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Question

Two particles A and B are released from rest from point O and are allowed to descend on two frictionless paths OO $_{1}$ and OO $_{2}$ in the vertical plane as shown in the figure. If the ratio of time taken by particles A and B to reach the circumference of the circle is $\frac{18}{x}$ then x is:

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Solution

The correct option is A 18
we will consider kinematics equation.
$S (= l) = u t + \frac{1}{2} a t^{2}$ , u=0 hence,
$l = \frac{1}{2} a t^{2}$
$t = \sqrt{\frac{2 l}{g}}$
$t = \sqrt{\frac{2 \times 2 R c o s θ}{g c o s θ}} = \sqrt{\frac{4 R}{g}}$ , which is independent of $θ$
hence $\frac{t_{2}}{t_{1}} = 1$
$⟹ 18 / x = 1$
$⟹ x = 18$

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Two particles A and B are released from rest from point O and are allowed to descend on two frictionless paths OO1 and OO2 in the vertical plane as shown in the figure. If the ratio of time taken by particles A and B to reach the circumference of the circle is 18x then x is:

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