An autmobile moves on road with a speed of `54 km//h`. The radius of its wheel is `0.45 m` and the moment of inertia of the wheel about its axis of rotation is `3 kg m^(2)`. If the vehicle is brought to rest in `15 s`, the magnitude of average torque tansmitted by its brakes to the wheel is :

Options

A. `2.86 kg m^(2) s^(-2)`
B. `6.66 kg m^(2) s^(-2)`
C. `8.58 kg m^(2) s^(-2)`
D. `10.86 kg m^(2) s^(-2)`

Answer

B

Explanation

Here, `u = 54 km//h = 15 m//s, r = 0.45m`,
`I = 3kg m^(2), t = 15 sec, omega = 0`
Now `u = r omega_(0) or omega_(0)= (u)/(r ) = (15)/(0.45) = (100)/(3)`
As `omega = omega_(0) + alpha t`
`0 = (100)/(3) + (-alpha) (15)` [As force is retarding]
or `alpha = (100)/(54)`, Also `tau = I alpha`
`= 3 xx (100)/(45) = 6.66 kg m^(2) s^(-2)`