A) \[\frac{4{{w}_{1}}{{w}_{2}}}{{{w}_{1}}+{{w}_{2}}}\]
B) \[\frac{{{w}_{1}}{{w}_{2}}}{{{w}_{1}}+{{w}_{2}}}\]
C) \[\frac{2{{w}_{1}}{{w}_{2}}}{{{w}_{1}}+{{w}_{2}}}\]
D) \[\frac{{{w}_{1}}+{{w}_{2}}}{2}\]
Correct Answer: A
Solution :
The free body diagram of the given situation is shown. Taking the upward and downward motions respectively, we get \[{{T}_{1}}-mg={{m}_{1}}(g+a)\] \[T-2mg={{m}_{1}}a\] for second weight \[{{m}_{2}}g-T={{m}_{2}}(g-a)\] \[2{{m}_{2}}g-T={{m}_{2}}a\] \[T=\frac{4{{m}_{1}}{{m}_{2}}g}{({{m}_{1}}+{{m}_{2}})}\] \[{{w}_{1}}={{m}_{1}}g,\,\,{{m}_{2}}={{m}_{2}}g\]hence switch \[T=\frac{4{{w}_{1}}{{w}_{2}}}{({{w}_{1}}+{{w}_{2}})}\]You need to login to perform this action.
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