A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)

The given rotational speed is 60 rpm (revolutions per minute). To convert this value to radians per second, use the following conversion: 1 revolution = \( 2\pi \) radians, 1 minute = 60 seconds. \(Rotational~speed~(in~radians~per~second) = \frac{60 ~revolutions}{minute} \times \frac{2\pi ~radians}{1~revolution} \times \frac{1 ~minute}{60~seconds} = \pi~radians~per~second\)

The initial angular velocity, denoted by \( \omega_i \), is the same as the rotational speed in radians per second, which we calculated in step 1. Therefore, \( \omega_i = \pi~rad/s \)

Since we want to stop the wheel, the final angular velocity, denoted by \( \omega_f \), should be equal to 0.

The time required to stop the wheel, denoted by \(t\), is given as one minute. We need to convert this time to seconds: \(t = 1 ~minute \times \frac{60 ~seconds}{1 ~minute} = 60~seconds\)

Now, let's use the equation of motion for angular velocity, which is \(\omega_f = \omega_i + \alpha t \), where \(\alpha\) is the angular acceleration. We already have \(\omega_f = 0\), \(\omega_i = \pi~rad/s\), and \(t = 60~s\). Plugging these values into the equation, we get: \(0 = \pi + \alpha \times 60\) Now, solve for \(\alpha\): \(\alpha = - \frac{\pi}{60}~rad/s^2\)

Finally, let's use the formula for torque: \( \tau = I\alpha \), where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. Given moment of inertia, \(I = 2 kg.m^2 \) and angular acceleration, \(\alpha = -\frac{\pi}{60}~rad/s^2\), we can find the torque: \(\tau = (2 ~ kg.m^2) \times\left(-\frac{\pi}{60}~ rad/s^2\right)\) \(\tau = -\left(\frac{\pi}{30}\right) ~ N.m\) Thus, the torque required to stop the wheel in one minute is \( -\left(\frac{\pi}{30}\right) ~ N.m \). This torque has a negative sign indicating that it acts in the opposite direction to the rotational motion of the wheel, which is needed to stop the wheel. Comparing this answer to the given options, the correct choice is: \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\)