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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2024 · 8 April · Evening Shift · Physics · Laws of Motion

A given object takes n times the time to slide down 45^ rough...

A given object takes n\mathrm{n} times the time to slide down 45∘45^{\circ} rough inclined plane as it takes the time to slide down an identical perfectly smooth 45∘45^{\circ} inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :A given object takes n times the time to slide down 45^ rough inclined plane as it takes the time to slide down an

Options

  1. A1−1n2\sqrt{1-\frac{1}{n^2}}
  2. B1−n21-n^2
  3. C1−1n21-\frac{1}{n^2}
  4. D1−n2\sqrt{1-n^2}

Correct answer

C. 1−1n21-\frac{1}{n^2}

Step-by-step solution

A given object takes n times the time to slide down 45^ rough inclined plane as it takes the time to slide down an
Case-1 : No friction a=gsin⁡θℓ=12( gsin⁡θ)t12t1=2ℓgsin⁡θ\begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta \\ & \ell=\frac{1}{2}(\mathrm{~g} \sin \theta) \mathrm{t}_1^2 \\ & \mathrm{t}_1=\sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \end{aligned} Case-2 : With friction a=gsin⁡θ−μgcos⁡θℓ=12( gsin⁡θ−μgcos⁡θ)t222ℓgsin⁡θ−μgcos⁡θ=n2ℓgsin⁡θμ=1−1n2\begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta \\ & \ell=\frac{1}{2}(\mathrm{~g} \sin \theta-\mu \mathrm{g} \cos \theta) \mathrm{t}_2^2 \\ & \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta-\mu g \cos \theta}}=\mathrm{n} \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \\ & \mu=1-\frac{1}{\mathrm{n}^2} \end{aligned}
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