A) \[\frac{{{m}_{1}}}{{{m}_{2}}}F\]
B) \[\frac{{{m}_{2}}}{{{m}_{1}}}F\]
C) \[\frac{{{m}_{1}}F}{{{m}_{1}}+{{m}_{2}}}\]
D) \[\frac{{{m}_{2}}F}{{{m}_{1}}+{{m}_{2}}}\]
Correct Answer: D
Solution :
We know that the forces gets distributed in the direct proportion of the masses. So force acting on block of mass\[{{m}_{2}}\], must be equal to force per unit mass times the mass of block\[2\]. \[=\frac{{{m}_{1}}{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}=\frac{9\times 10}{10}=9\,\,N\]
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