When an object is shot from the bottom of a long smooth inclined plane kept at an angle with horizontal, it can travel a distance along the plane. But when the inclination is decreased to and the same object is shot with the same velocity, it can trave

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Solution

Q.
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 [2019]

1 $1 : 2 \sqrt{3}$

2 $1 : \sqrt{2}$

3 $\sqrt{2} : 1$

4 $1 : \sqrt{3}$

Ans.
(4)

$v^{2} - u^{2} = 2 a x$

For case I : $v^{2} = u^{2} - 2 (g \sin θ_{1}) x_{1}$ $[∵ a = - g \sin θ]$

$x_{1} = \frac{u^{2}}{2 g \sin θ_{1}}$ $[∵ v = 0]$

For case II : $v^{2} = u^{2} - (2 g \sin θ_{2}) x_{2}$

$x_{2} = \frac{u^{2}}{2 g \sin θ_{2}}$

$∴$ $\frac{x_{1}}{x_{2}} = \frac{\sin θ_{2}}{\sin θ_{1}} = \frac{\sin 30 °}{\sin 60 °} = \frac{1}{\sqrt{3}}$

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