NEET Solved Paper 2019 Question 25
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{}^\circ $ and the same object is shot with the same velocity, it can travel $ x _2 $ distance. Then $ x _1:x _2 $ will be: [NEET 5-5-2019]
Options:
A) $ 1:\sqrt{3} $
B) $ 1:2\sqrt{3} $
C) $ 1:\sqrt{2} $
D) $ \sqrt{2}:1 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ v^{2}=u^{2}+2as $
$ 0=u^{2}2gsin\theta S $
$ S=\frac{u^{2}}{2g\sin \theta } $
$ S\propto \frac{1}{\sin \theta } $
$ \frac{x _1}{x _2}=\frac{{\sin_2}\theta }{{\sin_1}\theta }=\frac{\sin 30{}^\circ }{\sin 60{}^\circ }=\frac{1}{\sqrt{3}} $
BETA
