A motorcycle moving with a velocity of `72kmh^(-1)` on a float takes a turn on the road at a point where the radius of curvature of the road is 20m. The acceleration due to gravity is `10ms^(-2)`. In order to avoid skidding he must not behind with respect to the vertical plane by an angle greater than

Options

A. `theta=tan^(-1)(2)`
B. `theta =tan^(-1)(6)`
C. `theta=tan^(-1)(4)`
D. `theta=tan^(-1)(25.92)`

Answer

A

Explanation

Using the formula for motor cycle, not to skid
`theta=tan^(-1)((v^(2))/(rg))`
When `t=20m, v=72km//h=72xx5/18=20m//s`
`theta=tan^(-1)((20xx20)/(20xx10))=tan^(-1)(2)`