ}$
answered Jun 25, 2013 by meena.p

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