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In the figure, the wedge is moved with an acceleration of $10 \sqrt{3} m / s^{2}$ . It is seen that the block starts climbing up on the smooth inclined face of the wedge. The time taken by the block to reach the top is

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$\frac{4}{\sqrt{5}} s$

$\frac{1}{\sqrt{5}} s$

$\frac{2}{\sqrt{5}} s$

$\frac{3}{\sqrt{5}} s$

answer is A.

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Detailed Solution

For block, this experiences pseudo force with respect to wedge
Question Image
Hence $15 m - 5 m = ma, a = 10 m / s^{2}$
$S = 1 m = \frac{1}{2} {at}^{2} \Rightarrow t = \frac{1}{\sqrt{5}} \sec$

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