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NEET PHYSICSEasy

When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60∘60^\circ with horizontal, it can travel a distance x1x_1 along the plane. But when the inclination is decreased to 30∘30^\circ and the same object is shot with the same velocity, it can travel x2x_2 distance. Then x1:x2x_1 : x_2 will be:

A

1: 2\sqrt{2}

B

2:1\sqrt{2}: 1

C

1: 3\sqrt{3}

D

1: 232\sqrt{3}

Step-by-Step Solution

(Stopping distance)x1=u22gsin⁡60∘\text{(Stopping distance)} \quad x_1 = \frac{u^2}{2g\sin 60^\circ}

(Stopping distance)x2=u22gsin⁡30∘\text{(Stopping distance)} \quad x_2 = \frac{u^2}{2g\sin 30^\circ}

⇒x1x2=sin⁡30∘sin⁡60∘=1×22×3=1:3\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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