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. T is the tension in strings,\[{{w}_{1}}\]and\[{{w}_{2}}\]are the weights acting downwards. Taking the upward and downward motion respectively, we get: \[T-{{m}_{1}}g={{M}_{1}}(g+a)\] \[\Rightarrow \] \[T-2{{m}_{1}}g={{m}_{1}}a\] ...(1) For second weight \[{{m}_{2}}g-T={{m}_{2}}(g-a)\] \[\Rightarrow \] \[2{{m}_{2}}g-T={{m}_{2}}a\] ...(2) On solving Eqs. (1) and (2), we get \[T=\frac{4{{m}_{1}}{{m}_{2}}g}{({{m}_{1}}-{{m}_{2}})}\] Given,\[{{w}_{1}}={{m}_{1}}g,{{w}_{2}}={{m}_{2}}g\]hence, we get \[T=\frac{4{{w}_{1}}+{{w}_{2}}}{({{w}_{1}}+{{w}_{2}})}\]You need to login to perform this action.
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