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Physics · PYQ NEET UG 2019(1)
हिन्दीWhen 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 \( 30^\circ \) and the same object is shot with the same velocity, it can travel \( x_2 \) distance. Then \( x_1 : x_2 \) will be:
Correct answer · Option 1
\( 1 : \sqrt{3} \)
Explanation
Retardation on incline is \( a = g \sin\theta \). Distance \( x = \frac{v^2}{2a} = \frac{v^2}{2g\sin\theta} \implies x \propto \frac{1}{\sin\theta} \). Thus, \( \frac{x_1}{x_2} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \).
