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(B) Let the length of both grooves be $L$. The acceleration of a particle sliding down a smooth groove inclined at an angle $	heta$ with the vertical

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$1 : \sqrt{2}$

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(B) Let the length of both grooves be $L$. The acceleration of a particle sliding down a smooth groove inclined at an angle $	heta$ with the vertical is $a = g \cos 	heta$.For the vertical chord $AB$,the angle with the vertical is $	heta_1 = 0^\circ$. Thus,$a_{AB} = g \cos 0^\circ = g$.The time taken is $t_{AB} = \sqrt{\frac{2L}{a_{AB}}} = \sqrt{\frac{2L}{g}}$.For the chord $CD$,the angle with the vertical is $	heta_2 = 60^\circ$. Thus,$a_{CD} = g \cos 60^\circ = g/2$.The time taken is $t_{CD} = \sqrt{\frac{2L}{a_{CD}}} = \sqrt{\frac{2L}{g/2}} = \sqrt{\frac{4L}{g}} = 2\sqrt{\frac{L}{g}}$.Taking the ratio,$\frac{t_{AB}}{t_{CD}} = \frac{\sqrt{2L/g}}{2\sqrt{L/g}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

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