question_answer

Two blocks A and B of masses \[{{m}_{A}}=1kg\]and \[{{m}_{B}}=3kg\]are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : (Take \[g=10\text{ }m/{{s}^{2}}\])

[JEE Main 10-4-2019 Afternoon]

A) 16 N

B) 40 N

C) 12 N

D) 8 N

Correct Answer: A

Solution :

[a]

\[{{a}_{A\max }}=\mu g=2m/{{s}^{2}}\]

\[F-8=4\times 2\]

\[F=16N\]