A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is:

  • Option 1)

    20 ms-1

  • Option 2)

    30 ms-1

  • Option 3)

    5 ms-1

  • Option 4)

    10 ms-1

 

Answers (1)

As we learnt in

Bending a Cyclist -

From figure. 

Rsin	heta=frac{mv^{2}}{r}                            (i)

Rcos	heta=mg                            (ii)

(i) & (ii)

tan	heta=frac{v^{2}}{rg}

	heta=tan^{-1}left( frac{v^{2}}{rg} 
ight )

V = velocity

r = radius of track

	heta= angle with which cycle leans

- wherein

*    radius of curve is small.

*    velocity of the cyclist is large.

 

 

 

 

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

\theta = 45^{\circ}

r=90m

g=10m/s^{2}

\tan 45^{\circ}=\frac{v^{2}}{90\times 10}

V=30m/s


Option 1)

20 ms-1

Incorrect

Option 2)

30 ms-1

Correct

Option 3)

5 ms-1

Incorrect

Option 4)

10 ms-1

Incorrect

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