A) \[g/8\]
B) \[g/4\]
C) \[\frac{g}{2}\]
D) \[g\]
Correct Answer: B
Solution :
According to the question, \[T=2{{m}_{1}}g\] Thus, \[{{m}_{2}}\]will not be lifted. Hence, its acceleration\[{{a}_{2}}=0\] For\[{{m}_{1}}\], \[T-{{m}_{1}}g={{m}_{1}}{{a}_{1}}\] \[\Rightarrow \] \[{{a}_{1}}=\frac{2{{m}_{1}}g-{{m}_{1}}g}{{{m}_{1}}}=g\] \[{{a}_{cm}}=\frac{{{m}_{1}}{{a}_{1}}+{{m}_{2}}{{a}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{{{m}_{1}}{{a}_{1}}+0}{{{m}_{1}}+{{m}_{2}}}=\frac{{{m}_{1}}g}{{{m}_{1}}+3{{m}_{1}}}=\frac{g}{4}\]You need to login to perform this action.
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