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

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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.

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