Angular speed of a particle increases from 2rads-1 to 4 rads-1 across any two diameterically opposite positions. Its angular acceleration will be:

Explanation

The correct answer is Option C:The following is the solution:We can use the equation of motion in angular form, 2as = ω₂² - ω₁², where s is the angular displacement, ω₁ is the initial angular velocity, and ω₂ is the final angular velocity.If the angular speed of the particle increases from 2 radians per second to 4 radians per second across any two diametrically opposite positions, then the change in angular velocity is:Δω = ω₂ - ω₁ = 4 rad/s - 2 rad/s = 2 rad/sLet's assume that the particle travels a distance of π radians across the two diametrically opposite positions. Then, we can use the equation of motion in angular form to calculate the angular displacement as follows:2as = ω₂² - ω₁²2(π)α = (4 rad/s)² - (2 rad/s)²πα = 6 rad/s²Therefore, the angular acceleration of the particle when its angular speed increases from 2 radians per second to 4 radians per second across any two diametrically opposite positions is 6/π radians per second squared..The other options are incorrect because they do not follow the correct formula and calculations for angular acceleration in this scenario.