When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with hor

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Question

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 x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1 : x2 will be:

Accepted Answer

Correct Answer: D — 1 : √3Key Concept: When an object projected up a smooth inclined plane decelerates due to the component of gravity along the plane. Using the kinematic equation v² = u² − 2as with final velocity v = 0:Formula: s = u² / (2a), where a = g sinθ (deceleration along the incline)Step-by-step Solution:Since the object comes to rest, applying v² = u² − 2as:0 = u² − 2(g sinθ)ss = u² / (2g sinθ)Therefore, s ∝ 1/sinθ (for same initial velocity u)x1/x2 = sin 30° / sin 60° = (1/2) / (√3/2) = 1/√3Thus x1 : x2 = 1 : √3Why other options are wrong:Options A and B arise from incorrect proportionality. Option C (2:1) would be the inverse ratio, applicable if the angle were reversed.

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