A) \[{{\mu }_{k}}=1-\frac{1}{{{n}^{2}}}\]
B) \[{{\mu }_{k}}=\sqrt{1-\frac{1}{{{n}^{2}}}}\]
C) \[{{\mu }_{s}}=1-\frac{1}{{{n}^{2}}}\]
D) \[{{\mu }_{s}}=\sqrt{1-\frac{1}{{{n}^{2}}}}\]
Correct Answer: A
Solution :
When friction is absent \[{{a}_{1}}=g\sin \theta \] \[\therefore \] \[{{S}_{1}}=\frac{1}{2}{{a}_{1}}t_{1}^{2}\] ...(i) When friction is present \[{{a}_{2}}=g\sin \theta -{{\mu }_{k}}g\cos \theta \] \[{{S}_{1}}=\frac{1}{2}{{a}_{2}}t_{2}^{2}\] ...(ii) When friction is present \[{{a}_{2}}=g\sin \theta -{{\mu }_{k}}g\cos \theta \] \[{{S}_{2}}=\frac{1}{2}{{a}_{2}}t_{2}^{2}\] From Eqs. (i) and (ii), we get \[\frac{1}{2}{{a}_{1}}t_{1}^{2}=\frac{1}{2}{{a}_{2}}t_{2}^{2}\] \[{{a}_{1}}t_{1}^{2}={{a}_{2}}{{(n{{t}_{1}})}^{2}}\] \[{{a}_{1}}={{n}^{2}}{{a}_{2}}\] \[\frac{{{a}_{2}}}{{{a}_{1}}}=\frac{g\sin \theta -{{\mu }_{k}}g\cos \theta }{g\sin \theta }=\frac{1}{{{n}^{2}}}\] \[\frac{g\sin 45{}^\circ -{{\mu }_{k}}\cos 45{}^\circ }{g\sin 45{}^\circ }=\frac{1}{{{n}^{2}}}\] \[1-{{\mu }_{k}}=\frac{1}{{{n}^{2}}}\]or \[{{\mu }_{k}}=1-\frac{1}{{{n}^{2}}}\]
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