The time taken by an object to slide down 450 rough inclined plane is n times as it takes to slide down a perfectly smooth 450 incline plane. The coefficient of kinetic friction between the object and the incline plane is.

l=12g(sin⁡θ−μcos⁡θ)t12=g22(1−μ)t12.....(i)l=12gsin⁡θt22=g22t22……….(ii)From equation (i) & (ii)(1−μ)t12=t22 as t1=nt2we get, n2=11−μ ⇒μ=1−1n2

The time taken by an object to slide down 450 rough inclined plane is n times as it takes to slide down a perfectly smooth 450 incline plane. The coefficient of kinetic friction between the object and the incline plane is.

l=12g(sin⁡θ−μcos⁡θ)t12=g22(1−μ)t12.....(i)l=12gsin⁡θt22=g22t22……….(ii)From equation (i) & (ii)(1−μ)t12=t22 as t1=nt2we get, n2=11−μ ⇒μ=1−1n2

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The time taken by an object to slide down 45⁰ rough inclined plane is n times as it takes to slide down a perfectly smooth 45⁰ incline plane. The coefficient of kinetic friction between the object and the incline plane is.

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$\sqrt{1 - \frac{1}{n^{2}}}$

$1 - \frac{1}{n^{2}}$

$\sqrt{\frac{1}{1 - n^{2}}}$

$1 + \frac{1}{n^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} l = \frac{1}{2} g (\sin θ - μcos θ) t_{1}^{2} = \frac{g}{2 \sqrt{2}} (1 - μ) t_{1}^{2} . . . . . (i) \\ l = \frac{1}{2} gsin {θt}_{2}^{2} = \frac{g}{2 \sqrt{2}} t_{2}^{2} \dots \dots \dots . (ii) \end{array}$
From equation (i) & (ii)
$(1 - μ) t_{1}^{2} = t_{2}^{2} as t_{1} = {nt}_{2}$
$we get, n^{2} = \frac{1}{1 - μ} \Rightarrow μ = 1 - \frac{1}{n^{2}}$