When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60 °$ with horizontal, it can travel a distance $x_{1}$ along the plane. But when the inclination is decreased to $30 °$ and the same object is shot with the same velocity, it can travel $x_{2}$ distance. Then $x_{1} : x_{2}$ will be

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$1 : \sqrt{2}$

$\sqrt{2} : 1$

$1 : \sqrt{3}$

$1 : 2 \sqrt{3}$

answer is D.

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Detailed Solution

$v^{2} - u^{2} = 2 ax$

$For case I : v^{2} = u^{2} - 2 ({gsinθ}_{1}) x_{1} [∵ a = - gsinθ] \begin{array}{l} x_{1} & = \frac{u^{2}}{2 {gsinθ}_{1}} & [∵ v = 0] \end{array} For case II : v^{2} = u^{2} - (2 {gsinθ}_{2}) x_{2} x_{2} = \frac{u^{2}}{2 {gsinθ}_{2}} ∴ \frac{x_{1}}{x_{2}} = \frac{{sinθ}_{2}}{{sinθ}_{1}} = \frac{\sin 30 °}{\sin 60 °} = \frac{1}{\sqrt{3}}$