(i) They are transverse in nature, and travel with the speed of light in vacuum
(ii) The selfless attitude of the scientist, service mindedness and modesty.

(a) The values shown by Shalini are :
(i) high degree of general awareness,
(ii) ability to take quick decisions,
(iii) concern for her aunt,
(iv) helping and caring nature.
(b) In microwave oven, microwaves of required wavelength get strongly absorbed by water due to its energy which heated the water. Since every food items has lot of water, so they can be heated and cooked quickly in a microwave over. With this, the microwaves heat the food item completely, not simply from outside as in case of normal oven.

(a) X-rays are produced by bombarding a metal target by high energy electrons.
(b) To study the atomic structures, treatment for certain forms of cancer.
(c) Presence of mind. alertness, taking initiative, helpful, caring.

(a) The value displayed by Moni are:
(i) High degree of general awareness
(ii) Ability to convince someone
(iii) Amicable nature.
(b) A global nuclear war would make the substantial parts of the sky getting covered by very minute particles. They will be responsible for preventing sun rays from reaching the earth for a very long time, resulting in 'nuclear winter'.

(i)  Jayant is helpful, caring and has presence of mind to deal with adverse situation.
(ii) X-ray is produced by the bombardment of a metal target by high energy electrons.
(iii) It is used in the determination of structure of solids by X-rays diffraction method.

(a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long-distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
Hence, satellites are helpful in reflecting TV signals. Also, they help in long-distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.

Here, R = 10 cm = 0.1 cm;
C = 200 pF = 200 x 10^-12 F = 2 x 10^-10 F;
E_rms = 200 V; \(\omega\) = 200 rad s^-1 ;
r = 2.0 x 10^-2 m.
(a) \({ I }_{ rms }=\frac { { E }_{ rms } }{ 1/\omega C } =\omega C{ E }_{ rms }\)
= 200 x (2 x 10^-10) x 200
(b) Yes, because I_D = 1
(c) \({ I }_{ 0 }=\sqrt { 2 } { I }_{ rms }=\sqrt { 2 } \times 8\times { 10 }^{ -6 }\)
= 11.312 x 10^-6 A
(d) Consider a loop of radius r between two circular plates of parallel plate capacitor placed coaxially with them. The area of this loop \({ A }^{ \prime }=\pi { r }^{ 2 }\)
By symmetry, the magnetic field \(\overrightarrow { B } \) is equal in magnitude and is tangentially to the circle at every point. In this case, only a part of displacement current I_D will cross the loop of area \({ A }^{ \prime }\) . Therefore, the current passing through the area \({ A }^{ \prime }\)
\({ I }^{ \prime }=\frac { { I }_{ D } }{ \pi { R }^{ 2 } } \times \pi { r }^{ 2 }=\frac { { I }_{ D } }{ { R }^{ 2 } } { r }^{ 2 }\)
Using Ampere's Maxwell law we have, \(\oint { \overrightarrow { B } .\overrightarrow { dl } } ={ \mu }_{ 0 }\times \)(total current through the area \({ A }^{ \prime }\)) or \(B=\frac { { \mu }_{ 0 }{ I }_{ 0 }r }{ 2\pi { R }^{ 2 } } =\frac { 4\pi \times { 10 }^{ -7 }\times 11.312\times { 10 }^{ -6 }\times 2\times { 10 }^{ -2 } }{ 2\pi \times { \left( 0.1 \right) }^{ 2 } } \)
or \(2\pi rB={ \mu }_{ 0 }\frac { { I }_{ 0 } }{ { R }^{ 2 } } { r }^{ 2 }\)
= 4.525 x 10^-12 T

Total energy falling on the surface,
U = 18 x 20 x 30 x 60 J = 6.48 x 10⁵ J
Total momentum delivered to the surface is
\(p=\frac { U }{ c } =\frac { 6.48\times { 10 }^{ 5 } }{ 3\times { 10 }^{ 8 } } =2.16\times { 10 }^{ -3 }kg{ ms }^{ -1 }\)
The average force exerted on the surface is
\(F=\frac { p }{ t } =\frac { 2.16\times { 10 }^{ -3 } }{ 30\times 60 } =1.2\times { 10 }^{ -6 }N\)
It the surface is a perfect reflector, the change of momentum will be = p - (- p)
= 2 p = 2 x 2.16 x 10^-3 kg ms^-1
Now average force,
\(F=\frac { 2\times 2.16\times { 10 }^{ -3 } }{ 30\times 60 } =2.4\times { 10 }^{ -6 }N\)

Consider an electric charge at rest so that at a point P some distance away, we have electric field but no magnetic field. Let, at time \(t=0\) , an impulse be given to the charge such that it starts moving with some finite velocity. For a moving charge, we expect at P both electric and magnetic fields, but we cannot immediately decide whether the magnetic field at P will change from zero to finite value instantaneously at \(t=0\) or after some time.
Instantaneous change means infinite rate of change. If the change is instantaneous at all points then considering any loop, we will conclude from Faraday's law that an infinite e.m.f. and infinite electric field is set up. This in turn would imply an infinite magnetic field as seen from the result. Fields are always finite away from charges and clearly the situation just described is inconsistent with known laws of electricity and magnetism.
\(\oint { \overset { \rightarrow }{ B } } .\overset { \rightarrow }{ dl } ={ \mu }_{ 0 }{ \varepsilon }_{ 0 }\frac { d\phi _{ e } }{ dt } \)
The moving charge sets up a magnetic field in its neighbourhood which in turn creates an electric field in the neighbourhood. The process continues since both time-varying electric and magnetic fields act as sources of each other. Thus an electromagnetic wave is started when a charge is accelerated. It is only when the wave reaches the point P that the magnetic field at P changes.
This shows that an accelerated charge emits an electromagnetic wave. It can also be shown that the electromagnetic wave and the oscillator will have the same frequency.

Energy of photon, \(\mathrm{E}=\mathrm{hv}\)
This implies,\(\mathrm{E}=\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Where, \(\mathrm{h}=6.62 \times 10^{-34} \mathrm{js}\)
\( \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\)
If wave length \(\lambda\) is in meter and energy is in joule then, we will divide E by \(1.6 \times 10^{-19}\) to convert into eV (Electron volt).
\(\therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda \times 1.6 \times 10^{-19}} \mathrm{eV}\)
(1) For y - rays wave length ranges from  to less that \(10^{-14} \mathrm{~m}\)
Therefore,  \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \times 10^3 \mathrm{eV} \approx 10^4 \mathrm{eV}\)
Thus, \( \lambda=10^{-10} \mathrm{~m}, \text { energy }=10^4 \mathrm{eV} \text { and }\) \( \lambda=10^{-14} \mathrm{~m} \text {, energy }=10^8 \mathrm{eV}\)
Energy of y - rays ranges between 10⁴ to \(10^8 \mathrm{eV}\)
(2) For X - rays wave length ranges from \(10^{-8} \mathrm{~m}\) to \(10^{-7} \mathrm{~m}\) For \(\lambda=10^{-8}\)
Therefore, \(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-8} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=12.4 \approx 10^2 \mathrm{eV}\)
\( \lambda=10^{-13} \mathrm{~m} \text {, energy }=10^7 \mathrm{eV}\)
(3) For violet radiation \(\lambda\) ranges from \(4 \times 10^{-7}\) to \(6 \times 10^{-10}\)
Therefore, for \(\lambda=4 \times 10^{-7}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV} =3.1 \mathrm{eV} \approx 10^{10} \mathrm{eV}\)
\(\lambda=6 \times 10^{-10} \mathrm{~m} \text {, Energy }=10^3 \mathrm{eV}\)
Energy of ultraviolet radiation vary between \(10^{10}\) to \(10^3 \mathrm{eV}\).
(4) For visible radiations wave length range from \(4 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-7} \mathrm{~m}\)
Therefore,
For \(\lambda=4 \times 10^{-7} \mathrm{~m}\), and Energy \(=10^{10} \mathrm{eV}\)
\(\text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{7 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{VV} \)
\(=1.77 \mathrm{eV} \approx 10^{\circ} \mathrm{eV}\)
(5) For infrared radiation $\lambda$ range from \(7 \times 10^{-7} \mathrm{~m}\) to \(7 \times 10^{-14} \mathrm{~m}\)
Therefore, \(\lambda=7 \times 10^{-7} \text {, energy }=10^{\circ} \mathrm{eV}\)
For \(\lambda=7 \times 10^{-4} \text {, energy }=\frac{1}{1000} \text { times }\)
the other order of  \(10^{-3}\)eV
(6) For micro waves $\lambda$ ranges from 1 mm to 0.3 m
For \(\lambda=1 \mathrm{~mm}\) or \(10^{-3}\)
energy is equal to \( \text { Energy }=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^{-3} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-3} \mathrm{eV} \approx 10^{-3} \mathrm{eV}\)
For \(\lambda=0.3 \mathrm{~m} \text {, Energy }=4.1 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV} \text {. }\)
(7) For Radio waves $\lambda$ ranges from 1 m to few km For $\lambda=1 \mathrm{~m}$
For λ=1m
Energy is equal to
\(=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{10^0 \times 1.6 \times 10^{-19}} \mathrm{eV} \)
\(=1.24 \times 10^{-6} \mathrm{eV} \approx 10^{-6} \mathrm{eV}\)
Energy for λ of the order of few km≈10⁻⁶eV
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source

Y₁ Microwaves
Microwave oven, Aircraft Navigator or any other
Y₂ Ultraviolet waves
Sterilize surgical instruments, food preservation or any other

Infrared region.

Source: Infrared waves are produced by hot objects and vibration of atoms and molecules.
Uses:
(i) In remote control of TV, VCR, etc.
(ii) In photography in thin mist.

(i) Microwaves are produced by special vacuum tubes like the klystron, / Magnetron / Gunn diode.
The frequency of microwaves is selected to match the resonant frequency of water molecules, so that energy is transferred efficiently to the kinetic energy of the molecules.
(ii) (a) Associated with the green house effect.
(b) In remote switches of household electrical appliances.

(i) \(2.21\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(ii) \(2.21\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(iii) \(4.42\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(iv) \(3.3\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)

The electromagnetic waves in the order of increasing wavelength are 
(a) \( \gamma \) rays
(b) x-rays 
(c) Microwaves
​​​​​(d) Radiowaves

If \(\theta \) is the angle which the direction of propagation of c.m. wave with y-axis, then
\(cos \ \theta =\frac { \overset { \rightarrow }{ r } .\hat { j } }{ r } =\frac { (\hat { i } +\hat { j } +\hat { k } ) }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \ or \ \theta ={ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) \)

(i) Microwave
(ii) Infrared waves are used to treat muscular strain.
(iii) X-rays are used as a diagnostic tool in medicine. X-rays are produced by X-ray tubes or inner shell electrons.