एक कार R त्रिज्या की वक्रीय सड़क पर चल रही है। सड़क θ कोण पर आनत है। कार के टायरों और सड़क के बीच घर्षण गुणांक μs है। इस सड़क पर अधिकतम सुरक्षित वेग है:

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AIIMS BSc NURSING 2024 Memory-Based Paper
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  1. \(\sqrt{g R^{2} \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}\)
  2. \(\sqrt{g R \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}\)
  3. \(\sqrt{\frac{g}{R} \frac{\mu_{s}+\tan \theta}{1-\mu_{2} \tan \theta}}\)
  4. \(\sqrt{\frac{g}{R^{2}} \frac{\mu_{\mathrm{s}}+\tan \theta}{1-\mu_{\mathrm{s}} \tan \theta}}\)

Answer (Detailed Solution Below)

Option 2 : \(\sqrt{g R \frac{\mu_{s}+\tan \theta}{1-\mu_{s} \tan \theta}}\)
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AIIMS BSc NURSING 2024 Memory-Based Paper
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गणना

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उपरोक्त आरेख से, हमारे पास है

N = mg cosθ + (mv² / R) sinθ (1)

fmax = μN ⇒ fmax = μs mg cosθ + (μs mv² / R) sinθ

mg sinθ + fmax = (mv² / R) cosθ (2)

मान रखने पर,

mg sinθ + μs mg cosθ + (μs mv² / R) sinθ = (mv² / R) cosθ

g sinθ + μs g cosθ = (v² / R)(cosθ − μs sinθ)

gR [ (tanθ + μs) / (1 − μs tanθ) ] = v²

v = √[ gR (tanθ + μs) / (1 − μs tanθ) ]

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