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Q.

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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a

b

2mμ+1

c

10m1-μ

d

10mμ-1

answer is .

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Detailed Solution

FBD of AIf the acceleration of ‘C’ is aFor block ‘A’ N = 8ma -------(1)          8mg=μN = μ8ma-------(2)    a= gμand acceleration a can be written by the equation of system (A+B+C)m1g = (10 m + m1)a----(3)8mg = μ8m(m1g10m + m1)10m +m1 = μm110m = (μ-1)m1⇒m1 = 10mμ-1
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