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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A car moves at a speed of $36 \;km\;h^{-1}$ on a level road, The coefficient of fraction between the tyres and the road is $0.8 $. The car negotiates a curve of radius R. If $g =10 ms^2$ the car will skid (or slip) while negotiating the curve if value of R is

$\begin{array}{1 1} 10\; m \\ 13\;m \\ 14\;m \\ 16 \;m \end{array} $

Can you answer this question?

1 Answer

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Answer : $10\; m$
Speed of car $(v) =36 \;km\;h^{-1} =10 ms^{-1}$
The maximum centripetal force that friction can provide is
$f_{max} =\mu \;mg =\large\frac{mv^2}{R}$
$R_{min} =\large\frac{v^2}{\mu g}= \frac{10 \times 10}{0.8 \times 10}$
$\qquad= 12.5\;m$
This is the minimum radius the curve must have for the car to negotiate it with out sliding at a speed of $10 \;ms^{-1}.$
Hence the correct choice is (a)
answered Aug 26, 2014 by meena.p

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