A) \[\frac{{{a}_{2}}}{R}\]
B) \[\frac{({{a}_{2}}+g)}{R}\]
C) \[\frac{2({{a}_{2}}+g)}{R}\]
D) None of these
Correct Answer: C
Solution :
[c] \[{{m}_{2}}g-T={{m}_{2}}{{a}_{2}}\] ? (i) \[TR=\frac{{{m}_{2}}{{R}^{2}}}{2}{{\alpha }_{2}}\] ? (ii) \[{{\alpha }_{1}}R={{a}_{2}}-{{\alpha }_{2}}R\] ? (iii) \[TR=\left( \frac{{{m}_{1}}{{R}^{2}}}{2} \right){{\alpha }_{1}}\] ? (iv) \[{{\alpha }_{2}}=\frac{2T}{{{m}_{2}}R}=\frac{2}{{{m}_{2}}R}.({{m}_{2}}{{a}_{2}}+{{m}_{2}}g)\] \[=2\left( \frac{{{a}_{2}}+g)}{R} \right)\]You need to login to perform this action.
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