A) \[\frac{{{M}_{2}}(\sin \beta )g}{{{M}_{1}}+{{M}_{2}}}\]
B) \[\frac{{{M}_{1}}(\sin \alpha )g}{{{M}_{1}}+{{M}_{2}}}\]
C) \[\left( \frac{{{M}_{2}}\sin \beta -{{M}_{1}}\sin \alpha }{{{M}_{1}}+{{M}_{2}}} \right)g\]
D) zero
Correct Answer: C
Solution :
Let \[{{M}_{2}}\] moves downwards with an acceleration a and the tension in the string is T. \[\therefore \] \[{{M}_{2}}g\,\sin \beta -T={{M}_{2}}a\] ?.(i) \[T-{{M}_{1}}g\,\sin \alpha ={{M}_{1}}a\] ?...(ii) Adding (i) and (ii), we get \[{{M}_{2}}g\,\sin \beta -{{M}_{1}}g\,\sin \alpha =({{M}_{2}}+{{M}_{1}})\,a\] \[a=g\frac{({{M}_{2}}\,\sin \beta -{{M}_{1}}\sin \alpha )}{({{M}_{2}}+{{M}_{1}})}\].You need to login to perform this action.
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