X is resistor, because current and voltage are in phase.
Y is inductor, because voltage is ahead of current in phase Z is capacitor, because current is ahead of voltage in phase.
When all the three elements are connected in series, the phasor diagram is as shown below

We assume that circuit is capacitive in nature from the impedance triangle OQP, we get
\( V_{m}^{2} =V_{m R}^{2}+\left(V_{m C}-V_{m L}\right)^{2} \)
\(\text {where } V_{m R} =I_{m} R ; V_{m C}=I_{m} X_{C} \text { and } \)
\(V_{m L} =I_{m} X_{L} \)
\( \therefore V_{m} =I_{m} \sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}} \)
\(\frac{V_{m}}{I_{m}} =\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}} \)
The impedance of the circuit is given by

\(Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}\)
From the graph we come to know about the quality factor of the circuit, i.e. its selectivity.

(i) Torque acting on the loop is given by
\(\tau=M B \sin \theta\)
As the angle between the magnetic field vector and the dipole moment vector is zero.
\(\tau=M B \sin (0)=0 \mathrm{Nm}\)
(ii) Magnitude of force is given by
\(F=\frac{2 \mu_{0} I_{1} I_{2} l}{4 \pi}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]\)
where \( l=10 \times 10^{-2} \mathrm{~m}, I_{1}=2 \mathrm{~A}, I_{2}=5 \mathrm{~A} r_{1}=1 \times 10^{-2} \mathrm{~m}, r_{2}=5 \times 10^{-2} \mathrm{~m} \)
\(F=\left[2 \times 10^{-7} \times 2 \times 5 \times 10^{-1}\right] \quad\left[\frac{1}{10^{-2}}-\frac{1}{5 \times 10^{-2}}\right] \)
\( F =20 \times 10^{-8}\left[1-\frac{1}{5}\right] \times \frac{1}{10^{-2}} \)
\(=\frac{20 \times 10^{-6} \times 4}{5}=16 \times 10^{-6} \mathrm{~N}\)
The net force is attractive because the arm of the loop carrying current in the same direction as the Direction of current in the wire is nearer.

The force, on a wire of length I,carrying a current I, in a magneticfield \(\overrightarrow { B } \) is given by \(\overrightarrow { F } =\left( \overrightarrow { l } \times \overrightarrow { B } \right) \) For a rectangular loop, places as shown, in a magnetic field \(\overrightarrow { B } \)

|Force on arm BCI = IForce on arm DAI = l/Bsin \(\alpha\)
Where \(\alpha\)=angle between side BC and \(\overrightarrow { B } \)
These two forces add up to zero as they are collinear (along the axis of the coil) and act in opposite directions
|Force on arm ABI = IForce on arm CDI = IbB
These two equal and opposite forces are not collinear. The perpendicular distance between their lines of action is, as shown
\(2\times\frac{a}{2}sin\theta=a sin\theta\)
Torque acting on the coil, has a magnitude \(\tau \) where
\(\tau \)=(lbB)x(a sin \(\theta\))=IAB sin \(\theta\) (A=ab=Area of the coil)
In vector form, \(\overrightarrow { \tau } =\overrightarrow { A } \times \overrightarrow { B } \)
But \(I\overrightarrow { A } \)=\(\overrightarrow { m } \), as given
\(\overrightarrow { \tau } =\overrightarrow { m } \times \overrightarrow { B } \)

Since the galvanometer has 50 divisions, so current for full scale deflection is
\({ I }_{ g }=\frac { 1 }{ 10 } \times 50 \ mA=5mA=5\times { 10 }^{ -3 }A,\) 
\(G=60\Omega ,S=2.5\Omega .\)
Let I be the maximum current which a galvanometer can read when shunted with resistance S, then
\({ I }_{ g }=\frac { IS }{ G+S } \ or \ I=\frac { { I }_{ g }(G+S) }{ S } \) 
\( =\frac { (5\times { 10 }^{ -3 })(60+2.5) }{ 2.5 } =125\times { 10 }^{ -3 }A\)
= 125 mA

Given  \(\phi =5{ t }^{ 3 }+4{ t }^{ 2 }+2t-5,\quad t=2s,\)
\(R=10\Omega \)
\(\therefore \)  Magnitude of induced e.m.f.
\(e=\left| e \right| =\frac { d\phi }{ dt } \)
or \(e=\frac { d }{ dt } \left[ 5{ t }^{ 3 }+4{ t }^{ 2 }+2t-5 \right] \)
\(=15{ t }^{ 2 }+8t+2\)
\(=15\times 4+8\times 2+2\)
or  \(e=78V\)
\(\therefore \)  \(I=\frac { e }{ R } =\frac { 78 }{ 10 } =7.8A\)

(a) (i)0.33J
(ii) 0.66 J
(b) (i) \(\tau\) = 0.33Nm
(ii) zero

(i) From graph (a), it is clear that resistance (opposition to current) is not changing with frequency, i.e. resistance does not depend on frequency of applied voltage, so the circuit element here is pure resistive (R), From graph (b), it is clear that resistance  increase linearly with frequency, so the circuit element here in induction in nature.
Inductive resistance, X_L = 2πvL.
⇒ X_L ∝ v
(ii) Impedance offered by the series combination of resistance (R) and inductor (L).
\(z=\sqrt { { R }^{ 2 }+{ { X }_{ L } }^{ 2 } } =\sqrt { { R }^{ 2 }+\left( 2\pi vL \right) ^{ 2 } } \)
In L-R circuit, the applied voltage leads the current by phase  \(\phi \) where tan \(\phi =\frac { { X }_{ L } }{ R } \)

Force per unit length is
\(F=\frac { \mu _{ 0 }2I^{ 2 } }{ 4\pi r } \left[ \because I_{ 1 }=I_{ 2 }=I \right] \quad \quad\)
If r increased to 2r and I is reduced to \(\frac { I }{ 2 } \), then new
force per unit length is \(\begin{aligned} F^{\prime} & =\frac{\mu_0}{4 \pi} \times \frac{2(I / 2)^2}{2 r} \end{aligned}\)
\(​​\begin{aligned} =\frac{1}{8}\left(\frac{\mu_0}{4 \pi} \cdot \frac{2 I^2}{r}\right) \Rightarrow F^{\prime}=\frac{F}{8} \end{aligned}\)
\(\therefore\) Force per unit length between them is\(\frac { F }{ 8 } \)      

8 x 10^-5 N

One ampere of current can be defined as the amount of current which when flows through two infinitely long parellel wires seperated by one metere produces an attractive foce 2 x 10^-7N/m.

The current carrying loop is equivalent to a magnetic dipole. The magnetic dipole does not experience any net force in a uniform magnetic field.

As current through element X leads the alternating voltage applied by \(\pi /2\) radian, therefore X is a pure capacitance. Its reactance is \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi v C} .\)

Force on a current carrying conductor in magnetic field, \(F=IlB{ sin }\theta \) |
 (i) Force F will be minimum if \({ sin }\theta =0\) or \(\theta ={ 0 }^{ o }\) or 180^o ,  i.e.,the linear conductor carrying current is parallel or antiparallel to the direction of magnetic field.
(ii) ) Force F will be maximum if \({ sin }\theta ={ 1 }\) or \(\theta ={ 90 }^{ o }\)  ,  i.e.,the linear conductor carrying current is perpendicular to the direction of uniform magnetic field.

(i) The principle of working of a moving coil galvanometer is based on the fact that if a current carrying coil is placed in a uniform radial field, it experiences a torque.
A galvanometer cannot be used as such to measures current in a circuit due to following two reasons:
(a) Galvanometer is very sensitive. It gives full scale deflection with a quite small current (nearly few micro ampere).
(b) In order to measrue the current, the galvanometer has to be connected in series of circuit. As its resistance is large, its presence in the circuit will decrease the effective current in the circuit.
(ii)  Need for a radial magnetic field : The relation between the current (i) flowing through the galvanometer coil, and the angular deflection (ϕ) of the coil (from its equilibrium position), is ϕ = [NABI sin θ/ k]
where θ is the angle between the magnetic field vector B and the equivalent magnetic moment vector μm of the current carrying coil. Thus I is not directly proportional to ϕ . We can ensure this proportionality by having θ = 90°. This is possible only when the magnetic field vector B, is a radial magnetic field. In such a field, the plane of the rotating coil is always parallel to vector B .To get a radial magnetic field, the pole pieces of the magnet, are made concave in shape. Also a soft iron cylinder is used as the core.

According to the question,

(i) Let RS moves with speed v rightward and also RS is at distances x₁ and x₂ from PQ at instants t₁ at t₂, respectively
∴ At t₁ flux linked with loop 1, i.e. PQRS, φ₁ = B(lx₁) Similarly, at instant t₂, flux linked with loop 2, i.e. PQR'S',
\(\phi_{2}=B\left(l x_{2}\right)\)
\(\therefore \text { Change in flux, } \Delta \phi=\phi_{2}-\phi_{1}=B l\left(x_{2}-x_{1}\right)=B l \Delta x\)
\(\Rightarrow \quad \frac{\Delta \varphi}{\Delta t}=B l \frac{\Delta x}{\Delta t}=B l v\) \(\left[\because v=\frac{\Delta x}{\Delta t}\right]\)
By Faraday's law, magnitude of induced emf, e = vB!.
(ii) If resistance of loop is R, then \(I=\frac{\nu B l}{R}\)
\(\therefore \text { Magnetic force }=I B l \sin 90^{\circ}=\left(\frac{v B l}{R}\right) B l\)
\(=\frac{v B^{2} l^{2}}{R} \quad\left[\because \sin 90^{\circ}=1\right]\)
∴ External force must be equal to magnetic force and in opposite directions.
\(\therefore \text { External force }=\frac{v B^{2} l^{2}}{R}\)
(iii) As, \(P=I^{2} R=\left(\frac{v B l}{R}\right)^{2} \times R=\frac{v^{2} B^{2} l^{2}}{R^{2}} \times R\)
\(\therefore \quad P=\frac{v^{2} B^{2} l^{2}}{R}\)

(ii) Magnetic field lines due to a current carrying loop is given as below:

(iii) Magnetic field due to straight part
\(B=\int {\frac{\mu_0}{4\pi} \frac{Idl \times r}{r^{3}}}\)
For point 0, dl and r for each element of the straight segments AB and DE are parallel. Therefore, dIｘr = 0. Hence, magnetic field due to straight segments is-zero.
Magnetic field at the centre due to circular point
= \(\frac{Magnetic \ field \ at \ the \ centre \ of \ circular \ coil }{2}\)   [∵ Here, coil is stay]
\(=\frac{1}{2}(\frac{\mu_0 I}{2r})=\frac{\mu_0 I}{4r}\)
\(\Rightarrow B=\frac{\mu_0 I}{4r}=\frac{(4\pi \times10^{-7}\times 12}{4\times 2 \times 10^{-2}}=6\pi \times 10^{-5}T\) 

Principle: It works on the principle of mutual induction.
Working: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links with both primary and secondary windings. Let cp be the flux in each turn in the core at a time due to current in the primary when a voltage Vp is applied to it.
\({ V }_{ s }={ E }_{ s }={ N }_{ s }\frac { d\phi }{ dt } \)
The alternating flux also induces an emf, called back emf in the primary given by
\({ V }_{ p }={ E }_{ p }=-{ N }_{ p }\frac { d\phi }{ dt } \)
But \({ E }_{ p }={ V }_{ p }\)
and \({ E }_{ s }={ V }_{ s }\)
So,
\({ V }_{ s }=-{ N }_{ s }\frac { d\phi }{ dt }\)
\( \\ { V }_{ p }=-{ N }_{ p }\frac { d\phi }{ dt } \)
\(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { { N }_{ s } }{ { N }_{ p } } \)
For a step-up transformer, \(\frac { { V }_{ s } }{ { V }_{ p } } >1\)
So,
\(\frac { { V }_{ s } }{ { V }_{ p } } >1\)
(ii) Sources of energy loss in transformer (any two) Flux leakage / Joule's loss in the resistance of windings / Loss due to eddy currents / Hysteresis loss / Humming loss.
(iii) A step-up transformer steps up the voltage while it steps down the current. So the input and output power remain the same (provided there is no loss). Hence there is no violation of the principle of energy conservation.

Phase difference between voltage and current,
\(tan\ \phi={{{X}_{L}-{X}_{C}}\over{R}}\)
and I₀ = \({ { {V}_{0} }\over{Z } }={ { {V}_{0} }\over{ \sqrt{{({X}_{L}-{X}_{C})}^{2}} +{R}^{2}} }\)
\(\therefore\)  Expression of AC,
I = I₀ sin \((\omega t-\phi)\)
Conditions for resonanceInductive reactance must be equal to capacitive reactance
i.e. X_L= X_C
As, X_L= X_C
\(\Rightarrow\)  \({\omega}_{0}L={{1}\over{{\omega}_{0}C}}\)
\(\Rightarrow\)   \({ \omega }_{ 0 }^{ 2 }={ { 1 }\over{ LC } }\Rightarrow {\omega}_{0}={ { 1 }\over{ \sqrt{LC} } }\)
where ,\({\omega}_{0}\)= resonant angular frequency Impedance becomes minimum and equal to ohmic resistance
i.e. Z = Z. minimum = R
AC becomes maximum
\(\therefore \)  \({I}_{max}={{{V}_{max}}\over{{Z}_{min}}}={{{V}_{max}}\over{R}}\)
Voltage and current arrives in same phase.
Power factor
i.e. cos \(\phi={{{P}_{av}}\over{{V}_{rms}{I}_{rms}}}={{True\ Power}\over{Apparent\ Power}}\)
Also, \(cos\ \phi={{R}\over{Z}}={{R}\over{\sqrt{{R}^{2}+({X}_{L}-{X}_{C})^{2}}}}\)
The power factor is maximum
i.e. cos \(\phi\).= + 1, in L-C-R series AC circuit when circuit is in resonance. The power factor is minimum when phase angle between V and I is 90°, i.e. either pure inductive circuit or pure capacitive AC circuit.

(a) In one revolution 
Change of area, \(dA=\pi l^2\)
\(\therefore\) change of magnetic flux
\(d\phi =\vec { B } .\vec { dA } =B.dA{ \cos { 0 } }^{ \circ }\)
\(=B\pi l^{ 2 }\)
(i) Induced emf, \(\varepsilon \) \(=B\pi l^{ 2 }/T=B\pi l^{ 2 }/v\)
(ii) Induced current in the rod, \(I=\frac{\varepsilon }{R}=\frac{\pi vB l^2}{R}\)
(b) Force acting on the rod, F = IlB
\(=\frac{\pi vB^2 l^3}{R}\)
The external force required to rotate the rod opposes the Lorentz force acting on the rod/ external force acts in the direction opposite the Lorentz force
(c) Power required to rotate the rod
Power = Force X velocity
P = Fx v
\(=\frac{\pi vB^2 l^3}{R}\times v\)

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(a): Both the assertion and reason are true and reason is the correct explanation of assertion. When the magnet falls, the magnetic flux through the copper ring increases and induced e.m.f. is produced in the ring. The induced emf so produced, opposes the motion of falling magnet. Therefore, the acceleration of the falling magnet will be less than that due to gravity.

(i) (c) : Here \(L=0.12 \mathrm{H}, C=480 \mathrm{nF}^{\prime}=480 \times 10^{-9} \mathrm{~F}\)
\(R=23 \Omega, V=230 \mathrm{~V}\)
\(V_{0}=\sqrt{2} \times 230=325.22 \mathrm{~V}\)
\(I_{0}=\frac{V_{0}}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}\)
At resonance, \(\omega L-\frac{1}{\omega C}=0\)
\(\omega=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{0.12 \times 480 \times 10^{9}}}=4166.67 \mathrm{rad} \mathrm{s}^{-1}\)
\(v_{R}=\frac{4166.67}{2 \times 3.14}=663.48 \mathrm{~Hz}\)
(ii) (a) : Current \(I_{0}=\frac{V_{0}}{R}=\frac{325.22}{23}=14.14 \mathrm{~A}\)
(iii) (b) : Maximum power \(P_{\max }=\frac{1}{2}\left(I_{0}\right)^{2} R\)
\(=\frac{1}{2} \times(14.14)^{2} \times 23=2299.3 \mathrm{~W}\)
(iv) (d) : Quality factor \(Q=\frac{X_{L}}{R}=\frac{\omega_{r} L}{R}\)
\(=\frac{4166.67 \times 0.12}{23}=21.74\)
(v) (b)

(i) (c) :Ohm is the unit of capacitive reactance.
(ii) (a): Capacitive reactance \(X_{C}^{1}=\frac{1}{\omega C}=\frac{1}{2 \pi v C}\)
\(=\frac{1}{2 \pi \times 10^{6} \times 5 \times 10^{-6}}=0.032 \Omega\)
(iii) (b): In capacitive circuit, resistance to the flow of current is offered by the capacitor.
(iv) (b)
(v) (a): Current \(I_{v}=\frac{E_{v}}{X_{C}}=\frac{E_{v}}{1 / 2 \pi v C}=(2 \pi v C) E_{v}\)
\(I_{v}=2 \times 3.14 \times 50 \times 10^{-6} \times 200=6.28 \times 10^{-2} \mathrm{~A}\)