A) \[\frac{{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}\]
B) \[\sqrt{\frac{2d({{m}_{1}}+{{m}_{2}})}{{{m}_{2}}g}}\]
C) \[\sqrt{\frac{2{{m}_{2}}d}{({{m}_{1}}+{{m}_{2}})g}}\]
D) \[\frac{4d({{m}_{1}}+{{m}_{2}})}{3g{{m}_{1}}}\]
Correct Answer: B
Solution :
[b]: Let a be common acceleration of the system. The free body diagrams of two blocks are as shown in the figure. Their equations of motion are \[T={{m}_{1}}a\] ...(i) \[{{m}_{2}}g-T={{m}_{2}}a\] ...(ii) From (i) and (ii), we get\[a=\frac{{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}\] Using, \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[\Rightarrow \]\[d=0\times t+\frac{1}{2}\frac{{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}{{t}^{2}}\] (Using (iii)) or\[t=\sqrt{\frac{2d({{m}_{1}}+{{m}_{2}})}{{{m}_{2}}}}\]You need to login to perform this action.
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