A) \[T=2\frac{m_{1}^{2}+m_{2}^{2}}{{{m}_{1}}+{{m}_{2}}}g\]
B) \[T=\frac{2{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\]
C) \[T=\frac{2m_{2}^{2}+m_{1}^{2}}{{{m}_{1}}+{{m}_{2}}}g\]
D) \[T=2({{m}_{1}}+{{m}_{2}})g\]
Correct Answer: B
Solution :
: Let tension in string = T acceleration\[=a\] \[\therefore \] For greater mass, \[{{m}_{2}}g-T={{m}_{2}}a\] For smaller mass, \[T-{{m}_{1}}g={{m}_{1}}a\] Divide them to eleminate a \[\therefore \]\[\frac{{{m}_{2}}g-T}{T-{{m}_{1}}g}=\frac{{{m}_{2}}a}{{{m}_{1}}a}\] Or \[{{m}_{1}}{{m}_{2}}g-{{m}_{1}}T={{m}_{2}}T-{{m}_{1}}{{m}_{2}}g\] Or \[2{{m}_{1}}{{m}_{2}}g={{m}_{1}}T+{{m}_{2}}T=T({{m}_{1}}+{{m}_{2}})g\] Or \[T=\frac{2{{m}_{1}}{{m}_{2}}}{({{m}_{1}}+{{m}_{2}})}g\]
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