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In the arrangement shown in the figure mass of the block B and A are 2 m,,8 m respectively. Surface between B and floor is smooth. The block B is connected to block C by means of a pulley. If the whole system is released then the minimum value of mass of the block C so that the block A remains stationary with respect to B is: (Coefficient of friction between A and B is $μ$ and pulley is ideal)

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$\frac{10 m}{μ - 1}$

$\frac{2 m}{μ + 1}$

$\frac{m}{μ}$

$\frac{10 m}{1 - μ}$

answer is .

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Detailed Solution

FBD of A

If the acceleration of ‘C’ is a

For block ‘A’ N = 8ma -------(1)

8mg= $μN = μ 8 ma - - - - - - - (2)$ a= $\frac{g}{μ}$

and acceleration a can be written by the equation of system (A+B+C)

$m_{1} g = (10 m + m_{1}) a - - - - (3)$

$8 mg = μ 8 m (\frac{m_{1} g}{10 m + m_{1}})$

10m + $m_{1} = {μm}_{1}$

10m = $(μ - 1) m_{1}$

$\Rightarrow m_{1} = \frac{10 m}{μ - 1}$

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