Fusion reaction that takes place in the cores of Sun and other stars results in an enormous amount of energy which is called as stellar energy.

(i) The minimum mass of fissile material necessary to sustain chain reaction is called "critical mass".
(ii) It depends on the nature, density and the size of fissile material.

 d = 4.54 km = 4500 m
 t = 3 s
v = ?
\(v=\cfrac { d }{ t } =\cfrac { 4500 }{ 3 } \)  
v = 1500 m/s

According to Ohm's law, at a constant temperature, the steady current 'I' flowing through a conductor is directly proportional to the potential difference 'V' between the two ends of the conductor.
\(I\alpha V \Rightarrow\)  V = IR

In the cases given below there will be no Doppler effect.
(i) When source (S) and listener (L) both are at rest.
(ii) When source (S) and listener (L) are moving in mutually perpendicular directions.
(iii) When source (S) and listener (L) move in such a way that distance between them remains constant.
(iv) If the source (S) is situated at the center of the circle along which the listener (L) is moving.

Change in volume \((\Delta V)=0.3-0.25 =0.05\mathrm{~m}^{3}\)
Original volume \(\left(V_{0}\right)=0.3 \mathrm{~m}^{3}\)
Change in Temperature \(\Delta \mathrm{T}=50 \mathrm{~K}\)
\(a_{\mathrm{V}} =?  \)
\(\frac{\Delta V}{V_{0}} =a_{\mathrm{V}} \Delta \mathrm{T}\)
\(a_{\mathrm{V}} =\frac{0.05}{0.3 \times 50}=\frac{0.05}{15}=0.0033 \mathrm{~K}^{-1}\)
Co-efficient of cubical expansion of Zinc bar 0.0033

\(\text {Given } m_{1} : m_{2}=3: 4  \)          
\(a_{2} =12 \mathrm{~m} \mathrm{~s}^{-2}  \)
\(a_{1} =?\)
Using Newton's second law,
\(\mathrm{F} =\mathrm{m \times a}  \)
\(\mathrm{F} =\mathrm{m}_{1} \times a_{1}=3 a_{1}  \)
\(\mathrm{~F} =\mathrm{m}_{2} \times a_{2}=4 \times 12=48 \mathrm{~N}  \)
\(3 a_{1} =48  \)
\(a_{1} =48 / 3=16 \mathrm{~m} \mathrm{~s}^{-2}  \)
\(a_{1} =16 \mathrm{~ms}^{-2}\)
The acceleration produced on the other body is \(16 \mathrm{~m} \mathrm{~s}^{-2}\).

(i) Chemical Equilibrium: It is state of a reversible chemical reaction in which no change in the amount of reactants and products takes place.
(ii) At equilibrium, Rate of forward reaction = Rate of backward reaction
Characteristics of equilibrium:
(i) In chemical equilibrium, the rates of forward and backward reactions are equal.
(ii) The observable properties such as pressure, concentration, color, density, viscosity etc., of the system remain unchanged with time.
(iii) The chemical equilibrium is a dynamic equilibrium, because both the forward and backward reactions continue to occur even though it appears static externally.
(iv) In physical equilibrium, the volume of all the phases remain constant.

(i) In a decomposition reaction, a single compound splits into two or more simpler substances by the help of heat. It is called 'Thermolysis'.
(ii) In this type of reaction, the reactant is decomposed by applying heat.
(iii) For example, It is a class of compound to element/element decomposition. i.e. a compound (HgO) is decomposed into two elements (Hg and Oxygen).

(iv) When calcium carbonate is heated, it breaks down in to calcium oxide and carbon dioxide.
(v) It is a type of compound to compound/compound decomposition.
(vi) In thermal decomposition reaction, heat is supplied to break the bonds.
(vii) Such reactions, in which heat is absorbed, are called 'Endothermic reactions'.
\(\mathrm{CaCO}_{3(\mathrm{~s})} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2}(\mathrm{~g})\)

(i) Copper gets covered with a green layer of basic copper carbonate in the presence of CO₂ and moisture.
(ii) \(2 \mathrm{Cu}+\mathrm{O}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2}\)
(iii) Copper reacts with con. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to from copper sulphate and SO₂.
(iv)  \(\mathrm{Cu}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CuSO}_{4}+\mathrm{SO}_{2} \uparrow+2 \mathrm{H}_{2} \mathrm{O}\)
(v)  So, A is-Copper (Cu)
\(\mathrm{B}\ is \ -\mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2}\) Basic copper carbonate.
\(\mathrm{C}\ is \ -\mathrm{CuSO}_{4}\) (copper sulphate)
\(\mathrm{D}\ is \ -\mathrm{SO}_{2}\) (Sulphur oxide)

a) Wire is cut into 5 equal parts. Since all dimensions are same, resistance of each wire is equal and has a value = \(\frac {R}{5}\)
b) Formula for finding the effective resistance when connected in parallel is
\(\frac{1}{R_{p}^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}\)
Here, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}=\mathrm{R}_{4}=\mathrm{R}_{5}=\frac{\mathrm{R}}{5}\)
\(\frac{1}{R_{p}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}\)
\(\frac{1}{R_{p}}=\frac{25}{R}  \)
\(R_{p}=\frac{R}{25} \Omega\)
c) If the resistors are connected in series, then the effective resistance will be
\(\mathrm{R}_{\mathrm{s}}=\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5} \)
\(\mathrm{R}_{\mathrm{s}}=\frac{5 \mathrm{R}}{5}=\mathrm{R}\)
Ratio of effective resistance in series connection to that of the parallel connection is
\(\frac{R_{s}}{R_{p}}=\frac{R}{R / 25}=\frac{25}{1}\Rightarrow R_s:R_p=25:1\)

(i) An atom is no longer indivisible.
(ii) Atoms of the same element may have different atomic masses (isotopes ₁₇Cl³⁵, ₁₇Cl³⁷)
(iii) Atoms of different elements may have the same atomic masses (isobars ₁₈Ar⁴⁰, ₁₇Ca⁴⁰)
(iv) Atoms of one element can be transmuted into atoms of other elements. So atom is no longer indestructible. It is called artificial transmutation.
(v) Atoms may not always combine in a simple whole number ratio [Eg: Glucose C₆H₁₂O₆, sucrose C₁₂H₂₂O₁₁)
(vi) Atom is the smallest particle that takes part in a chemical reaction.
(vii) The mass of an atom can be converted into energy (E = mc²).

Mass of 1 mole of nitrogen \(\left(N_{2}\right)=2 \times N=2 \times 14=28 \mathrm{~g}\)
Mass of 3 moles of hydrogen \(\left(\mathrm{H}_{2}\right)=3[2 \times \mathrm{H}]\)
=3[2 x 1]=3[2]
=6 g
Mass of 2 moles of ammonia \(\left(\mathrm{NH}_{3}\right)=2\left[\mathrm{NH}_{3}\right]\)
=2[1 x N + 3 x H] 
=2[1 x14 + 3 x 1] 
=2[14 + 3] 
=2[17] = 34 g
\(\left.\begin{array}{c}1 \text { mole of nitrogen }(28 \mathrm{~g}) \\ + \\ 3 \text { mole of hydrogen }(6 \mathrm{~g})\end{array}\right\}=\text {2 moles of ammonia} (34 \mathrm{~g})\)