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A ball of mass mis released from the top of an inclined plane of inclination `theta` as shown. It strikes a rigid surface at at a distance `(3l)/(4)` from top elastically. Impulse imparted to ball by the rigid surface is
image
A. `msqrt(3/2)gh`
B. `msqrt(3gh)`
C. `2msqrt(3gh)`
D. `msqrt(6gh)`

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Correct Answer - D
Loss in PE = gain in KE
`mgh_(1) = 1/2mv^(2)`
`mg xx 3/4h = 1/2mv^(2) rArr v=sqrt(3gh)/2`
Now, impulse imparted, `J = 2mv = 2msqrt(3gh)/2` = `msqrt(6gh)`

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