A) \[g\,\,\sin \,\theta \]
B) \[g\,\,\cos \,\theta \]
C) \[g\sqrt{3{{\cos }^{2}}\theta +1}\]
D) \[g\sqrt{3{{\sin }^{2}}\theta +1}\]
Correct Answer: D
Solution :
[d] \[{{a}_{t}}=g\sin (90{}^\circ -\theta )=g\cos \theta \] \[v=\sqrt{2gh}=\sqrt{2gl\sin \theta }\] \[{{a}_{c}}=\frac{{{v}^{2}}}{l}=\frac{2gl\sin \theta }{l}=2g\sin \theta \] \[a=\sqrt{{{a}_{c}}^{2}+{{a}_{t}}^{2}}=g\sqrt{3{{\sin }^{2}}\theta +1}\]
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