A car starts moving from rest on a horizontal circular road of radius of curvature R at . The speed of the car is increasing at a constant rate a. The coefficient of friction between the road and the tyres of the car is The car starts slipping at time`t = t_(0)` . Then `t_(0)` is equal to:

Options

A. `(.^(4)sqrt(mu^(2) g^(2) R^(2)))/(a)`
B. `(.^(4)sqrt(R^(2) (mu^(2) g^(2) - a^(2))))/(a)`
C. `(.^(4)sqrt(R^(2) (a^(2) - mu^(2) g^(2))))/(a)`
D. `(.^(4)sqrt(R^(2) (mu^(2) g^(2) + a^(2))))/(a)`

Answer

B

Explanation

At instant of slipping velocity `v = at_(0) ((v^(2))/(R ))^(2) + a^(2) = mu^(2) g^(2)`
Solving we get `t_(0) = (.^(4)sqrt(R^(2) (mu^(2) g^(2) - a^(2))))/(a)`