A uniform rod of length $200 \,\mathrm{~cm}$ and mass $500 \,\mathrm{~g}$ is balanced on a wedge placed at $40\,cm$ mark. A mass of $2\, \mathrm{~kg}$ is suspended from the rod at $20\, \mathrm{~cm}$ and another unknown mass $'m'$ is suspended from the rod at $160\, \mathrm{~cm}$ mark as shown in the figure. Find the value of $'m'$ such that the rod is in equilibrium. $\left(\mathrm{g}=10 \,\mathrm{~m} / \mathrm{s}^{2}\right)$

The moment of inertia of a solid flywheel about its axis is $0.1\,kg-m^2$. A tangential force of $2\,kg\,wt$. is applied round the circumference of the flyweel with the help of a string and mass arrangement as shown in the figure. If the radius of the wheel is $0.1\,m,$ find the angular acceleration of the solid fly wheel (in $rad/sec^2$)

A massless string is wrapped round a disc of mass $M$ and radius $R$. Another end is tied to a mass $m$ which is initially at height $h$ from ground level as shown in the fig. If the mass is released then its velocity while touching the ground level will be

A uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$, the initial angular acceleration of the rod will be 

In an experiment with a beam balance on unknown mass $m$ is balanced by two known mass $m$ is balanced by two known masses of $16\, kg$ and $4\, kg$ as shown in figure. The value of the unknown mass $m$ is ....... $kg$.

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-intertial fiame of reference. The relationship between the force $\vec{F}_{\text {rot }}$ experienced by a particle of nass in moving on the rotating disc and the force $\vec{F}_{\text {in }}$ experienced by the particle in an inertial frame of reference is

$\vec{F}_{\text {rot }}=\vec{F}_{\text {in }}+2 m\left(\vec{v}_{\text {rot }} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega},$

where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\bar{r}$ is the position vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the $x$-axis along the slot, the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$. A sm a $1$ block of mass $m$ is gently placed in the slot at $\vec{r}=(R / 2) \hat{i}$ at $t=0$ and is constrained to move only along the slot.

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($1$) The distance $r$ of the block at time $t$ is

($A$) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$   ($B$) $\frac{R}{2} \cos \omega t$   ($C$) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$

($D$) $\frac{F}{2} \cos 2 \omega t$

($2$) The net reaction of the disc on the block is

($A$) $\frac{1}{2} m \omega^2 R\left(e^{2 \omega t}-e^{-2 \omega t}\right) \hat{j}+m g \hat{k}$

($B$) $\frac{1}{2} m \omega^2 R\left(e^{\omega t}-e^{-a t t}\right) j+m g k$

($C$) $-m \omega^2 R \cos \omega t \hat{j}-m g \hat{k}$

($D$) $m \omega^2 R \sin \omega t \hat{j}-m g \hat{k}$

Give the answer quetioin ($1$) ($2$)