A) \[g\sqrt{{{l}^{2}}-1}\]
B) \[g({{l}^{2}}-1)\]
C) \[\frac{g}{\sqrt{{{l}^{2}}-1}}\]
D) \[\frac{g}{{{l}^{2}}-1}\]
Correct Answer: C
Solution :
Here,\[\sin \theta =\frac{1}{l}\] When horizontal acceleration is imparted to the inclined plane, [hen a pseudo acceleration acts on the object. For the object to remain stationary relative to the incline, we have \[a=g\tan \theta =g\frac{1}{\sqrt{{{l}^{2}}-1}}\] or \[a=\frac{g}{\sqrt{{{l}^{2}}-1}}\]You need to login to perform this action.
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