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Question

A solid cube is placed on a horizontal surface. The coefficient of friction between them is μ, where μ<1/2. A variable horizontal force is applied on the cube's upper face, perpendicular to one edge and passing through the mid-point of the edge, as shown in figure. The maximum acceleration with which it can move without toppling is


A
g(12μ)
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B
g(1+2μ)
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C
g2(12μ)
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D
g2(1+2μ)
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Solution

The correct option is C g2(12μ)
FBD of the block given in the question, viewed from the side is shown below.


By using principle of moments, we balance clockwise and anticlockwise torque about O.
Thus, we get
P×a=(mg)×a2 (where, a side of cube)
P=mg2 ... (1) [for just toppling the block]
Now, applying Newton's 2nd law on the cube:
PμN=mamax
mg2μmg=mamax
amax=g2(12μ)
Thus, option (c) is the correct answer.

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