When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60⁰ with horizontal, it can travel a distance x₁ along the plane. But when the inclination is decreased to 30⁰ and the same object is shot with the same velocity, it can travel x₂ distance. Then x₁ : x₂ will be

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1 : 2 $\sqrt{3}$

$\sqrt{2}$ : 1

1 : $\sqrt{3}$

1 : $\sqrt{2}$

answer is A.

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Detailed Solution

(Stopping distance) $x_{1} = \frac{u^{2}}{2 gsin 60^{0}}$
(Stopping distance) $x_{2} = \frac{u^{2}}{2 gsin 30^{\circ}}$
$\Rightarrow \frac{x_{1}}{x_{2}} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}} = \frac{1 \times 2}{2 \times \sqrt{3}} = 1 : \sqrt{3}$

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