From  \(b=\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { Z{ e }^{ 2 }cot \ { \theta }/{ 2 } }{ E } \)
For given value of b, when E is increased, \(cot\frac { \theta }{ 2 } \)increases; \(tan\frac { \theta }{ 2 } \) decreases; \(\theta \) decreases.

\(\frac { { E }^{ 0 } }{ { B }^{ 0 } } =c\)

Useful Intensity,
\(I=\frac { power }{ area } =\frac { 100\times \left( 2.5/100 \right) }{ 4\pi { \left( 3 \right) }^{ 2 } } =\frac { 2.5 }{ 36\pi } W{ m }^{ -2 }\)
Half of this intensity (I) belongs to electric field and half of that to magnetic field. Therefore,
\(\frac { I }{ 2 } =\frac { 1 }{ 4 } { \varepsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c \ or \ { E }_{ 0 }=\sqrt { \frac { 2I }{ { \varepsilon }_{ 0 }c } } \)
\(=\sqrt { \frac { 2\times \left( 2.5/36\pi \right) }{ \left( \frac { 1 }{ 4\pi \times 9\times { 10 }^{ 9 } } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } =4.08V{ m }^{ -1 }\)
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.08 }{ 3\times { 10 }^{ 8 } } =1.36\times { 10 }^{ -8 }T\)

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
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.75\times { 10 }^{ 8 } }{ 3\times { 10 }^{ 8 } } =1.58T\)

(i) This wavelength (21 cm) corresponds to the radio waves.
(ii) This frequency (1057 MHz) also corresponds to the radio waves (short wavelength).
(iii) As, T = 2.7 K
Using the formula,
\(\lambda\)_m T = b = 0.29 cm-K
\(\lambda_m=\frac{0.29}{2.7}=0.11 \mathrm{~cm}\)
This wavelength corresponds to the microwaves region of the electromagnetic spectrum.
(iv) This wavelength lies in the visible region of the electromagnetic spectrum.
(v) Energy, E = 14.4 keV
= 14.4 \(\times\) 10³ \(\times\) 1.6 \(\times\) 10^-19 J
Frequency of wave,
\(\begin{aligned} v & =\frac{E}{h}=\frac{14.4 \times 1.6 \times 10^{-16}}{6.6 \times 10^{-34}} \end{aligned}\)
= 3.5 \(\times\) 10¹⁸ Hz
This frequency lies in the X-ray region of the electromagnetic spectrum.

Suppose natural boron contains x
As atomic mass of natural boron = weighted average of masses of two isotopes.
\( \therefore 10.81=\frac { x\times 10.003+(100-x)11.009 }{ 100 } \)
\(1081=-0.996xd+1100.9;\)
\(x=19.98\)
Hence, relative abundance of \( \ _{ 5 }{ { \beta }^{ 10 } }=19.98%, \ and \ relative \ abundance \ of \ _{ 5 }{ { \beta }^{ 11 } }\)
\( =100-19.98=80.02%\)
\(=80.02%\)

\(1 \ eV=1.602\times { 10 }^{ -19 }joule\)
\( 1.602\times { 10 }^{ -19 } \ joule=1 \ eV\)
\(1 \ joule=\frac { 1 }{ 1.602\times { 10 }^{ -19 } } eV\)
\(1 \ joule=6.242\times { 10 }^{ 18 }eV\)
\((ii) \ From \ E={ mc }^{ 2 }\)
\(m=\frac { E }{ { c }^{ 2 } } \therefore 1\ a.m.u.=\frac { 931\times 1.602\times { 10 }^{ -13 } }{ { \left( 3\times { 10 }^{ 8 } \right) }^{ 2 } }\)
\(=1.66\times { 10 }^{ -27 }kg\)
Now, definition, Mass of \(\ _{ 6 }{ { C }^{ 12 } }=12a.m.u.\)
\(=12\times 1.657\times { 10 }^{ -27 }\ kg=1.99\times { 10 }^{ -26 }kg\)

(i) x-rays have the wavelength range,\({ 10 }^{ 13 } \ m \ to \ 3\times { 10 }^{ -8 }m\) which is smaller than that of ultraviolet rays of a wavelength range \(6\times { 10 }^{ -9 }m \ to \ 4\times { 10 }^{ -7 }m, \ i.e., \ { \lambda }_{ x }<{ \lambda }_{ uv }\)
since energy. \(E=\frac { hc }{ \lambda } \ or \ E\propto \frac { 1 }{ \lambda } ; \ so \ \frac { { E }_{ x } }{ { E }_{ uv } } =\frac { { \lambda }_{ uv } }{ \lambda _{ x } } >1 \ or \ { E }_{ x }>{ E }_{ uv }\) Thus x-rays are more energetic than ultraviolet rays
(ii) The wavelength of radio waves is of range. \(0.3m \ to \ 6\times { 10 }^{ 2 }m\) which is very large as compared to the size of molecules of our blood, flesh and bones etc. As the energy of radio waves is quite small so these radio waves can not penetrate the blood of our body. That is why we can not use the radio waves to detect the fracture in bones
(iii) Just as different em waves have different application depending on their wavelength, in the same way, every person has different qualities, which make him suitable for different purpose/fields.
The aim of a teacher or manager is to is identify the quality in different children/persons and put them to their best use accordingly.

(a) Presence of mind, knowledge of subject.
(b) Used in telecommunication.
(c) \(\lambda =e/v=(3\times 10^{ 8 })/(3\times 10^{ 11 })={ 10 }^{ -3 }\)

Not necessarily, The basic requirement is that the total electric flux through the area enclosed by the loop should vary with time. The flux change may arise from any portion of the area.
Elsewhere E or dE/dt may be zero. In particular. there need be no electric field at points which make the loop.

As Number of atoms in \( _{ 38 }{ { Sr }^{ 90 } }=6.023\times { 10 }^{ 23 }\)
Number of atoms in 15mg of  \(\ _{ 38 }{ { Sr }^{ 90 } }=\frac { 6.023\times { 10 }^{ 23 } }{ 90 } \times \frac { 15 }{ 1000 } ,\ i.e.,\ N=1.0038\times { 10 }^{ 20 }\)
Rate of disintegration \( \ \frac { dN }{ dt } =\lambda N=\frac { 0.693 }{ T } N=\frac { 0.693\times 1.0038\times { 10 }^{ 20 } }{ 28\times 3.154\times { 10 }^{ 7 } }\)
\( =7.877\times { 10 }^{ 10 }Bq\)

Given, v = 2 x 10⁸ m / s and \(\mu _{ r }\)
The speed of electromagnetic waves in  medium is given by 
v = \(\frac { 1 }{ \sqrt { \mu \varepsilon } } \)
Where, \(\mu \) and \(\varepsilon \) are absolute permeability and absolute permittivity of the medium.
Now,  \(\mu \) = \(\mu _{ 0 }\mu _{ r }\)
and  \(\varepsilon =\varepsilon _{ 0 }\varepsilon _{ r }\)
Eq. (i) becomes, v = \(\frac { 1 }{ \sqrt { \mu _{ 0 }\mu _{ r }\varepsilon _{ 0 }\varepsilon _{ r } } } \)
= \(\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } \) x  \(\frac { 1 }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)
v = \(\frac { c }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)  \(\left[ c=\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } \right] \)
On squaring both sides, we get 
\(\varepsilon _{ r }=\frac { c^{ 2 } }{ v^{ 2 }\mu _{ r } } \)
= \(\frac { (3\times10^{ 8 })^{ 2 } }{ (2\times10^{ 8 })^{ 2 }\times1 } \) = 2.25

Angular momentum of an electron in nth orbit  = \(\frac { nh }{ 2\pi } \)                      
By Bohr's hypothesis, We have
    \(mvr=\frac { nh }{ 2\pi } \quad \Rightarrow \quad v=\frac { nh }{ 2\pi mr } \)
Time period to complete a revolution in an orbit,
\(T=\frac { 2\pi r }{ v } =\frac { 2\pi r\left( 2\pi mr \right) }{ h } =\frac { 4{ \pi }^{ 2 }{ mr }^{ 2 } }{ nh }\)
Since, radius of the orbit is proportional to \( { n }^{ 2 } \)  , hence        
\(r\propto { n }^{ 2 } \)
\(r={ a }_{ 0 }{ n }^{ 2 } \)
\(T=\frac { 4{ \pi }^{ 2 }{ m }_{ 0 }^{ 2 }\quad { n }^{ 4 } }{ nh } =\frac { 4{ \pi }^{ 2 }{ m }_{ 0 }^{ 2 }\quad { n }^{ 3 } }{ h } \)
Number of orbits completed in \({ 10 }^{ -8 }s \)               
  \(=\frac { { 10 }^{ -8 } }{ T } =\frac { { 10 }^{ -8 }\times h }{ 4{ \pi }^{ 2 }{ m }_{ 0 }^{ 2 }\quad { n }^{ 3 } } \)
\(=\frac { { 10 }^{ -8 }\times 66{ \times 10 }^{ -34 } }{ 4\times { \left( 3.14 \right) }^{ 2 }\times 9.1\times { 10 }^{ -31 }\times { \left( 53\times { 10 }^{ -11 } \right) }^{ 2 }\times { \left( 2 \right) }^{ 3 } } \)
\(=8\times { 10 }^{ 6 }\)