Here, V₁ = 5.5MHz = 5.5 x 10⁶ Hz
v₂ = 16 MHz = 16 x 10⁶Hz
\(\therefore \quad \lambda_{1}=\frac{c}{v_{1}}=\frac{3 \times 10^{8}}{5.5 \times 10^{6}}\)
\(=0.545 \times 10^{2} \mathrm{~m}=54.5 \mathrm{~m}\)
\(\Rightarrow \quad \lambda_{2}=\frac{c}{v_{2}}=\frac{3 \times 10^{8}}{16 \times 10^{6}}=0.1875 \times 10^{2}\)
\(=18.75 \mathrm{~cm}\)
Hence, corresponding wavelengths of above frequencies are 54.5 m and 18.75 m.

For maximum current through the external resistance, external resistance = total internal resistance of cells
\(\text { or } \ R=\frac{n r}{m}\)
∴ \(1.5=\frac{n \times 0.5}{\frac{12}{n}}\)  [∵ mn = 12]
or 36 = n²
 ∴ n = 6 and m = 2

Energy, \(E=10^{-3} \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)
\(E=10^{-3} \times 9 \times 10^{16}=9 \times 10^{13} \mathrm{~J}\)
Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

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

As, electron is revolving clockwise, therefore conventional current due to the motion of electron will be in anti-clockwise direction.
So, according to right hand rule,magnetic field at point A will be in outward direction to the plane of circular loop.

As we know E = V + Ir and \(V=\frac{E}{2}, r=1 \Omega\)
\(\therefore E=\frac{E}{2}+I r \ \Rightarrow \ I=\frac{E}{2 r}=\frac{1.5}{2 \times 1}=0.75 \mathrm{~A}\)

\(\text { Given } E_{y}=4 \times 10^{5} \cos \left(3.14 \times 10^{8} t-1.57 x\right) \mathrm{N} / \mathrm{C}\)
∵ General equation of electric field is given by
\(E_{y}=E_{0} \cos (\omega t-k x) \mathrm{N} / \mathrm{C}\)
\(\therefore E_{0}=4 \times 10^{5} \mathrm{~N} / \mathrm{C}, \omega=3.14 \times 10^{8} \mathrm{rad} \mathrm{s}^{-1} \)
\(k=1.57 \mathrm{rad} \cdot \mathrm{m}^{-1} \)
\((i) \lambda=\frac{2 \pi}{k}=\frac{2 \times 3.14}{1.57} \mathrm{~m}=4 \mathrm{~m} \)
\((ii) \mathrm{v}=\frac{\omega}{2 \pi}=\frac{3.14 \times 10^{8}}{2 \times 3.14} \mathrm{~Hz}=5 \times 10^{7} \mathrm{~Hz}\)
(iii) The direction of propagation is +x direction.
\( \text { (iv) } v=\frac{\omega}{k}=\frac{3.14 \times 10^{8}}{1.57} \mathrm{~m} / \mathrm{s}=2 \times 10^{8} \mathrm{~m} / \mathrm{s} \)
\(\text { (v) } \mu=\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=1.5 \)
\(\text { (vi) } \because \frac{E_{0}}{B_{0}}=c \)
\(\Rightarrow B_{0}=\frac{E_{0}}{c}=\frac{4 \times 10^{5}}{3 \times 10^{8}} \mathrm{~T}=1.33 \times 10^{-3} \mathrm{~T} \)

We have the expression for nuclear radius as:
R = R₀A¹/³
Where,
R₀ = Constant.
A = Mass number of the nucleus
Nuclear matter density, \(\rho \ =\frac{Mass\ of\ the\ nucles}{Volume\ of\ the\ nucles}\)
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
\(\therefore \rho=m \frac{A}{\frac{4}{3} \pi R^{3}}=\frac{3 \mathrm{~mA}}{4 \pi\left(R_{0} A^{\frac{1}{3}}\right)^{3}}=\frac{3 m A}{4 \pi R_{0}^{3} A}=\frac{3 \mathrm{~m}}{4 \pi R_{0}^{3}}\)
Hence, the nuclear matter density is independent of A. It is nearly constant.

Here, B=5X10^-3T, v=5X10⁷ ms^-1,
         e=1.6X10^-19C, m=9.1X10^-31 kg
(i) Radius of the circular path,
    \(r=\frac { mv }{ Be } =\frac { \left( 9.1\times { 10 }^{ -31 } \right) \times \left( 5\times { 10 }^{ 7 } \right) }{ \left( 5\times { 10 }^{ -3 } \right) \times \left( 1.6\times { 10 }^{ -19 } \right) } \) 
               = 5.7X10^-2 m
               = 5.7cm
(ii) Frequencyof revolution,
            \(v=\frac { eB }{ 2\pi m } =\frac { \left( 1.6\times { 10 }^{ -19 } \right) \times \left( 5\times { 10 }^{ -3 } \right) }{ 2\times 3.14\times \left( 9.1\times { 10 }^{ -31 } \right) } \) 
                = 1.4X10⁸ Hz 

If A is the area of cross-section of the wire, then number of electrons per unit volume of copper wire is
n = \(\frac{2 \times 10^{21}}{A\times(5\times10^{-2})}m^{-3}\)
= nAev_d
= \(\frac{2\times10^{21}}{A\times(5\times10^{-2})} \times A \times(1.6 \times 10^{-19})\times (0.25 \times 10^{-3})\)
= 1.6 A

Since impact parameter is given by:
\(b={Ze^2cot{\theta\over2}\over4\pi\epsilon_0 E}\)
when b = 0, \(cot{\theta\over2}=0\)
or \({\theta\over2}=90^0\)
or  \(\theta =180^0\)
i.e. \(\alpha\) -particle retrace its path as its angle of scattering is 180⁰

Since \(\varepsilon _{ E }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ { E }^{ 2 } }\) 
and \(\varepsilon _{ B }=\frac { { B }^{ 2 } }{ 2\mu _{ 0 } } \) 
\(\therefore \) Total energy density \(u={ u }_{ E }+{ u }_{ B }=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }_{ 2 }+\frac { 1 }{ 2 } \frac { { B }^{ 2 } }{ { \mu }_{ 0 } } \) 
For a plane em wave, Band E are related as
\({ B }_{ v }=\frac { { E }_{ V } }{ C } \) 
\(u=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }_{ v2 }+\frac { 1 }{ 2\mu _{ 0 } } \left( \frac { { E }_{ v } }{ c } \right) ^{ 2 }\) 
\(={ E^{ 2 } }_{ v }\frac { \left( { \mu }_{ 0 }\varepsilon _{ 0^{ c2 } }+1 \right) }{ 2{ \mu }_{ 0^{ c2 } } } ={ E }^{ 2 }_{ v }\) 
or      \(u=\frac { { E }^{ 2 }_{ v }(1+1) }{ 2{ \mu }_{ 0 }\frac { 1 }{ { \mu }_{ 0 }\varepsilon _{ 0 } } } \)   \(\left[ \therefore { c }^{ 2 }=\frac { 1 }{ { \mu }_{ 0 }\varepsilon _{ 0 } } \right] \) 
\(=\varepsilon _{ 0 }{ E }^{ 2 }_{ v }\) 
So     \(I=\frac { Energy/Time }{ Area } \) 
or
\(I=\frac { Energy \ density/Time }{ Area } \times volume\) 
\( =Energy\quad density\times \frac { length }{ Time } \)
\( =u\times c\)
Using eqn.(i), we get
\(\\ I=\varepsilon _{ 0 }{ E }^{ 2 }_{ v }\)

Given
\({ R }_{ 1 }=1\Omega ,{ R }_{ 2 }=2\Omega ,{ R }_{ 3 }=3\Omega \)
(a) Total resistance of series combination
\({ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\)
\({ R }_{ s }\) = 1 + 2 + 3 = \(6\Omega \)

(b) Since E = I (R + r)
I = \(\frac { E }{ { R }_{ s }+0 } =\frac { E }{ { R }_{ s } } =2A\)
\({ V }_{ 1 }={ IR }_{ 1 }=2\times 1=2V\)
\({ V }_{ 2 }={ IR }_{ 2 }=2\times 2=4V\)
\({ V }_{ 3 }={ IR }_{ 3 }=2\times 3=6V\)

Given, resistance of heating element at temperature 27°C,   
R₂₇ = 100 \(\Omega\)
Resistance of heating element at temperature t°C,
R_t = 117 \(\Omega\)
\(\alpha\)=1.70 \(\times\)10^-4 °C^-1, t = ?
By using the formula of temperature coefficient of resistance,
\(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\)   ...(i)
Here, R₁ = R_t, R₁ = R₂₇, t₂ = t and t₁ = 27°C
Such that, \(\alpha=\frac{R_t-R_{27}}{R_{27}(t-27)}\)
 Substituting given values in Eq. (i), we get
\(\begin{aligned} 1.70 \times 10^{-4} & =\frac{117-100}{100(t-27)} \\ \end{aligned}\)
\(\begin{aligned} \text { or } \quad t-27 & =\frac{17}{100 \times 1.70 \times 10^{-4}} \end{aligned}\)
or t = 1000 + 27 = 1027°C 

Magnetic field strength, B = 6.5 x 10⁻⁴ T
Charge of the electron, e = 1.6 x 10⁻¹⁹ C
Mass of the electron, m_e = 9.1 x 10⁻³¹ kg
Velocity of the electron, v = 4.8 x 10⁶ m/s
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = ν
Angular frequency of the electron = ω = 2πν
Velocity of the electron is related to the angular frequency as:
v = rω
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
\(e v B=\frac{m v^{2}}{r}\)
\(e B=\frac{m}{r}(r \omega)=\frac{m}{r}(2 \pi r v)\)
\(v=\frac{B e}{2 \pi m}\)
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
\(V=\frac{6.5 \times 10^{-4} \times 1.6 \times 10^{-19}}{2 \times 3.14 \times 9.1 \times 10^{-31}}\)
= 18.2 x 10⁶ Hz
\(\approx\) 18 MHz
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.

Length of the solenoid, l = 80 cm = 0.8 m
There are five layers of windings of 400 turns each on the solenoid.
∴ Total number of turns on the solenoid, N = 5 x 400 = 2000
Diameter of the solenoid, D = 1.8 cm = 0.018 m
Current carried by the solenoid, I = 8.0 A
Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,
\(B=\frac{\mu_{0} N I}{l}\)
Where,
μ₀ = Permeability of free space = 4π x 10^–7 T m A^–1
\(B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}\)
\(=8 \pi \times 10^{-3}=2.512 \times 10^{-2} T\)
Hence, the magnitude of the magnetic field inside the solenoid near its centre is 2.512 x 10^–2 T.

Atomic mass of \(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\),m₁ = 55.934939 u
\(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\) nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus,Δm = 26 x m_H + 30 x m_n − m₁
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm = 26 x 1.007825 + 30 x 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c²
∴Δm = 0.528461 x 931.5 MeV/c²
The binding energy of this nucleus is given as:
E_b1 = Δmc²
Where,
c = Speed of light
∴E_b1 = 0.528461 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 492.26 MeV
Average binding energy per nucleon =\(\frac{492.26}{56}=8.79 \mathrm{MeV}\)
Atomic mass of \(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\), m₂ = 208.980388 u
\(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\) nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 x m_H + 126 x m_n − m₂
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm' = 83 x 1.007825 + 126 x 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c²
∴Δm' = 1.760877 x 931.5 MeV/c²
Hence, the binding energy of this nucleus is given as:
E_b2 = Δm'c²
= 1.760877 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 1640.26 MeV
Average bindingenergy per nucleon \(=\frac{1640.26}{209}=7.848 \mathrm{MeV}\)

While drawing the plot. we have to keep in mind that first binding energy will increase sharply and then it will be constant almost.
For plot of binding energy per nucleon as the function of mass number A
Following are the two conclusions that can be drawn regarding the nature of the nuclear force.
The force is attractive and strong enough to produce a binding energy of few MeV per nucleon.
The two important conclusions regarding the nature of nuclear force are given below.
(i) The nuclear force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the wide range of mass number 30 < A < 170 indicate that nuclear force is a short-range force.
(b) (i) According to the binding energy curve, a very heavy nucleus (A > 170), has lower binding energy per nucleon compared to nuclei of middle mass number (30 < A < 170).
Thus, if a heavy nucleus breaks into two nuclei of mass number between 30 and 170, nucleons get more tightly bound. This implies energy would be released in the process. (nuclear fission)
(ii) When two light nuclei (A < 10) join to form a heavier nucleus, the binding energy per nucleon of fused heavier nucleus increases.
Again it indicates that energy would be released in the process (nuclear fusion).
(c) The basic nuclear process of neutron undergoing β-decay is given as
\(n \rightarrow p+e^{-}+\bar{v}\)
Here \(\bar{v}\) is antinutrino.
Neutrino and antineutrino both are neutral particles with very small (possibly, even zero) mass compared to the electrons. They have only weak interaction with other particles. Therefore, the detection of neutrinos is found very difficult.

Given, E = 10 V, r = 3 \(\Omega\), I = 0.5 A
As, \(\begin{aligned} I & =\frac{E}{R+r} \end{aligned}\)
\(\begin{aligned} R & =\frac{E}{I}-r \end{aligned}\)
\(\begin{aligned} =\frac{10}{0.5}-3=17 \Omega \end{aligned}\)
and terminal voltage, V = IR = 0.5 \(\times\)17 = 8.5 V
 

n = 300 turns/m ;
I = 5 A ; l = 0.5 m, r = 10^-2 m.
\({ B }={ \mu }_{ o }nI=\left( 4\pi \times { 10 }^{ -7 } \right) \times 300\times 5\) 
= 1.9 x 10^-3 T.

(b): At every instant the ratio of the magnitudes of the electrip field to the magnetic field of an em wave is given by E/B = c.
From this equation, the magnitude of electric vector is much greater than the magnitude of magnetic vector. Also electromagnetic waves, carry energy which is equally shared by electric and magnetic field.

(a): Experimentally, it is found that the average radius of a nucleus is given by
\(R=R_{0} A^{1 / 3} \text { where } R_{0}=1.1 \times 10^{-15} \mathrm{~m}=1.1 \mathrm{fm}\)
and A = mass number
The volume of a nucleus is \(V=\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R_{0}^{3} A\)
Now as the masses of a proton and a neutron are roughly equal, say m, the mass of a nucleus is also roughly proportional to the mass number A, M = mA
Hence density within a nucleus \(\rho=\frac{M}{V}=\frac{m A}{\frac{4}{3} \pi R_{0}^{3} A}\)
\(=\frac{m}{\frac{4}{3} \pi R_{0}^{3}}\) is independent of the mass number A.

(c): An ammeter is a low resistance galvanometer. It is used to measure the current in amperes. To measure the current of a circuit, the ammeter is connected in series in the circuit so that the current to be measured must pass through it. Since, the resistance of ammeter is low, so its inclusion in series in the circuit does not change the resistance and hence the main current in the circuit.

(a): Drift velocity of free electrons is given by \(v_{d}=\frac{e E}{m} \tau\)
where,\(E=\frac{\text { Potential difference }}{\text { length }}=\frac{V}{l}\)
\(\therefore \quad v_{d}=\frac{e V}{m l} \tau \text { i.e., } v_{d} \propto 1 / l \text { where, } \frac{e V \tau}{m}\)
It mean if I is doubled, the drift velocity will become half of the original value.

(i) (a) :Magnetic field due to a long current carrying wire at r
\(B=\frac{\mu_{0}}{2 \pi} \frac{I}{r}\)
(ii) (d): Let I' be the current in region r < R
Then, \(I^{\prime}=\frac{I}{\pi R^{2}} \pi\left(r^{2}\right) \text { or } I^{\prime}=\frac{I r^{2}}{R^{2}}\)
So, magnetic. field \(B=\frac{\mu_{0} I^{\prime}}{2 \pi r}=\frac{\mu_{0} I r^{2}}{2 \pi R^{2} r}=\frac{\mu_{0} I r}{2 \pi R^{2}}\)
(iii) (a): Magnetic field due to a long straight wire of radius a carrying current I at a point distant r from the centre of the wire is given as follows

\(B=\frac{\mu_{0} I r}{2 \pi a^{2}} \quad \text { for } \quad r<a\)
\(B=\frac{\mu_{0} I}{2 \pi a} \quad \text { for } r=a\)
\(B=\frac{\mu_{0} I}{2 \pi r} \quad \text { for } r>a\)
The variation of magnetic field B with distance r from the centre of wire is shown in the figure.
(iv) (d): Let the magnetic fields due to a long straight wire of radius R carrying a steady current I at a distance r from the centre of the wire are
\(B_{1}=\frac{\mu_{0} I r}{2 \pi R^{2}} \quad(\text { For } r<R)\)
and \(B_{2}=\frac{\mu_{0} I}{2 \pi R} \quad(\text { For } r>R)\)
So, the magnetic field at \(r=\frac{R}{2} \text { is } B_{1}=\frac{\mu_{0} I}{2 \pi R^{2}}\left(\frac{R}{2}\right)=\frac{\mu_{0} I}{4 \pi R}\)
and at \(r=2 R \text { is } B_{2}=\frac{\mu_{0} I}{2 \pi(2 R)}=\frac{\mu_{0} I}{4 \pi R}\)
\(\therefore\) Their corresponding ratio is \(\frac{B_{1}}{B_{2}}=\frac{\left(\mu_{0} I / 4 \pi R\right)}{\left(\mu_{0} I / 4 \pi R\right)}=1\)
(v) (c)