Two blocks of masses $M_{1}$ and $M_{2}$ are connected with a string passing over a pulley as shown in the figure. The block $M_{1}$ lies on a horizontal surface. The coefficient of friction between the block $M_{1}$ and the horizontal surface is $μ$ . The system accelerates. What additional mass $m$ should be placed on the block $M_{1}$ so that the system does not accelerate?

\frac{M_{2} - M_{1}}{μ}

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\frac{M_{2}}{μ} - M_{1}

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M_{2} - \frac{M_{1}}{μ}

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(M_{2} - M_{1}) μ

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Solution

The correct option is B $\frac{M_{2}}{μ} - M_{1}$
For the system to be in equilibrium:
For block of mass $M_{1} :$
Frictional force, $f =$ tension in the string, $T$
where $f = μ (m + M_{1}) g$ ... (1)
$∴ T = μ (m + M_{1}) g$ ... (2)
For the equilibrium of block of mass $M_{2} :$
$T = M_{2} g$ ... (3)
Hence, $μ (m + M_{1}) g = M_{2} g$
$∴ m = \frac{M_{2}}{μ} - M_{1}$
Hence, the correct answer is option (b).

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