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In the arrangement shown in figure, there is a friction force between the blocks of masses m and 2m. The mass of the suspended block is m. The block of mass m is stationary with respect to block of mass 2m. The minimum value of coefficient of friction between m and 2m is

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$\frac{1}{\sqrt{2}}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{3}$

answer is C.

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Detailed Solution

Maximum acceleration due to friction of mass mover mass 2m can be μg. Now, for the whole system

$a = \frac{Net pulling force}{Total mass}$

$∴ μ g = \frac{m g}{4 m}$

or $μ = \frac{1}{4}$

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