A wheel is rotating with an angular speed of 20 rad / sec . It is stopped to rest by applying a constant torque in 4s . If the moment of inertia of the wheel about its axis is 0.20 kg-m², then the work done by the torque in two seconds will be

The correct option is 30 J.

Here is the step-by-step explanation of the solution:

1. Identify the given parameters:
Initial angular speed of the wheel,
${\omega }_0=20\text{rad/s}$
Final angular speed (since it comes to rest),
$\omega =0\text{rad/s}$
Time taken to come to rest,
$t=4\text{s}$
Moment of inertia of the wheel,
$I=0.20{\text{kg-m}}^2$

2. Calculate the angular acceleration ($\alpha$):
Using the first equation of rotational motion:
$\omega ={\omega }_0+\alpha t$
Substitute the values into the equation:
$0=20+\alpha \left(4\right)$
$4\alpha =-20$
$\alpha =-5{\text{rad/s}}^2$
The negative sign indicates a constant retarding angular acceleration.

3. Find the angular speed at $t=2\text{seconds}$:
Using the equation of motion again to find the angular velocity (${\omega }_2$) after 2 seconds:
${\omega }_2={\omega }_0+\alpha t$
${\mathrm{��}}_2=20+(-5)\left(2\right)$
${\omega }_2=20-10=10\text{rad/s}$

4. Apply the Work-Energy Theorem:
According to the work-energy theorem for rotational motion, the work done by the torque is equal to the change in the rotational kinetic energy of the wheel:
$W=\Delta K=K_f-K_i$
Where the initial kinetic energy is:
$K_i=\frac{1}{2}I{\omega }_0^2=\frac{1}{2}\left(0.20\right){\left(20\right)}^2=0.10\times 400=40\text{J}$
And the final kinetic energy after 2 seconds is:
$K_f=\frac{1}{2}I{\omega }_2^2=\frac{1}{2}\left(0.20\right){\left(10\right)}^2=0.10\times 100=10\text{J}$

5. Calculate the work done:
$W=10\text{J}-40\text{J}=-30\text{J}$
The work done by the retarding torque is $-30\text{J}$ (the negative sign indicates energy is removed from the system). The magnitude of the work done is 30 J.