A body moves on a horizontal circular road of radius `r`, with a tangential acceleration `a_(t)`. The coefficient of friction between the body and the road surface Is `mu`. It begins to slip when its speed is `v`.
(i) `v^(2)=murg`
(ii) `mug=(v^(4)/(r^92))+a_(t))`
(iii) `mu^(2)g^(2)=(v^(4)/(r^(2)+a_(t)^(2))`
(iv) The force of friction makes an angle `tan^(-1)(v^(2)//a_(t)r)` with the direction of motion at the point of slipping.

Options

A. `(i),(ii)`
B. `(ii),(iii)`
C. `(i),(iv)`
D. `(iii),(iv)`

Answer

D

Explanation

`sqrt(F_(c)^(2)+F_(t)^(2))lemumg`
`((mv^(2))/(r ))^(2)+(ma_(t))^(2)le(mu^(2)m^(2)g^(2))`
`(v^(4))/(r^(2))+a_(t)^(2)lemu^(2)g^(2)`
`tan theta=(F_(c))/(F_(t))=(mv^(2)//r)/(ma_(t))=(v^(2))/(a_(t)r)`
`theta=tan^(-1)((v^(2))/(a_(t)r))`