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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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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

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

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


answered Jul 4, 2013 by meena.p
edited Jan 25, 2014 by meena.p

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