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A uniform rod of length ‘L’ and mass ‘M’ has been placed on a rough horizontal surface. The horizontal force ‘F’ applied on the rod such that the rod is just in the state of rest. If the coefficient of friction varies according to the relation $μ = K x$ . Where ‘K’ is $a + ve$ constant. Then

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Tension at

x = \frac{3L}{4} is \frac{9F}{16}

Tension at mid point is

F/4

Tension at mid point is

F/3

Tension at

x = \frac{L}{4} is \frac{F}{16}

answer is B, C, D.

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Detailed Solution

$- \int_{0}^{T} dF = \int_{o}^{x} \frac{M}{L} d x gk x$

$T = \frac{Mg}{L} \frac{x^{2}}{2}$ $If x = \frac{L}{2}, T = \frac{F}{4}$

$If x = L, F = \frac{Mg}{L} \frac{L^{2}}{2} = \frac{MgL}{2}$

$If x = \frac{L}{4}, T = \frac{Mg}{2L} \frac{L^{2}}{16} = \frac{F}{16}$

$If x = \frac{3L}{4}, T = \frac{Mg}{2L} \frac{{9L}^{2}}{16} = \frac{9F}{16}$

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