A wheel of moment of inertia 2.5 kg `m^2` has an initial angular velocity of 40 rad `s^(-1)`. A constant torque of 10 N m acts on the wheel. The time during which the wheel is accelerated to 60 rad `s^(-1)` is

Options

A. 4.5 s
B. 6 s
C. 5 s
D. 2.5 s

Answer

C

Explanation

Here , I = `2.5 kg m^(2) , omega_(0) = 40 rad s^(-1)`
`tau = 10 N m , omega = 60 rad s^(-1)`
As `tau = I alpha or alpha = (tau)/(I) = (10)/(2.5) = 4 rad s^(-2)`
Using `omega = omega_(0) + alpha t , t = (omega - omega_(0))/(alpha)`
Substituting the values , we get `t = (60 - 40)/(4 ) = 5`s