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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A car of mass $2 \times 10 ^3 kg$ is to travel over a turn of radius $0.75 km$ and slope of angle $5^{\circ}$ If coefficient of friction between wheel and road is 0.5 maximum speed the car can travel with out slipping is $(\tan 5^{\circ}=0.087)$

\[(a)\;9.8 m/s \quad (b)\;67 m/s \quad (c)\;0.75 m/s \quad(d)\;1 m/s \]
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$ R \cos \theta- f\sin \theta=mg$
$ R \sin \theta+ f\cos \theta=\large\frac{mv^2}{r}$
$f=\mu R$
$=>\large\frac{v^2}{rg}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}$
$=>\large\frac{\tan \theta+\mu}{1-\mu \sin \theta}$
$v=\sqrt {\large\frac{750 \times 10(0.087 +0.5)}{1-(0.5 \times 0.087)}}$
$\quad= 67 \;m/s$
Hence b is the correct answer.


answered Jul 5, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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