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Exam
NEET
Subject
Physics
Chapter
Laws of Motion

NEET · Physics · Laws of Motion

Statement I: A cyclist is moving on an unbanked road with a speed...

Statement I: A cyclist is moving on an unbanked road with a speed of 7kmh−1andtakesasharpcircularturnalongapathoftheradiusof7km{h}^{- 1} and takes a sharp circular turn along a path of the radius of2mwithoutreducingthespeed.Thestaticfrictioncoefficientiswithout reducing the speed. The static friction coefficient is0 . 2.Thecyclistwillnotslipandpassthecurve(g=9.8ms−2). The cyclist will not slip and pass the curve \left(g = 9 . 8m{s}^{- 2}\right)

Statement II : If the road is banked at an angle of 45°45 °, cyclist can cross the curve of 2m2m radius with the speed of 18.5kmh−118 . 5km{h}^{- 1} without slipping. In the light of the above statements, choose the correct answer from the options given below.

Options

  1. AStatement I is incorrect and statement II is correct
  2. BStatement I is correct and statement II is
    incorrect
  3. CBoth statement I and statement II are false
  4. DBoth statement I and statement II are true

Correct answer

D. Both statement I and statement II are true

Step-by-step solution

Statement I{v}_\max = \sqrt{\muRg} = \sqrt{\left(0 . 2\right) \times 2 \times 9 . 8}

{v}_\max = 1 . 97m{s}^{- 1}

7kmh−1=1.944ms−17km{h}^{- 1} = 1 . 944m{s}^{- 1}

Speed is lower than {v}_\max, hence it can take safe
turn.

Statement II 

{v}_\max = \sqrt{Rg \left[\dfrac{tan \theta + \mu}{1 - \mu tan \theta}\right]}

=2×9.8[1+0.21−0.2]=5.42ms−1= \sqrt{2 \times 9 . 8\left[\dfrac{1 + 0 . 2}{1 - 0 . 2}\right]} = 5 . 42m{s}^{- 1}

18.5kmh−1=5.14ms−118 . 5km{h}^{- 1} = 5 . 14m{s}^{- 1}

Speed is lower than {v}_\max, hence it can take safe turn.

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