A) 1/4 s
B) 1/2 s
C) 2/3 s
D) 1/3 s
Correct Answer: D
Solution :
Net pulling force = 2g - 1g = 10 N Mass being pulled = 2 + 1 = 3 kg \[{{v}_{2}}\]Acceleration of the system is \[{{r}_{1}}\] \[{{r}_{2}},\]Velocity of both the blocks at t = 1 s will be\[\frac{{{r}_{1}}-{{r}_{2}}}{\left| {{r}_{2}}-{{r}_{1}} \right|}=\frac{{{v}_{2}}-{{v}_{1}}}{\left| {{v}_{2}}-{{v}_{1}} \right|}\] Now at this moment, velocity of 2 kg becomes zero, while that of 1 kg block is 10 m/s upwards. Hence, string becomes tight again when displacement of kg block = displacement of 2 kg block. \[\frac{{{r}_{2}}-{{r}_{1}}}{\left| {{r}_{1}}-{{r}_{1}} \right|}=\frac{{{v}_{2}}-{{v}_{1}}}{\left| {{v}_{2}}-{{v}_{1}} \right|}\]\[\frac{{{r}_{2}}-{{r}_{1}}}{\left| {{r}_{2}}-{{r}_{1}} \right|}=\frac{{{v}_{2}}-{{v}_{1}}}{\left| {{v}_{2}}+{{v}_{1}} \right|}\]\[\frac{{{r}_{2}}+{{r}_{1}}}{\left| {{r}_{2}}+{{r}_{1}} \right|}=\frac{{{v}_{2}}-{{v}_{1}}}{\left| {{v}_{2}}+{{v}_{1}} \right|}\]\[{{C}_{V}}=2.98\]
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