A) \[\frac{aL}{2g}\]
B) \[\frac{gL}{2a}\]
C) \[\frac{gL}{a}\]
D) \[\frac{aL}{g}\]
Correct Answer: D
Solution :
Newton's equations are A\[\Delta p\sin \theta =ma\]...(i) and A\[\Delta p\cos \theta =mg\]...(ii) By Eqs. (i) and (ii), we get \[\tan \theta =\frac{a}{g}=\frac{h}{L}\] or\[h=\frac{aL}{g}\]You need to login to perform this action.
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