Two blocks of mass M₁ and M₂ are connected with a string which passes over a smooth pulley. The mass $M_{1}$ is placed on a rough inclined plane as shown in fig. The coefficient of friction between the
block and the inclined plane is µ. What should be the maximum mass M₂ so that block M₁ slides downwards ?

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$M_{2} = M_{1} (\sin θ + μcos θ)$

$M_{2} = M_{1} (\sin θ - μcos θ)$

$M_{2} = \frac{M_{1}}{(\sin θ + μcos θ^{'})}$

$M_{2} = \frac{M_{1}}{(\sin θ - μcos θ)}$

answer is B.

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Detailed Solution

For downward acceleration,

$\begin{array}{l} (M_{1} gsin θ - {μM}_{1} gcos θ) \geq M_{2} g \\ M_{1} (\sin θ - μcos θ) \geq M_{2} \\ ∴ {(M_{2})}_{max} = M_{1} (\sin θ - μcos θ) . \end{array}$

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