A) 30 %
B) 40 %
C) 70%
D) 50%
Correct Answer: B
Solution :
Since each mass is moving 50 m in 5 s, therefore using the relation \[s=ut+\frac{1}{2}a{{t}^{2}},\]we have \[50=0\times 5+\frac{1}{2}\times a\times {{5}^{2}}\] or \[a=\frac{100}{25}=4\,m{{s}^{2}}\] Let mass of one become\[{{m}_{1}}\]and that of other \[{{m}_{2}},\]where\[{{m}_{1}}>{{m}_{2}}\]. As\[{{m}_{1}}\]moves downwards with acceleration \[a=4\,m{{s}^{-2}}\] \[\therefore \] \[a=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)g\] So, \[4=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)(10)\] \[\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)=\frac{4}{10}=\frac{2}{5}\] \[\therefore \]% of mass transferred \[=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\times 100\] \[=\frac{2}{5}\times 100\] \[=40%\]
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