In the arrangement shown in Fig. there is no friction between the block of mass $2 m$ and ground, but there is friction between of the blocks of masses m and $2 m$ . The block of mass m is stationary with respect to block of mass $2 m$ . The value of coefficient of friction between m and $2 m$ is:

\frac{1}{2}

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

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\frac{1}{4}

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\frac{1}{3}

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Solution

The correct option is B $\frac{1}{4}$
REF.Image.
Net acceleration = $\frac{D r i v i n g F o r c e}{T o t a l M a s s}$
$a = \frac{m g}{4 m} \Rightarrow a = g / 4$
As mass m B stationary w.r.t 2 m, friction
should provide this effect.
$m m g = m g / 4$
$m = 1 / 4$

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