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A car of mass $m$ is moving on a horizontal circular path of radius $r$ . At an instant its speed is $v$ and is increasing at a rate $a$

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the friction coefficient between the ground and the car is $μ = \tan^{- 1} (\frac{v^{2}}{r g})$

the friction coefficient between the ground and the car is not less than $a / g$

the acceleration of the car is towards the center of the path

the magnitude of the frictional force on the car is greater than $\frac{m v^{2}}{r}$

answer is B, C.

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Detailed Solution

$F_{t} = m a$ and $F_{c} = \frac{m v^{2}}{r}$

$∴ F$ = Frictional force = $\sqrt{F_{c}^{2} + F_{t}^{2}} > \frac{m v^{2}}{r}$

Also $μ = \tan θ = \frac{F_{t}}{m g} = \frac{a}{g}$

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