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The coefficient of static friction, $μ_{s}$ between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. $(g = 10 m / s^{2})$ .

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0.4 kg

2.0 Kg

4.0 Kg

0.2 Kg

answer is A.

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Detailed Solution

$m_{B} g = μ_{s} m_{A} g {$ $∵ m_{A} g = μ_{s} m_{A} g\}$

$\Rightarrow m_{B} = μ_{s} m_{A}$

$Or, m_{B} = 0.2 \times 2 = 0.4 kg$

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