The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$. The torque on the wheel becomes zero at $t = $ ...... $s$.

$A$ disc of mass $10 \ kg$ and radius $0.1 \ m$ is rotating at $120 \ rpm$. $A$ retarding torque brings it to rest in $10 \ s$. If the same torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force is: (in $\pi \ N$)

$A$ hollow sphere of mass $1 \,kg$ and radius $10 \,cm$ is free to rotate about its diameter. If a force of $30 \,N$ is applied tangentially to it,its angular acceleration is (in $rad/s^2$):

$A$ torque of $30 \ Nm$ is applied for $15 \ s$ on a stationary wheel of mass $5 \ kg$. The moment of inertia of the wheel is $2 \ kg \ m^2$. The angular displacement of the wheel in $10 \ s$ is ....... $rad$.

$A$ constant torque of $1000 \; Nm$ turns a wheel of moment of inertia $200 \; kg \cdot m^2$ about an axis through its centre. The wheel is at rest initially. Its angular velocity after $3 \; s$ is

Two circular discs of radius $10 \ \text{cm}$ each are joined at their centres by a rod of length $30 \ \text{cm}$ and mass $600 \ \text{g}$ as shown in the figure. If the mass of each disc is $600 \ \text{g}$ and the applied torque between the two discs is $43 \times 10^{5} \ \text{dyne cm}$, the angular acceleration of the discs about the given axis $AB$ is . . . . . . $\text{rad/s}^{2}$.