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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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An inclined plane makes an angle $30 ^{\circ}$ with horizontal. A groove $OA=5 m$ cut on the plane makes an angle $30 ^{\circ}$ with $OX$. A short smooth cylinder is free to slide down the groove under the influence of gravity. The time taken by the cylinder to reach from A to O is $(g=10 m/s^2)$

 

\[(a)\;4\;s \quad (b)\;2\;s \quad (c)\;2 \sqrt 2 \;s \quad(d)\;1\;s \]

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1 Answer

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The plane is inclined at $30 ^{\circ}$ to horizontal and the path of the cylinder is at $30 ^{\circ}$ to OX
Therefore the acceleration of cylinder
$a=(g \sin 30 ^{\circ}) (\sin 30^{\circ})$
$\quad= 10 \bigg(\large\frac{1}{2}\bigg) \bigg(\frac{1}{2}\bigg)$
$\quad=2.5 m/s^2$
time taken $=t=\sqrt {\large\frac{2s}{a}}$
$\qquad=\sqrt {\large\frac{2 \times 5}{2.5}}$
$\qquad=2s$
Hence b is the correct answer.

 

answered Jul 5, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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