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A car is taking turn on a circular path ...

A car is taking turn on a circular path of radius R. If the coefficient of friction between the tyres and road is `mu`, the maximum velocity for no slipping is

A

`muRg`

B

`2muRg`

C

`(muRg)^(1//2)`

D

`(2muRg)^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(mv^(2))/(R)lemumgrArrv_("max")=sqrt(muRg)`
`v_("max")` : maximum velocity for no slipping
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