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If the coefficient of friction between the wedge ‘A’ and block ‘B’ shown in the figure is ' $μ$ ', then the maximum possible horizontal acceleration of ' $A$ ' for which ' $B$ ' doesn’t slip is [angle of inclination of wedge $= 45^{0}$ ]

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$g (\frac{1 - μ}{1 + μ})$

$μ g$

$\frac{g}{μ}$

$g (\frac{1 + μ}{1 - μ})$

answer is B.

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Detailed Solution

For max acceleration the block is just about to move up the incline.

$m a \cos θ = m g \sin θ + f_{k}, f_{k} = μ N N = m g \cos θ + m a \sin θ m a \cos θ = m g \sin θ + μ (m g \cos θ + m a \sin θ) a = g \frac{(\sin θ + μ \cos θ)}{(\cos θ - μ \sin θ)}$

$\cos θ = 45^{0} \Rightarrow a = \frac{g (1 + μ)}{(1 - μ)}$

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