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When an object is shot from the bottom o...

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 `x_(1)` along the plane. But when the inclination is decreased to `30^(@)` 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 : sqrt 2`

B

`sqrt 2 :1`

C

`1 : sqrt 3`

D

`1 : 2 sqrt 3`

Text Solution

Verified by Experts

The correct Answer is:
C


(Stopping distance ) `xx_(1)=(u^(2))/(2g sin 60^(@))`
(Stopping distance) `x_(2)=(u^(2))/(2g sin 30^(@))`
`implies (x_(1))/(x_(2))=(sin 30^(@))/(sin 60^(@))=(1xx 2)/(2 xx sqrt 3)= 1 : sqrt 3`
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