Physics

Yes. Angular Velocity is a vector quantity i.e. it has directions. The direction of rotation can influence the sign of angular velocity's value. If the movement of the object is anticlockwise, the sign will be positive. Similarly, if the movement of the object is clockwise, the sign will be negative.

For resonance frequency

$\begin{array}{rr}& \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}\\ & \Rightarrow \omega =\frac{1}{\sqrt[]{\mathrm{L}\mathrm{C}}}=\frac{1}{\sqrt[]{10*{10}^{-3}*1*{10}^{-6}}}\\ & =\frac{1}{\sqrt[]{{10}^{-8}}}={10}^4\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}\\ & \mathrm{f}=\left(\frac{\omega }{2\pi }\right)=\frac{{10}^4}{2\pi }=1.59\mathrm{k}\mathrm{H}\mathrm{z}\end{array}$

Magnetic energy stored in an inductor $=\frac{1}{2}{\mathrm{L}\mathrm{I}}^2$

$\begin{array}{rr}& =\frac{1}{2}*4*{10}^{-6}*(2{)}^2\\ & =8\mu \mathrm{J}\end{array}$

${\text{?}\mathrm{}}_{\mathrm{s}}\stackrel{?}{\mathrm{B}}\cdot \mathrm{d}\stackrel{?}{\mathrm{A}}=0\mathrm{}$ (No monopole exist)

Even if the torque is same, the rotation of a body can be slow or fast depending on the distribution of mass along the surface area. If the mass of a body is more on the axis area, it's speed of rotation will be faster as compared to the body whose mass is accumulated away from the axis due to which the body will resist the angular acceleration.

For conductors α $\alpha$  is (+ve)

For semiconductors $\alpha$  is (-ve)

In simple words, a push or pull applied on a body is known as a force. Force is a concept used in terms of linear motion. On the other hand, the force which causes a body to spin around an axis instead of moving from one point to aother is called torque. Torque is a theory which comes into account while studying rotational motion and is considered as the rotational analogue of force.

$\mathrm{R}=22*{10}^3\pm 5\mathrm{\%}$ First band  Red

2nd band $=$  Red

3rd band =Orange (Multiplier)

4th band = Gold (Tolerance)

$\mathrm{d}\stackrel{?}{\mathrm{B}}=\frac{{\mu }_0(\mathrm{I}\stackrel{?}{\mathcal{l}}\stackrel{?}{\mathcal{l}}*\stackrel{?}{\mathrm{r}})}{4\pi {\mathrm{r}}^3}$

The expression for magnetic field depends on current carrying element $\mathrm{I}\mathrm{d}?\stackrel{?}{\mathcal{l}}$ $\stackrel{}{}$ , which is a vector quantity, therefore, statement-I is correct and statement-II is wrong.

No current through '  $G$  '

So potential difference across $\mathrm{R}$ is $2\mathrm{V}$

$\begin{array}{rr}& \mathrm{i}=\frac{8}{400}\\ & \mathrm{R}=\left(\frac{2}{8}*400\right)=100\mathrm{\Omega }\end{array}$