(a) The direction of the flow of the induced current is always such that it will oppose the change that produces it.
Polarity A ⟶ (+ve); B⟶  (-ve)
(b) (i) Here B = vertical component of Earth's magneticfield
B = (5 x 10^-4 sin 30^o)T
= 2.5 x 10^-4 T
∴ V = \(\left[ 2.5\times { 10 }^{ -4 }25\times \frac { 1800\times { 10 }^{ 3 } }{ 60\times 60 } \right] volt\)
 = 3.125 volt
(ii) Now B = horizontal component of Earth's magnetic field
 = B cos 30^o = \(\frac { B\sqrt { 4 } }{ 2 } \)
V' = \(\sqrt { 3 } \) V =1.732 x 3.125 volt
= 5.4 volt

(i) Presence of mind, knowledge of subject, concern for his grandmother, empathy helping and caring.
(ii) The net charge of the capacitor in defibrillator is given by
Q = CV  [where, C = capacitance  V = voltage]
⇒ Q = 70\(\times\)10^-6\(\times\)5000 = 35\(\times\)10^-2C

Here, B = 4 x 10^-3 weber/m²;
 E = 2 x 10⁴ V/m.
As the path of moving electron is undeviated, so force on moving electron due to electric field is equal and opposite to the force on moving electron due to magnetic field, i.e., eE = evB
\(or \ v=\frac { E }{ B } =\frac { 2\times { 10 }^{ 4 } }{ 4\times { 10 }^{ -3 } } =5\times { 10 }^{ 6 } \ m/s\) 
When electron moves perpendicular to magnetic field, the radius r of circular path traced by electron is
\(or \ r=\frac { mv }{ eB } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 5\times { 10 }^{ 6 } \right) }{ \left( 1.6\times { 10 }^{ -19 } \right) \times 4\times { 10 }^{ -3 } } \) 
\(=7.11\times { 10 }^{ -3 } \ m=7.11 \ mm\) 

Here , \(E=3\times 10^3\hat i NC^{-1}\)
A = (0.1m)² = 10^-2m²
(i) As normal to area is in the direction of electric field, therefore, \(\theta =0^o\)
\(\phi=EA cos\theta=3\times 10^3\times10^{-2}cos ^o\)
\(=30Nm^2C^{-1}\)
(ii) In this case, \(\theta=60^o\)
\(\phi'=EA cos\theta=3\times 10^3\times 10^{-2}cos 60^{o}\)
\(=15 Nm^2C^{-1}\)

\(Here,\ I=5sin\ 100\pi t\)
\( Compare\ with\ I={ I }_{ 0 }sin\ \omega t\)
\({ I }_{ 0 }=5A,\ \omega =2\pi v=100\pi\)
\(v=\frac { 100\pi }{ 2\pi } =50 \ Hz\)
\(\ { I }_{ v }=\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 5 }{ \sqrt { 2 } } =\frac { 5\sqrt { 2 } }{ 2 } =3.54A\)

Here, wingspan, l = 40m, Speed, v = 1080 km hr^-1
= \(\frac { 1080\times 1000 }{ 60\times 60 } { ms }^{ -1 }\)= 300ms^-1. B = 1.75\(\times\)10^-5 T, e = ?
As  e = Blv
\(\therefore \)  e = 1.75\(\times\)10^-5\(\times\)40\(\times\)300 = 0.21 volt

Here, Maximum magnetic field,
B₀ = 5 x 10^-9T;
charge on alpha particle, q = + 2e
= 2 x 1.6 x 10^-19 = 3.2 x 10^-19C, 
v = 3 x 10⁷ ms^-1 
Maximum electric field,
E₀ = cB₀ = (3 x 10⁸) x (5 x 10^-19) = 150 Vm^-1
Maximum force on alpha particle due to electric field = q x E₀ = (3.2 x 10^-19) x 150
= 4.80 x 10^-17 N
Force on alpha particle due to magnetic field = qvB₀ = (3.2 x 10^-19) x ( 3 x 10⁷) x (5 x 10^-9) 
= 4.80 x 10^-19 N

Here, v = 6 x 10¹⁴ Hz, E₀ = 2 V m^-1
(i) Average energy density of the electric field
\({ u }_{ E }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }=\frac { 1 }{ 4 } \times \left( 8.85\times { 10 }^{ -12 } \right) \times { 2 }^{ 2 }\)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)
(ii) Average energy density of magnetic field
\({ u }_{ B }=\frac { { B }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { \left( { E }_{ 0 }/c \right) }^{ 2 } }{ { \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { E }_{ 0 }^{ 2 } }{ 4{ \mu }_{ 0 }{ c }^{ 2 } } \)
\(=\frac { 1 }{ 4 } \times \frac { { 2 }^{ 2 } }{ \left( 4\pi \times { 10 }^{ -7 } \right) \times { \left( 3\times { 10 }^{ 8 } \right) ^{ 2 } } } \)
\(=8.85\times { 10 }^{ -12 }J{ m }^{ -3 }\)

Uses of electromagnetic waves
Following are some important uses of electromagnetic waves:
1. Radio waves. Radio waves are electromagnetic waves and are used in radio and television communication systems.
2. Microwaves. Microwaves are used in radar and other communication systems.
3. Infrared radiations. Infrared rays are used in:
(a) solar water heater and solar cooker.
(b) weather forecasting
(c) taking photographs during fog, smoke etc.
(d) dehydrating fruits.
(e) the treatment of muscular strain.
(f) greenhouse to keep the plants warm.
4. Ultraviolet rays.  ultraviolet rays are used:
(a) for checking mineral samples by making use of its property of causing fluorescence and also used for the study of molecular structure.
(b) for sterilizing the surgical instruments because U.V.-rays destroy bacteria.
(c) in food preservations
(d) in detection of invisible writing 
5. X-rays. X-rays are used:
(a) in surgery
(b) in radiotherapy.
(c) in medical diagnosis to detect the fracture in bones etc.
(d) in detective departments to detect gold, diamond etc concealed in bags etc without opening them.
(e) in scientific research to study the crystal structure etc.
6. \(\gamma \)-rays. Gamma rays are used to get information of structure of atomic nucleus.

Given, resistance, \({ R }_{ 1 }=10\Omega ,\ { R }_{ 2 }=10\Omega \)
Voltage, V = 1V
Time t = 5 x 60 s = 300 s
Since, effective resistance in parallel combination will be
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } =\frac { 10\times 10 }{ 10+10 } =5\Omega \)        
Heat produced = \(\frac { { v }^{ 2 }t }{ { R }_{ p } } =\frac { { 1 }^{ 2 } }{ 5 } \times 5\times 60=\frac { 60 }{ 4.2 } cal=14.3\ \)cal

400 V

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

A Wheatstone bridge is said to be sensitive if it shows large deflection in the galvanometer when ever a small change in resistance is taking place in resistance arm.