Ncert Solutions Physics Class 12th

1.15 (a) Electric field intensity, E = 3 $*{10}^3$ î N/C

Magnitude of electric field intensity, $\left|E\right|$$=3*{10}^3$ = 
 N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = $s^2$ = 0.01 $m^2$

The plane of the square is parallel to the y-z plane, hence the angle between the unit vector normal to the plane and electric field, $\theta$= 0 $°$

Flux ( $\phi )$ through the plane is given by the relation,  $\phi$ = $\left|E\right|A\cos ?\theta$ = $3\mathrm{}*{10}^3*0.01*\cos ?0°=30$
${\mathrm{Nm}}^2/$C

(b) When the normal to its plane make a 60 $°$ angle with x-axis,$\theta$  = 60 
 . From the equation $\phi$ =$\left|E\right|A\cos ?\theta$we get  = $\phi$ $3\mathrm{}*{10}^3*0.01*\cos ?60°=15$ ${\mathrm{Nm}}^2/$C

1.14 Since the charges 1 & 2 are attracted towards + ve, their charges will be – ve. The charge 3 is attracted towards – ve, hence its charge will be +ve.

The charge to mass ratio (emf) is directly proportional to the displacement, charge 3 will have the highest charge to mass ratio.

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.”