A) \[\frac{2Mg}{2m+M}\]
B) \[\frac{2Mg}{2M+m}\]
C) \[\frac{2mg}{2M+m}\]
D) \[\frac{2mg}{2m+M}\]
Correct Answer: D
Solution :
mg - T = ma \[RT=I\propto \] \[RT=\frac{M{{R}^{2}}}{2}.\frac{a}{R}\] \[T=\frac{Ma}{2}\] \[mg-\frac{Ma}{2}=ma\] \[mg=a\left( \frac{M}{2}+m \right)\] \[mg=a\left( \frac{M+2m}{2} \right)\] \[a=\frac{2mg}{M+2m}\]You need to login to perform this action.
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