Class 12th

8.15 Long distance radio broadcasts use shortwave bands because only these bands can be refracted by ionosphere.

It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus these signals are not reflected by ionosphere. Hence, satellites are helpful in reflecting TV signals. Also they help in long distance TV transmissions.

With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.

The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiation present in sunlight and prevents from reaching Earth's surface.

In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth decreases rapidly, making it chilly and difficult for human survival.

A global nuclear war on the surface of the Earth would have disastrous consequences. Post nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light to reach the atmosphere. Also, it will lead to the depletion of the ozone layer.

8.13 A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Plank's law. It can be given by the relation,

T = 0.29 cmK

${\mathrm{}\lambda }_m=\frac{0.29}{T}$ cm K, where ${\mathrm{}\lambda }_{\mathit{m}}$

= maximum wavelength and T = temperature

Thus, the temperature for different wavelengths can be obtained as:

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${10}^{-4}$ cm, T = 2900 $°K$

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${5*10}^{-5}$ cm, T = 5800 $°K$

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${10}^{-6}$  cm, T = 290 $*{10}^3$ $°K$ and so on

The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.

1.32 (a) Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss's law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.

(b) Two charges of same magnitude and same sign are placed at a certain distance. The mid-point of the joining line of the charges is the null-point. When a test charge is displaced along the line, it experiences a restoring force, If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.

1.31 A proton has 3 quarks. Let there be n 'up quarks', then number of 'down quarks' = 3-n

Charge due to n 'up quarks' = $(\frac{2}{3}$ e) n

Charge due to (3-n) 'down quarks' =$-$$\frac{1}{3}e\left)\right(3-n)$ 

Total charge on a proton = +e = $(\frac{2}{3}$ e) n +  ( -$\frac{1}{3}e\left)\right(3-n)$ 

e =  $\frac{2ne}{3}$ +$\frac{ne}{3}$-e

2e = $\frac{3ne}{3}$

n = 2

Number of 'up quark' = 2 and number of 'down quark' = 1. Therefore a proton can be represented as 'uud'.

A neutron has 3 quarks. Let there be n 'up quark' in a neutron and (3-n) 'down quark'

Charge due to n 'up quark' = +$(\frac{2}{3}$ e)n

Charge due to (3-n) 'down quark' = 
- (

Since total charge of a neutron is zero, we get

+ $(\frac{2}{3}$e)n -($\frac{1}{3}e\left)\right(3-n)$ = 0

$(\frac{2}{3}$ e)n = ( $\frac{1}{3}e\left)\right(3-n)$

en = e or n = 1

Hence number of 'up quark' in neutron = 1

Number of 'down quark' in neutron = 3-1 = 2

Therefore a neutron can be represented as 'udd'.

1.30 

Take a long thin wire XY of uniform linear charge density $\lambda$. Consider a point A at a perpendicular distance l from the midpoint O of the wire. Let E be the electric field at point A due to the wire XY. Consider a small length element dx on the wire section with OZ = x. Let q be the charge on this piece.

Q = $\lambda dx$

Electric field due to the piece,

dE =  = $\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda dx}{{\left(AZ\right)}^2}$  = $\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda dx}{(l^2+x^2)}$since AZ =$\text{exception 25:}$

The electric field is resolved into two rectangular components. dE $\cos ?\theta$ is the perpendicular component and dE$\sin ?\theta$ $$ When the whole wire is considered, the component dE $\sin ?\theta \mathrm{}\mathrm{i}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{e}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{n}\mathrm{t}.\mathrm{}$ is cancelled. Only the perpendicular component dE $\cos ?\theta$ affects point A.

Hence, effective electric field at point A due to element dx is$d{E}_1$ 

 ………………….(1)

In  , $?AZO$ ,  $\tan ?\theta =\frac{x}{l}$ ,x =$l\tan ?\theta$……….(2)

On differentiating equation (2), we get

$\frac{dx}{d\theta }=l{sec}^2\theta$ , dx = …………….(3)

From equation (2), we have

 $x^2$ +$l^2$ = $l^2{tan}^2\theta +l^2$ =$l^2{sec}^2\theta$…….(4)

Putting equations (3) and (4) in equation (1), we get

 $d{E}_1=\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda l{sec}^2\theta d\theta \cos ?\theta }{l^2{sec}^2\theta }$ =$\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda \cos ?\theta d\theta }{l}$…….(5)

The wire is so long that $\theta$tends from  $-\frac{\pi }{2}$to$\frac{\pi }{2}$

By integrating equation (5), we get

 ${\int }_{-\frac{\pi }{2}}^{\mathrm{}\frac{\pi }{2}}d{E}_{\mathrm{1=}}$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{1}{4\pi {\epsilon }_0}*\frac{\lambda \cos ?\theta d\theta }{l})$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{l}{\left[\sin ?\theta \right]}_{\mathrm{}-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{l}*2=\frac{\lambda }{2\pi {\epsilon }_{0}l}$

Therefore the electric field due to long wire is $\frac{\lambda }{2\pi {\epsilon }_{0}l}$