A) \[15\text{ }kg\]
B) \[10\text{ }kg\]
C) \[5\,kg\]
D) \[0\,kg\]
Correct Answer: A
Solution :
Let T be the tension in the string; f= frictional force between block A and table; m?= minimum mass of C. For the just motion of block A on table \[T=f=\mu R=\mu (m+m')g=0.2(10+m')g\] ?.(i) For the just motion of block B, \[T=\text{ }5g\] ....(ii) From (i) and (ii), \[5g=0.2(10+m')g\] or \[5=2+0.2m'\] or \[m'=\frac{5-2}{0.2}=15kg\]You need to login to perform this action.
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