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A turn of radius 20 m is banked  for the vehicle going to a speed 5 m/s

If the width of a road is 8 m then what should be the height (in m)  of outer edge w.r.t  inner edge of the road 

Option: 1

1


Option: 2

0.5


Option: 3

0.75


Option: 4

0.25


Answers (1)

best_answer

As we have learned

Banking of Road -

From figure

R\cos\theta=mg                                (i)

R\sin\theta=\frac{mv^{2}}{r}                                (ii)

tan\theta=\frac{v^{2}}{rg}

tan\theta=\frac{\omega^{2}r}{g}=\frac{V\omega}{g}=\frac{h}{l}

h = height of outer edge

l=length of rod

r = radius

\omega=angular velocity

 

 

-

 

 

  From figure 

N \sin \theta = \frac{mv^2}{R}

N \cos \theta = mg

\tan \theta = \frac{v^2 }{Rg }

For small angle  \theta \: \: \: \tan \theta \approx \sin \theta \approx \frac{h}{l}

\frac{h}{l} = \frac{v^2 }{Rg } \Rightarrow = \frac{5 \times 5 }{20 \times 10 }

\frac{h}{l} = \frac{1}{8}

\Rightarrow h = l/8 \Rightarrow 8/8 = 1 m

 

 

 

 

 

 

 

Posted by

vinayak

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