A) 2.5 N
B) 5 N
C) 7.84 N
D) 10 N
Correct Answer: C
Solution :
Limiting frictional force \[{{F}_{s}}={{\mu }_{s}}R\] \[={{\mu }_{s}}mg\] \[=0.4\times 2\times 9.8=7.84\text{ }N\]
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