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.

The ratio of the radius of electron’s orbit to the radius of nucleus is (10^–10 m) /(10^–15 m) = 10⁵, that is, the radius of the electron’s orbit is 10⁵ times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 10⁵ times larger than the radius of the sun, the radius of the earth’s orbit would be 10⁵ x 7 x 10⁸ m = 7 x 10¹³ m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun. It implies that an atom contains a much greater fraction of empty space than our solar system does.

Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T =  4.5 x 10^-2 J
The angle between the bar magnet and the external magnetic field, θ = 30°
Torque is related to magnetic moment (M) as:
\(T=M B \sin \theta \therefore M=\frac{T}{B \sin \theta}\)
\(=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 J T^{-1}\)
Hence, the magnetic moment of the magnet is 0.36  J T^-1.

The ray diagram for the formation of the image of the phone is shown in Fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B'C = BC. You can yourself realise why the image is distorted.

No, as every material has some resistance. Yes, we can coil a wire to have resistance with inductance.

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.

1u = 1.6605 x 10^–27 kg
To convert it into energy units, we multiply it by c² and find that energy equivalent = \(1.6605 \times 10^{-27} \times\left(2.9979 \times 10^{8}\right)^{2} \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}^{2}\)
\(=1.4924 \times 10^{-10} \mathrm{~J}\)
\(=\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}} \mathrm{eV}\)
\(=0.9315 \times 10^{9} \mathrm{eV}\)
\(=931.5 \mathrm{MeV}\)
or, \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
For, \({ }_{8}^{16} \mathrm{O}, \quad \Delta M=0.13691 \mathrm{u}=0.13691 \times 931.5 \mathrm{MeV} / \mathrm{c}^{2}\)
\(=127.5 \mathrm{MeV} / \mathrm{c}^{2}\)
The energy needed to separate \({ }_{8}^{16} \mathrm{O}\) into its constituents is thus 127.5 MeV/c².

we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^–11 m is 2.2 × 10^–6 m/s. Thus, the frequency of the electron moving around the proton is
\(v=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
\(\approx \) 6.6 × 10¹⁵ Hz
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵ Hz.

Net outward flux through the surface of the box, Φ = 8.0 × 10³ N m²/C
For a body containing net charge q, flux is given by the relation,
\(\phi=\frac{q}{\epsilon_{0}}\)
∈₀ = Permittivity of free space
= 8.854 × 10⁻¹² N⁻¹C² m⁻²
q = ∈₀Φ
= 8.854 × 10⁻¹² × 8.0 × 10³
= 7.08 × 10⁻⁸
= 0.07 μC
Therefore, the net charge inside the box is 0.07 μC.
(b) No
Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
\(tan2\theta =\frac { d }{ 1.5 } \)
∴ d = 1.5 x tan 70° = 0.184 m = 18.4 cm
Hence, the displacement of the reflected spot of light is 18.4 cm.

Capacitance, C = 30μF = 30 × 10⁻⁶F
Inductance, L = 27 mH = 27 × 10⁻³ H
Angular frequency is given as:

\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 27\times { 10 }^{ -3 }\times 30\times { 10 }^{ -6 } } } =\frac { { 10 }^{ 4 } }{ 9 } =1.1\times { 10 }^{ 3 }rad/s\)
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10³ rad/s.

\(Here,M=? \ dI=6-2=4A,dt=0.1 \ s,e=1V\)
\( From \ e=M \ dI/dt\)
\(M=\frac { e.dt }{ dI } =\frac { 1 \ times 0.1 }{ 4 } =0.025 \ H\)

The given nuclear reaction is:
\({ }_{1}^{1} H+{ }_{1}^{3} H \rightarrow_{1}^{2} H+{ }_{1}^{2} H\)
It is given that:
Atomic mass \(m\left({ }_{1}^{1} H\right)=1.007825 \mathrm{u}\)
Atomic mass \(m\left({ }_{1}^{3} H\right)=3.016049 \mathrm{u}\)
Atomic mass \(m\left({ }_{1}^{2} H\right)=2.014102 u\)
According to the question, the Q-value of the reaction can be written as:
\(Q=\left[m\left({ }_{1}^{1} H\right)+m\left({ }_{1}^{3} H\right)-2 m\left({ }_{1}^{2} H\right)\right] c^{2}\)
\(=[1.007825+3.016049-2 \times 2.014102] c^{2}\)
\(Q=-0.00433 \times 931.5=-4.0334 \mathrm{MeV}\)
The negativeQ-value of the reaction shows that the reaction is endothermic.
The given nuclear reaction is:
\(\text { (ii) }{ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He}\)
t is given that:
Atomic mass of \(m\left({ }_{6}^{12} C\right)=12.0 u\)
Atomic mass of \(m\left(\begin{array}{l} 20 \\ 10 \end{array}\right)=19.992439\)
Atomic mass of \(m\left({ }_{2}^{4} \mathrm{He}\right)=4.002603 \mathrm{u}\)
The Q-value of this reaction is given as:
\(Q=\left[2 m\left({ }_{6}^{12} C\right)-m\left({ }_{10}^{20} N e\right)-m\left({ }_{2}^{4} H e\right)\right] c^{2}\)
\(=[2 \times 12.0-19.992439-4.002603] c^{2}\)
\(=\left(0.004958 c^{2}\right) u\)
\(=0.004958 \times 931.5=4.618377 \mathrm{MeV}\)
The positive Q-value of the reaction shows that the reaction is exothermic.

Here, number of turns per unit length,
\(n=\frac{N}{l}=15\) turns/cm = 1500 turns/m
A = 2.0 cm² = 2 \(\times\)10^-4m²
\(\begin{aligned} \therefore \frac{d I}{d t}=\frac{4-2}{0.1} \text { or } \frac{d I}{d t}=20 \mathrm{As}^{-1} \\ \end{aligned}\)
\(\begin{aligned} \therefore|e|=\frac{d \phi}{d t}=\frac{d}{d t}(B A) \quad\left[\because B=\frac{\mu_0 N I}{l}\right] \\ \end{aligned}\)
\(= \frac{A d}{d t}\left(\mu_0 \frac{N I}{l}\right)=A \mu_0\left(\frac{N}{l}\right) \frac{d l}{d t}\)
= (2 \(\times\) 10^-4) \(\times\)4 \(\pi\)\(\times\)10^-7 \(\times\)1500 \(\times\)20 V
= 7.5 \(\times\)10^-6 V

(i) (a) On increasing H (= ni); intensity of magnetization I (magnetic dipole moments induced per unit volume) increases due to more and more alignment of spinning electrons with the H-field, when all the magnetic dipole moments are aligned along H-field, the intensity of magnetisation reaches its maximum value and specimen is saturated.
(b) \(\overrightarrow{\mathrm{B}}=\mu_{0} \overrightarrow{\mathrm{H}}+\mu_{0} \overrightarrow{\mathrm{I}}\) 
Here I = Intensity of Magnetization
(c) Graph 1 represents \(\mathrm{B} \propto \mu_{0} \mathrm{H}\)
Graph 2 represents variation of B with \(\mu\)₀I
Graph 3 represents variation B with \(\left(\mu_{0} \mathrm{H}+\mu_{0} \mathrm{I}\right)\)
(ii) At high temperature the randomness in paramagnetic substances increases according to formula.
\(\chi=\frac{\mathrm{I}}{\mathrm{H}} \propto \frac{1}{\text { Temperature }}\)
\(\therefore\) When a paramagnetic substance is cooled to extremely low temperature, then disorientation effect is highly reduced.

(i) (b)
(ii) (b): Fast neutrons are slowed down by elastic scattering with light nuclei as each collision takes away nearly 50% of energy.
(iii) (d): Reactions I and II represent fission of uranium isotope \({ }_{92}^{235} \mathrm{U}\), when bombarded with neutrons that breaks it into two intermediate mass nuclear fragments. However, reaction III represents two deuterons fuses together to from the light isotope of helium.
(iv) (c): On an average 2.5 neutrons are released per fission of the uranium atom.
The energy of the neutron released per fission of the uranium atom is 2 MeV.
(v) (a): In fission process, when a parent nucleus breaks. into daughter products, the some mass is lost in the form of energy. Thus,mass of fission products < mass of parent nucleus.
\(\Rightarrow \frac{\text { Mass of fission products }}{\text { Mass of parent nucleus }}<1\)