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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 $x = \frac{3 L}{4} i s \frac{9 F}{16}$

Tension at $x = \frac{L}{4} i s \frac{F}{16}$

Tension at midpoint is $\frac{F}{3}$

Tension at midpoint is $\frac{F}{4}$

answer is B, C, D.

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Detailed Solution

$d m = \frac{M}{L} d x d F = μ d m g = k x \frac{M}{L} d x g$

$\int_{}^{} d F = \frac{M}{L} g k \int_{0}^{L} x d x$

$F = \frac{M}{L} g k {[\frac{x^{2}}{2}]}_{0}^{L}$

$= \frac{M}{L} g k [\frac{L^{2}}{2}] = \frac{M g k L}{2}$

Imagine a break at (x,0). Now lets try to write equilibrium between tension at the break and friction on the left portion.

$T = f$

$T = \frac{M g k}{L} \int_{0}^{x} x d x = \frac{M g k x^{2}}{2 L} = \frac{F x^{2}}{L^{2}}$

At mid-point, $x = \frac{L}{2}$ So, $T = \frac{F}{4}$

$a t x = \frac{L}{4}$ , $T = \frac{F}{16}$

$a t x = \frac{3 L}{4}$ , $T = \frac{9 F}{16}$

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