A) \[{{T}_{1}}=20\,N,\,{{T}_{2}}=60\,N\]
B) \[{{T}_{1}}=60\,N,\,{{T}_{2}}=60\,N\]
C) \[{{T}_{1}}=30\,N,\,{{T}_{2}}=50\,N\]
D) \[{{T}_{1}}=20\,N,\,{{T}_{2}}=100\,N\]
Correct Answer: A
Solution :
\[{{T}_{2}}=\frac{({{m}_{1}}+{{m}_{2}})F}{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})}\] \[=\frac{(2+4)\times 120}{(2+4+6)}\] \[=\frac{6\times 120}{12}\] \[{{T}_{2}}=60\,N\] \[{{T}_{1}}=\frac{{{m}_{1}}F}{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})}\] \[=\frac{2\times 120}{12}\] = 20 NYou need to login to perform this action.
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