A block of mass m is placed on a smooth inclined wedge ABC of inclination \[\theta \] as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and \[\theta \] for the block to remain stationary on the wedge is [NEET - 2018] |
A) \[a=g\,\,\cos \theta \]
B) \[a=\frac{g}{\sin \theta }\]
C) \[a=\frac{g}{\text{cosec}\theta }\]
D) \[a=g\tan \theta \]
Correct Answer: D
Solution :
[d] |
In non-inertial frame, |
\[\text{N }sin\text{ }\theta =ma...(i)\] |
\[\text{N cos }\theta =mg...(ii)\] |
\[\tan \theta =\frac{a}{g}\] |
\[a=g\tan \theta \] |
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