A particle of mass m is initially situated at the point P inside a hemispherical surface of radius r as shown in figure. A horizontal acceleration of
magnitude a_o is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, then time taken by particle to touch the sphere again is
where a is the angle which OP makes with radius OC.

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$\sqrt{(\frac{4 rtan α}{a_{0}})}$

$\sqrt{(\frac{4 rsin α}{a_{0}})}$

$\sqrt{(\frac{4 rcot α}{a_{0}})}$

$\sqrt{(\frac{4 rcos α}{a_{0}})}$

answer is C.

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Detailed Solution

Let the particle touches the sphere at point A as shown in figure.

Let PA=x. Now PB-x/2
From $ΔOPB, \cos α = PB / PO = PB / r$
$∴ PB = rcos α$
Therefore, $\frac{x}{2} = rcos α or x = 2 rcos α$
But $x = \frac{1}{2} a_{0} t^{2}$
So, $t = \sqrt{(\frac{2 x}{a_{0}})} = \sqrt{(\frac{2 \times 2 rcos α}{a_{0}})}$
$t = \sqrt{(\frac{4 rcos α}{a_{0}})}$