NEETPhysicsLaws of MotionMCQ+4 / −1
The coefficient of static friction, 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/s2)
- A2.0 kg
- B4.0 kg
- C0.2 kg
- D0.4 kg
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Correct answer: D
We get equations
$T + ma = {f_\mu }$ or $$T = \mu {N_A}(for\,a = 0)$$
and T = ma + mg or T = mBg (for a = 0)
$ \therefore $ $\mu {N_A} = {m_B}g$
$$ \Rightarrow {m_B} = \mu {m_A} = 0.2 \times 2 = 0.4\,kg$$
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