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Q.

A block of mass m is lying on a wedge having inclination angle α=tan−1⁡15. Wedge is moving with a constant acceleration a=2ms−2. The minimum value of coefficient of friction μ so that m remains stationary w.r.t. wedge is

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a

2/9

b

5/12

c

1/5

d

2/5

answer is B.

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Detailed Solution

FBD of m in frame of wedge,         N=mg cos α−ma sin αNow f=μN=macos⁡α+mgsin⁡α         μ=acos⁡α+gsin⁡αgcos⁡α−asin⁡α=a+gtan⁡αg−atan⁡α=512
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A block of mass m is lying on a wedge having inclination angle α=tan−1⁡15. Wedge is moving with a constant acceleration a=2ms−2. The minimum value of coefficient of friction μ so that m remains stationary w.r.t. wedge is