A) \[\frac{2g}{3}\]
B) \[\frac{2g}{6}\]
C) \[\frac{2g}{9}\]
D) \[\frac{g}{3}\]
Correct Answer: C
Solution :
[c] \[\frac{m}{2}g-T=\frac{m}{2}a\] ?.(i) \[T\cos 60{}^\circ =\frac{ma}{\cos 60{}^\circ }\] ?(ii) Solving (i) and (ii), Acceleration of ring \[=\frac{2g}{9}\]You need to login to perform this action.
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