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B
2N
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C
5N
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D
7.5N
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Solution
The correct option is A0.1N The FBD of system & FBD for individual block of 1kg is presented below:
If both blocks move move together, then a=Fmsystem=10101≃0.1m/s2
Now for 1kg block, f=m×a=1×0.1=0.1N
Limiting value of friction fl=(0.5)(1)(g)=5N
We find that f≤fL ∴ frictional force on 1kg block is f= 0.1N