A) \[mg\,\cos \,\theta \]
B) \[mg\,\sin \,\theta \]
C) \[mg\]
D) \[mg/\,\cos \,\theta \]
Correct Answer: D
Solution :
[d] When the whole system is accelerated towards left, then pseudo force (ma) works on a block towards right. For the condition of equilibrium: \[mg\sin \theta =ma\cos \theta \] \[\Rightarrow a=\frac{g\sin \theta }{\cos \theta }\] Therefore, force exerted by the wedge on the block \[N=mg\cos \theta +ma\sin \theta \] \[=mg\cos \theta +m\left( \frac{g\sin \theta }{\cos \theta } \right)\sin \theta \] \[=\frac{mg(co{{s}^{2}}\theta +si{{n}^{2}}\theta )}{\cos \theta }=\frac{mg}{\cos \theta }\]
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