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# The time taken by a particle to slide down a smooth inclined plane is double the time it would take in falling down along a height equal to vertical height of the plane. The inclination of the plane with horizontal is

$(a)\; 30 ^{\circ}\quad (b)\;45 ^{\circ} \quad (c)\;60^{\circ} \quad (d)90^{\circ}$

$S=\large\frac{1}{2} g$$\sin \theta (2t)^2 and \large\frac{h}{s}$$=\sin \theta$
$\large\frac{h}{\sin \theta}=\large\frac{1}{2}$$g \sin \theta (2t)^2 Also since 't' is time taken to fall h height of h. h=\large\frac{1}{2}$$gt^2$
$\large\frac{\Large\frac{1}{2}gt^2}{\sin \theta}=\frac{1}{2} $$\;g \sin \theta\; 4t^2 \sin ^2 \theta=\large\frac{1}{4}$$=>\theta=30 ^{\circ}$
Hence a is the correct answer.

edited Jan 25, 2014 by meena.p

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