A block of mass m is lying on a wedge having inclination angle $α = \tan^{- 1} (\frac{1}{5})$ . Wedge is moving with a constant acceleration $a = 2 m s^{- 2}$ . The minimum value of coefficient of friction $μ$ so that m remains stationary w.r.t wedge is

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1/5

2/9

5/12

2/5

answer is B.

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Detailed Solution

FBD of m in frame of wedge,

$N = m g \cos α - m a \sin α N o w f = μ N = m a \cos a + m g \sin α μ = \frac{a \cos α + g \sin α}{g \cos a - a \sin α} = \frac{a + g \tan α}{g - a \tan α} = \frac{5}{12}$

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