A given object takes $\mathrm{n}$ times the time to slide down $45^{\circ}$ rough inclined plane as it takes…

A given object takes $\mathrm{n}$ times the time to slide down $45^{\circ}$ rough inclined plane as it takes the time to slide down an identical perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :
  1. $\sqrt{1-\frac{1}{n^2}}$
  2. $1-n^2$
  3. $1-\frac{1}{n^2}$
  4. $\sqrt{1-n^2}$

Solution


Case-1 : No friction $\begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta \\ & \ell=\frac{1}{2}(\mathrm{~g} \sin \theta) \mathrm{t}_1^2 \\ & \mathrm{t}_1=\sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \end{aligned}$ Case-2 : With friction $\begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta \\ & \ell=\frac{1}{2}(\mathrm{~g} \sin \theta-\mu \mathrm{g} \cos \theta) \mathrm{t}_2^2 \\ & \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta-\mu g \cos \theta}}=\mathrm{n} \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \\ & \mu=1-\frac{1}{\mathrm{n}^2} \end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 2)

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