Total power of motor,
P = VI = 50 x 15 = 750 W
Mechanical power of motor, P' = 150 W
Power dissipated across the winding of motor
 = 750 - 150 = 600 W
Efficiency of motor = \(\frac{P'}{P}\times 100=\frac{150}{750}\times 100\)
= 20 %

\(Use \ B={ \mu }_{ o }{ \mu }_{ r }ni={ \mu }_{ o }{ \mu }_{ r }\frac { N }{ 2\pi r } I\)

When we start the motor, there is no back emf as the motor is at rest. So, a large current flows through the coil. As the motor rotates, the back emf increases and intake of current decreases.

Total required capacitance, C = 2 µF
Potential difference, V = 1 kV = 1000 V
Capacitance of each capacitor, C₁ = 1 µF
Each capacitor can withstand a potential difference, V₁ = 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as \(\frac{1000}{400}=2.5\)
Hence, there are three capacitors in each row.
Capacitance of each row \(=\frac{1}{1+1+1}=\frac{1}{3} \mu \mathrm{F}\)
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
\(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\ldots \ldots \ldots \ldots \ldots \ldots . n \text { terms }\)
\(=\frac{n}{3}\)
However, capacitance of the circuit  is given as 2 \(\mu \) F
\(\therefore \frac{n}{3}=2\)
n = 6
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 x 3 i.e., 18 capacitors are required for the given arrangement.

(a) Values displayed by Kamal:
(i) Being educated, he knows about MRI
(magnetic resonance imaging).
(ii) Took prompt decisions to take the help of his family, friends and neighbours to arranged the cost of MRI.
(iii) He showed his empathy, helping attitude and caring nature for his uncle. 
(b) Magnetic force on moving charge particle in uniform magnetic field B can be given as
F = q(vｘB) or |F| = qvb sin\(\theta\)
(i) Maximum force ate = 90°
F = qvB = \(1.6\times10^{-19}\times10^{4}\times0.1\)
= \(1.6\times10^{-16}N\)
(ii) Minimum force at e = 0° and 180°
F = 0
i.e. force is minimum when the charge particle either move parallel or anti-parallel to the magnetic field lines.

(i) Shyam is intelligent and has clarity in explaining facts.He also aware about health.
(ii) UV-rays are produced by the sun and is trapped by ozone layer.
(iii) UV-rays can cause skin cancer, so it is carcinogenic.

Here,
n = 8.4 x 10²⁸ m^-3,
A = 10^-6 m²,
I = 5.4 A
v_d = \(\frac{I}{nAe}\)
= \(\frac{5.4}{(8.4 \times10^{28})\times(10^{-6})\times(1.6 \times10^{-19})}\)
= 0.4 x 10^-3 m/s
= 0.4 mm/s

Here, I₁ = 500\(\mu A\) = 500 x 10^-6 A,
 \({ \theta }_{ 1 }={ 60 }^{ o },{ I }_{ 2 }=?,{ \theta }_{ 2 }=\frac { \pi }{ 5 } rad.=\frac { { 180 }^{ o } }{ 5 } ={ 36 }^{ o }\) 
\({ I }_{ 1 }=\frac { k }{ nBA } { \theta }_{ 1 }\) and \({ I }_{ 2 }=\frac { k }{ nBA } { \theta }_{ 2 }\) 
\(\therefore \frac { { I }_{ 2 } }{ { I }_{ 1 } } =\frac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } or\) 
\({ I }_{ 2 }=\frac { { \theta }_{ 2 } }{ { \theta }_{ 1 } } { I }_{ 1 }=\frac { 36 }{ 60 } \times 500\times { 10 }^{ -6 }=300\times { 10 }^{ -6 }\quad A\) 
Current sensitivity   \(=\frac { { \theta }_{ 2 } }{ { I }_{ 2 } } =\frac { 36^{ o } }{ 300\times { 10 }^{ -6 } } \) 
= 0.12 x 10^-6 degree/A = 0.12 degree/\(\mu A\)

(a) When the armature gets jammed and cannot rotate, back e.m.f. E = 0.
\(\therefore\) When connected to the supply, current  \(I=\frac { V }{ R } =\frac { 200 }{ 8.5 } =23.53amp\)
Which exceeds the safe limit of  20amp. Therefore, the armature will burn out.
(b) The e.m.f. generated is obviously equal to the back e.m.f. E = 157.5 volt.

\(Here, \ R=200 \ ohm, \ C=15.0\mu F=15\times { 10 }^{ -6 }F\)
\({ E }_{ v }=220V, \ v=50Hz, \ { I }_{ v }=? \ { V }_{ R }=?, \ { V }_{ C }=?\)
\( Now \ { X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } =\frac { 1 }{ 2\times 3.14\times 50\times 15\times { 10 }^{ -6 } } =212.2\Omega \)
\((a) \ Impedance \ of \ the \ circuit,\)
\(Z=\sqrt { { R }^{ 2 }+{ X }_{ C }^{ 2 } } =\sqrt { { 200 }^{ 2 }+\left( 212.3 \right) ^{ 2 } } =291.7 \ ohm\)
\( \therefore \ \ Current \ in \ the \ circuit, \ { I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 220 }{ 291.7 } =0.75A\)
\( (b) \ { V }_{ R }={ I }_{ v }\times R=0.75\times 200=150.8 \ V\)
\({ V }_{ C }={ I }_{ v }{ X }_{ C }=0.75\times 212.3=159.2 \ V\)
\( { V }_{ R }+{ V }_{ C }=150.8+159.2=310V, \ which \ is \ more \ than \ the \ source \ voltage \ of \ 220 \ V.\)
This paradox is resolved by the fact that the two voltage are not in same phase. Therefore, they cannot be added like ordinary numbers. As V_R and V_C are out of phase by \({ 90 }^{ \circ }\), therefore
\({ V }_{ RC }=\sqrt { { V }_{ R }^{ 2 }+{ V }_{ C }^{ 2 } } =\sqrt { \left( 150.8 \right) ^{ 2 }+\left( 159.2 \right) ^{ 2 } }\)
\(=220V,\ the\ source\ voltage\)

The corresponding wavelength to the frequency 8 x 10¹⁴ Hz is
\({ \lambda }_{ 1 }=\frac { c }{ { v }_{ 1 } } =\frac { 3\times { 10 }^{ 8 } }{ 8\times { 10 }^{ 14 } } =3.75\times { 10 }^{ -7 }m\)
The corresponding wavelength to the frequency 5 x 10¹⁷ is
\({ \lambda }_{ 2 }=\frac { c }{ { v }_{ 2 } } =\frac { 3\times { 10 }^{ 8 } }{ 5\times { 10 }^{ 17 } } =6\times { 10 }^{ -10 }m\)
As, \({ \lambda }_{ m }T=2.9\times { 10 }^{ -3 } \ or \ T=\frac { 2.9\times { 10 }^{ -3 } }{ { \lambda }_{ m } } \)
For, \({ \lambda }_{ 1 }=3.75\times { 10 }^{ -7 }m;\)
\({ T }_{ 1 }=\frac { 2.9\times { 10 }^{ -3 } }{ 3.75\times { 10 }^{ -7 } } =7.73\times { 10 }^{ 3 }K\)
\({ \lambda }_{ 2 }=6\times { 10 }^{ -10 }m;\)
\({ T }_{ 2 }=\frac { 2.9\times { 10 }^{ -3 } }{ 6\times { 10 }^{ -10 } } =4.83\times { 10 }^{ 6 }K\)
Temperature range is \(7.73\times { 10 }^{ 3 }K \ to \ 4.83\times { 10 }^{ 6 }K\)

Here, I = 3.0 x 10¹⁴ M m^-2 , E₀ = ?, B₀ = ?
Intensity of the plane electromagnetic wave is
\(I={ u }_{ av }c=\frac { 1 }{ 2 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c\)
\(\therefore { E }_{ 0 }=\sqrt { \frac { 2I }{ { \epsilon }_{ 0 }c } = } \sqrt { \frac { 2\times 3\times { 10 }^{ 14 } }{ \left( 8.85\times { 10 }^{ -12 } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } \)
= 4.75 x 10⁸ V m^-1
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.75\times { 10 }^{ 8 } }{ 3\times { 10 }^{ 8 } } =1.58T\)

Consider a conductor of resistance R and let V be the potential difference applied across the ends of AB. If I is the current flowing in the conductor for time t, then the total charge flowing from A to B in time t is:
q = It
Work done in carrying a charge q across P.D. of V is
W = Vq
q = It
W = VIt
V = IR
W = (IR)It
​​​​​​\(W={ I }^{ 2 }Rt\)
\(Again, \ I=\frac { V }{ Rt } \)
\(W=V\frac { V }{ R } t\)
\( W=\frac { { V }^{ 2 } }{ R } t\)
If entire work is dissipated as heat, then the heat produced is given by
H = W = \({ I }^{ 2 }Rt\) joule
= \(\frac { { I }^{ 2 }Rt }{ 4.2 } calorie\)
which is Joule's law.
(b) Cause of heat produced in a conductor when current flows through it. When the electrons travel through a conductor, they collide with the positive ions in the crystal lattice.

Larger

V = V1 + V2
Q = C1V = 6 x 10^-6  x 2 = 12\(\mu \)C
As C₂ is in series same amount of charge will also flow through it now V₂ = Q/C₂ = (12 x 10^-6) /(12 x 10^-6) = 1 volt
Total Battery voltage, V = 2 + 1 = 3 Volt

In case, test charges is not vanishingly small, it will produces its own electric field and the measured value of electric field will be different from the actual value of an electric field at that point.