Physics

This is because the terms used in rotational motion are analogous to the terms used in linear motion and can interchange each other with variation in the topic since their meaning is exactly the same. For example, what you call force in linear motion can be called torque in terms of rotational motion. Just replace the linear terms with angular terms and you are done with the major process.

This is because using radians will simplify the procedure of solving various numerical problems and complex mathematical equations. They help consider the terms such as arc length, centre, radius etc. directly into account which are connected to a circular path which eases our task of computing the formulas.

This is the type of acceleration due to which a circular moment of the object is possible. This force constantly directs the motion of the body towards the centre of gravitation. Without this motion, the object will move away in a straight line. It ensures that the object stays in a circular path by changing the direction of object's velocity towards the centre.

Slope of $(x-t)$  graph, gives velocity

$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{\mathrm{t}\mathrm{a}\mathrm{n}?{\theta }_1}{\mathrm{t}\mathrm{a}\mathrm{n}?{\theta }_2}=\frac{\mathrm{t}\mathrm{a}\mathrm{n}?{30}^{?}}{\mathrm{t}\mathrm{a}\mathrm{n}?{45}^{?}}=\frac{1}{\sqrt[]{3}}$

Peak voltage is $\sqrt[]{2}$  times rms voltages in ac.

In (a) & (c) circuits, both the junctions are in same biasing conditions so offers equal resistances.

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{48}{3*{10}^8}=16*{10}^{-8}$

$=1.6*{10}^{-7}\mathrm{T}$

Capacitive reactance $=\frac{1}{\omega \mathrm{C}}={\mathrm{X}}_{\mathrm{c}}$ (say) on decreasing the operating frequency $\omega$  reduces

As $X_c$ is inversely proportional to $\omega$ the value of $X_c$  increase

$\therefore {\mathrm{I}}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{D}}$

$=\frac{{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{c}}}$

As ${\mathrm{X}}_{\mathrm{c}}$  increases, therefore displacement current Id decreases.

For transformer

$\left(\frac{i_p}{i_s}\right)=\left(\frac{V_s}{V_P}\right)=\left(\frac{N_s}{N_P}\right)$

For bulb, $\mathrm{V}\mathrm{s}=12\mathrm{V}$

$\begin{array}{rr}& {\mathrm{i}}_{\mathrm{s}}=\left(\frac{60}{12}\right)=5\mathrm{A}\\ & \Rightarrow \frac{{\mathrm{i}}_{\mathrm{p}}}{5}=\left(\frac{12}{220}\right)\\ & \mathrm{}{\mathrm{i}}_{\mathrm{P}}=\frac{60}{220}=\left(\frac{3}{11}\right)=0.27\mathrm{A}\end{array}$