Ncert Solutions Physics Class 12th

1.13 When the third uncharged sphere C is brought in contact with the sphere A, then the charge is shared and becomes half. Then

 $q_A$ = $\frac{q}{2}$and  = $q_C$= $\frac{q}{2}$

When the charged sphere C is brought in contact with charged sphere B, the charge between both the sphere is shared and becomes half

 $q_B,q_C=\frac{1}{2}$ (q + = $\frac{q}{2})$$\frac{3q}{4}$ 

Hence the force of repulsion between sphere A and B can be given as

F = 
$\frac{1}{4\pi {?}_0}$ $*\frac{q_1{q}_2}{r^2}$, where ${?}_0$  = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-\mathrm{1 \; \; }}$$m^{-2}$ = =$\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}$ = $*\frac{(\frac{q}{2}*\frac{3q}{4})}{{0.5}^2}$ =$\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}$$*\frac{3*q^2}{{8*0.5}^2}$ 
= $\frac{1}{4*\pi *8.854\mathrm{}*{10}^{-12}}$ $*\frac{3*{(6.5\mathrm{}*{10}^{-7})}^2}{{8*0.5}^2}$  = 5.695$*{10}^{-3}$N

1.12 (a) Charge on sphere A, $q_A$ = 6.5 $*{10}^{-7}$ C

Charge on sphere B, $q_B$ = 6.5$*{10}^{-7}$ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between two spheres

F = $\frac{1}{4\pi {?}_0}$ $*\frac{q_1{q}_2}{r^2}$ , where ${?}_0$= Permittivity of free space = 8.854$*{10}^{-12}$ 
  $C^2{N}^{-1}$$m^{-2}$Therefore, F =  $\frac{1}{4*\pi *{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{6.5\mathrm{}*{10}^{-7}*6.5\mathrm{}*{10}^{-7}}{{0.5}^2}$N = 0.0152 N = 1.52  $*{10}^{-2}$N

(b) Charge on sphere A, $q_A$ = 2 $*$6.5  $*{10}^{-7}$C = 1.3

Charge on sphere B, $q_B$ = 2 $*$ 6.5 $*{10}^{-7}$C = 1.3

Distance between the spheres, r =
$\frac{50}{2}$ = 25 cm = 0.25 m

Force of repulsion between two spheres

F =  $\frac{1}{4\pi {?}_0}$ $*\frac{q_1{q}_2}{r^2}$, where ${?}_0$ = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$Therefore, F = 
 $\frac{1}{4*\pi *{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{1.3\mathrm{}*{10}^{-6}*1.3\mathrm{}*{10}^{-6}}{{0.25}^2}$N = 0.243 N

Ans.1.12

(a) Charge on sphere A, $q_A$ = 6.5 $*{10}^{-7}$ C

Charge on sphere B, $q_B$ = 6.5$*{10}^{-7}$ C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between two spheres

F = $\frac{1}{4?{?}_0}$ $*\frac{q_1{q}_2}{r^2}$ , where ${?}_0$= Permittivity of free space = 8.854$*{10}^{-12}$ 
  $C^2{N}^{-1}$$m^{-2}$Therefore, F =  $\frac{1}{4*?*{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{6.5\mathrm{}*{10}^{-7}*6.5\mathrm{}*{10}^{-7}}{{0.5}^2}$N = 0.0152 N = 1.52  $*{10}^{-2}$N

(b) Charge on sphere A, $q_A$ = 2 $*$6.5  $*{10}^{-7}$C = 1.3

Charge on sphere B, $q_B$ = 2 $*$ 6.5 $*{10}^{-7}$C = 1.3

Distance between the spheres, r = $$ = 25 cm = 0.25 m

Force of repulsion between two spheres

F =  $\frac{1}{4?{?}_0}$ $*\frac{q_1{q}_2}{r^2}$, where ${?}_0$ = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$Therefore, F = 
 $\frac{1}{4*?*{8.854\mathrm{}*{10}^{-12}}_{}}$ $*\frac{1.3\mathrm{}*{10}^{-6}*1.3\mathrm{}*{10}^{-6}}{{0.25}^2}$N = 0.243 N

1.11

(a) When polythene rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene negatively charged.

Amount of charge on the polythene piece, q = -3 $*{10}^{-7}$ C.

Amount of charge of 1 electron e = -1.6 $*10-19$

So number of electron transferred from wool to polythene

=
-$\frac{3\mathrm{}*{10}^{-7}}{-1.6\mathrm{}*{10}^{-19}}$1.875$*{10}^{12}$

(b) Since electron has a mass, so there will be transfer of mass also.

Mass of single electron,  $<math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math>$ = 9.1 $*{10}^{-31}$ kg

Total mass transferred from wool to polythene = 1.875 $*{10}^{12}*$9.1 $*{10}^{-31}$ kg

= 1.706 $*{10}^{-18}$kg$?$ negligible

Hence a negligible amount of mass is transferred from wool to polythene.

The principle of mass-energy equivalence, given by Einstein's equation:

$E = mc^2$It states that mass and energy are interchangeable.

A small amount of mass can be converted into a large amount of energy because the speed of light $(c = 3 \times 10^8 \, \text {m/s})$ is very large and appears squared in the equation.

Significance in Nuclear Reactions:

• Helps in calculating the energy release in Nuclear Reactions
• Provides explanation for powering Nuclear Reactors and the Sun
• Explanation of Binding Energy of Nucleous

Nuclear Binding Energy is the amount of energy required to completely dismantle a nucleus into its individual protons and neutrons which is equivalent to the amount of energy to form a nucleous from its constituent nucleons (protons and neutrons).

Binding energy is calculated using Einstein's mass-energy equivalence relation:

$B.E. = \Delta m \cdot c^2$Where:

• $\Delta m$ = mass defect (in kg or amu)

• $c$ = speed of light $(3 \times 10^8 \, \text {m/s})$ (3*108m/s)

As per the NCERT Textbooks, Thomson proposed a Atomic Structure of Atom that tells" An atom consists of a positively charged sphere in which the electrons are embedded like the seeds are embedded in watermelon. This model is often compared to a pudding or watermelon with electrons distributed like raisins or plums, also known as “plum pudding model.”

The presesnt day theory for structure of Atom is developed through many discoveries and hypothesises. In class 12 Physics, Atom chapter includes development of the sturucture of atom, and theories in the path of development of present theory. Students can check the ordered points below;

1. Thomson's Model of the Atom
2. Rutherford's Nuclear Model
3. Bohr's Model of the Hydrogen Atom
4. De Broglie's Hypothesis
5. Energy Emission Spectrum

These throries have been used to introduced the current theory of structure of atom.

Students who are preparing for the class 12 board exams needs additional practice questions after completing the NCERT Exercises. We have provided practice questions with accurate and atep-by-step solutions for students to better prepare for the board exams. Students can check the below provided link to access our additional practice questions along with previous year questions.

Chapter 12 Atoms NCERT Solutions