A) \[{{\left[ \frac{({{m}_{1}}-{{m}_{2}})}{({{m}_{1}}+{{m}_{2}}){{R}^{2}}+I} \right]}^{\frac{1}{2}}}gh\]
B) \[{{\left[ \frac{2({{m}_{1}}-{{m}_{2}})gh}{({{m}_{1}}+{{m}_{2}}){{R}^{2}}+I} \right]}^{\frac{1}{2}}}\]
C) \[{{\left[ \frac{{{m}_{1}}+{{m}_{2}}}{({{m}_{1}}+{{m}_{2}}){{R}^{2}}+I} \right]}^{\frac{1}{2}}}gh\]
D) \[{{\left[ \frac{2({{m}_{1}}+{{m}_{2}})gh}{({{m}_{1}}+{{m}_{2}}){{R}^{2}}+I} \right]}^{\frac{1}{2}}}\]
Correct Answer: B
Solution :
\[\Delta K+\Delta U=0\] \[\frac{1}{2}{{m}_{1}}{{v}^{2}}+\frac{1}{2}{{m}_{2}}{{v}^{2}}+\frac{1}{2}I\frac{{{v}^{2}}}{{{r}^{2}}}=({{m}_{1}}-{{m}_{2}})gh\] \[v=\sqrt{\frac{2({{m}_{1}}-{{m}_{2}})gh}{{{m}_{1}}+{{m}_{2}}+\frac{1}{{{r}^{2}}}}}\] \[w=\frac{V}{r}\]You need to login to perform this action.
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