$A$ table rotates about a vertical axis passing through its center with an angular velocity of $20\ rad/s$. $A$ flywheel is mounted on it,which rotates about a horizontal axis with an angular velocity of $40\ rad/s$. The resultant angular velocity of the flywheel is:

The mass of the pulley is $m$ and its radius is $R$. Assume the pulley to be a uniform disc. The pulley is free to rotate about an axis passing through its centre and perpendicular to its plane. The string is massless and inextensible,and it is wrapped on the pulley. There is no slipping between the string and the pulley. The length of the string which is not wrapped on the pulley is $2R$. $A$ block of mass $m$ is released from the position as shown in the figure. The impulse exerted by the string on the pulley at the moment the string becomes taut is $J$. The value of $\frac{2m \sqrt{gR}}{J}$ is equal to:

$A$ uniform rod is rotating in a gravity-free region with a certain constant angular velocity $\omega$. The variation of tensile stress $\sigma$ with distance $x$ from the axis of rotation is best represented by which of the following graphs?

If $\vec{\omega} = (3\hat{i} + 4\hat{j} + 5\hat{k}) \text{ rad/s}$ and $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) \text{ m}$,then the linear velocity vector of the particle is:

For a particle,the angular velocity is $\vec{\omega} = \hat{i} - 2\hat{j} + 3\hat{k}$ and the position vector is $\vec{r} = \hat{i} + \hat{j} + \hat{k}$. What is the linear velocity $\vec{v}$?

$A$ door $1.6 \,m$ wide requires a force of $1 \,N$ to be applied at the free end to open or close it. The force that is required at a point $0.4 \,m$ distance from the hinges for opening or closing the door is (in $\,N$)