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Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle $$60^{\circ}$$ with horizontal, it can travel a distance $$x_1$$ along the plane. But when the inclination is decreased to $$30^o$$ and the same object is shot with the same velocity, it can travel $$x_2$$ distance. Then $$x_1 : x_2$$ will be :


A
1:2
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B
2:1
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C
1:3
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D
1:23
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Solution

The correct option is A $$1 : \sqrt{3}$$
(Stopping distance) $$x_1 = \dfrac{u^2}{2g sin 60^o}$$
(Stopping distance) $$x_2 = \dfrac{u^2}{2g sin 30^o}$$
$$\Rightarrow \dfrac{x_1}{x_2} = \dfrac{sin 30^o}{sin 60^o} = \dfrac{1 \times 2}{2 \times \sqrt{3}} = 1 : \sqrt{3}$$
1259383_1618835_ans_16eeb3ca84ce47b88254edba4f7ecc09.png

Physics

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