A block of mass m is placed on a wedge of mass M. Coefficient of friction between them is $μ > \cot θ$ . The minimum acceleration of the wedge at which block appears stationary relative to wedge is

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\frac{g (\cos θ - μ \sin θ)}{\sin θ + μ \cos θ}

\frac{g (\sin θ + μ \cos θ)}{\cos θ - μ \sin θ}

g (\sin θ + μ \cos θ)

\frac{g (\sin θ - μ \cos θ)}{\cos θ + μ \sin θ}

answer is B.

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Detailed Solution

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$N = m (g \cos θ + a \sin θ)$

$ma \cos θ = mg \sin θ + μ N$

$macos θ = mgsin θ + μ mg \cos θ + μ ma \sin θ$

$ma (\cos θ - μ \sin θ) = mg (\sin θ + μ \cos θ)$

$a = \frac{g (\sin θ + μ \cos θ)}{\cos θ - μ \sin θ}$

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