# CLS Aipmt 16 17 XII Phy Study Package 6 SET 2 Chapter 8

July 31, 2017 | Author: Kareena Gupta | Category: Electromagnetic Radiation, Wavelength, Waves, Light, Oscillation

aksssh...

#### Description

Chapter

8

Electromagnetic Waves Solutions SECTION - A Objective Type Questions 1.

According to modified Ampere's circuital law (iD = displacement current) d ⎞ ⎛   0 ⎜ iC   0 E ⎟ dt ⎠ ⎝

(1)

∫ B  dl

(3)

∫ B  dl  

0

i

(2)

∫ B  dl

(4)

∫B d l  

 0 0

0

d E dt

⎞ ⎛ d E  iD ⎟ ⎜ iC dt ⎠ ⎝

Sol. Answer (1) According to modified Ampere's circuital law.

∫ B  dl 2.

d ⎞ ⎛   0 ⎜ iC   0 E ⎟ dt ⎠ ⎝

Displacement current is set up between the plates of a capacitor when the potential difference across the plates is (1) Maximum

(2)

Zero

(3)

Minimum

(4)

Varying

Sol. Answer (4) Displacement current, I d  A 0

dE dt

So, the potential difference has to change with time 3.

A parallel plate capacitor with circular plates of radius R is being charged as shown. At the instant shown, R the displacement current in the region between the plates enclosed between and R is given by 2 + –

i

(1)

3 i 4

(2)

1 i 4

i

(3)

3i

(4)

4 i 3

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

112

Electromagnetic Waves

Solution of Assignment (Set-2)

Id   0A

dE dE   0  R 2  i dt dt

⎛ R 2 ⎞ dE I ' d   0 ⎜ R 2 – ⎟ ⎝ 4 ⎠ dt 3 dE  0 R 2 4 dt

I 'd 

4.

3 i 4

Figure shows a circular region of radius R in which uniform magnetic field B exists. The magnetic field is dB . The induced electric field at a distance r from the centre for r < R is dt

increasing at a rate

B R

dB R 2 dt 2r

(3)

dB dt

(1) Wavelength

(2)

Frequency

(3) Intensity

(4)

Medium, in which it travels

(1)

dB r dt 2

O

(2)

Zero

(4)

Sol. Answer (1) Rate of increase of B 

or

E in  2r  –

d dt

E  2r  – A

dB dt

E  2r  – r 2

5.

E

dB dt

dB dt

– r dB 2 dt

The speed of electromagnetic waves depends upon

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electromagnetic Waves

113

Sol. Answer (4) I Speed of wave  

I

 0 r  0  r

It depends upon the medium. 6.

If E and B represent the electric and magnetic field vectors of an electromagnetic wave, then the direction of propagation of the electromagnetic wave is in the direction of (1) E

(2)

B

(3)

E B

(4)

B E

  Direction of propagation is given by E  B

7.

If an electromagnetic wave propagating through vacuum is described by E y = E 0 sin (kx –  t); Bz = B0 sin (kx –  t), then (1) E0 k = B0 

E0 B0 =  k

E 0 B0 

 k

(3)

E0  = B0 k

(1) Electric field

(2)

Magnetic field

(3) Both (1) & (2)

(4)

Neither electric field nor magnetic field

(2)

(4)

Sol. Answer (1) Speed of wave   8.

E0   B0 k

kE 0  B 0

Electromagnetic wave is deflected by

Sol. Answer (4) Electromagnetic wave consists of unchanged particle called photons which are neither deflected by electric field nor by magnetic field. 9.

Out of the following, choose the ray which does not travel with the velocity of light (1) X-ray

(2)

Microwave

(3)

-rays

(4)

-rays

(4)

Amplitude

(4)

Infra-red ray

Sol. Answer (4) -rays do not travel with speed of light, as they are not em waves. 10. Red light differs from blue light in its (1) Speed

(2)

Frequency

(3)

Intensity

Sol. Answer (2) Frequency is different for each light as they have different wavelengths. 11. Which of the following has the largest wavelength? (1) Radio wave

(2)

X-ray

(3)

Ultraviolet ray

Sol. Answer (1) Radio waves : 10–2 m to 104 m. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

114

Electromagnetic Waves

Solution of Assignment (Set-2)

12. Velocity of electromagnetic waves in a medium is  1/2 (1) ( 0 0 )

(2)

( 0  r 0  r ) 1/2

(3)

108

m/s

(4)

⎛ 0 r ⎞ ⎜⎝   ⎟⎠ 0 r

 1/2

1  

Speed of light in medium 

1  0 r  0  r

   0 r  0  r 

1 2

13. Which of the following is incorrect about a plane electromagnetic wave? (1) The electric field and magnetic field have equal average values (2) The electric energy and the magnetic energy have equal average values (3) The electric field and magnetic field both oscillate in same phase (4) The electric field and magnetic field oscillate in opposite phase Sol. Answer (4)   E and B are in the same phase but oscillate in different planes which are perpendicular to each other. 14. In a plane electromagnetic wave, which of the following has/ have zero average value in one complete cycle? (a) Magnetic field

(b)

Magnetic energy

(c) Electric field

(d)

Electric energy

(1) (a), (c)

(2)

(b), (c)

(3)

(a), (d)

(4)

All of these

Sol. Answer (1)   Both E and B are sinusoidal, so over a complete cycle, its value will be zero. 15. An electromagnetic wave is propagating in vacuum along z-axis, the electric field component is given by Ex = E0 sin(kz – t), then magnetic component is (1) Bx =

E0 sin  kz – t  C

(2)

By =

B0 sin  kz – t  C

(3) By =

E0 sin  kz – t  C

(4)

By = B0C sin(kz – t)

Sol. Answer (3) Direction of propagation  Z E x  E 0 sin  kZ – t   B  will be in the y-direction

 By  B 0 sin  kZ – t  E0  B c 0 B0 

So, B y 

E0 c E0 sin  kZ – t  c

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electromagnetic Waves

115

16. The speed of electromagnetic wave in a medium (whose dielectric constant is 2.25 and relative permeability is 4) is equal to (1) 0.5  108 m/s

(2)

0.25  108 m/s

(3)

0.75  108 m/s

(4)

1  108 m/s

Sol. Answer (4) Er = 2.25, r = 4

V 

1  0 r  0  r 1 1   0 0 2.25  4

3  10 8 1.5  2 V = 108 m/s 

17. The magnetic field in a plane electromagnetic wave is given by = 2 × 10–7sin(0.5 × 103x + 1.5 × 1011t). This electromagnetic wave is (1) Visible light

(2)

Infrared

(3)

Microwave

(4)

(3)

[M L–1 T–2

(4)

[M L2 T–1

10–6 cm

(4)

10–7 cm

Sol. Answer (4) B  2  10 –7 sin  0.5  10 –3 x  1.5  10 11t 

f 

c , k  0.5  10 –3 

  1.5  10 11 V 

 1.5  10 11   3  10 14 k 0.5  10 –3

  f k

k

2 



2 2  22   1.25  10 4 m k 7  0.5  10 –3 Radio waves

18. The dimensional formula of

1  0 E 2 is 2

(E = electric field) (1) [M L T–1

(2)

[M L2 T–2

Sol. Answer (3) 1 E  2 –2   0E 2  Energy density   ML 3T 2 V L 

= ML–1T –2  19. Which of the following represents an infra-red wavelength? (1) 10–4 cm

(2)

10–5 cm

(3)

Sol. Answer (1) 10–6 m or 10–4 cm  Infrared Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

116

Electromagnetic Waves

Solution of Assignment (Set-2)

20. Which of the following is not transported by electromagnetic waves? (1) Energy

(2)

Momentum

(3)

Charge

(4)

Information

Sol. Answer (3) Only energy, momentum and information can be transferred with the help of em waves, not any matter. 21. Hertz experiment is used for (1) Production of electromagnetic wave

(2)

Detection of electromagnetic wave

(3) Both (1) & (2)

(4)

None of these

Sol. Answer (3) Hertz was the scientist who produced and detected the em waves. 22. Electromagnetic waves are produced due to (1) A charge at rest

(2)

A moving charge

(3) An accelerated charge

(4)

None of these

Sol. Answer (3) Em waves are produced by accelerated charged particles. 23. Ozone layer blocks the radiation of wavelength (1) Less than 4 × 10–7 m

(2)

Between 4 × 10–7 m to 8 × 10–7 m

(3) More than 8 × 10–7m

(4)

None of these

Sol. Answer (1) Radiation having wavelength less than 4  10–7 m are blocked by ozone layer 24. Which of the following can be used in cancer treatment? (1) X-rays

(2)

UV-rays

(3)

-rays

(4)

Both (1) & (3)

(4)

D>C>B>A

Sol. Answer (4) Both X-rays and -rays are used for the treatment of cancer. 25. The following can be arranged in decreasing order of wave number A. AM radio

B.

C. Microwave

D.

(3)

A>B>C>D

(1) A > B > D > C

(2)

C>D>B>A

SECTION - B Objective Type Questions 1.

E where the symbols have c their usual meanings and the energy in a given volume of space due to the electric field part is U, then the energy due to the magnetic field part will be

If the electric field and magnetic field of an electromagnetic wave are related as B =

(1)

U c

(2)

U c2

(3)

U 2

(4)

U

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electromagnetic Waves

117

B

E C

Energy due to E 

1  E2 2 0

B0 2 B  Energy due to 2 0

E part 

U 2

Bpart 

U 2

E part  B part

2.

The direction of poynting vector represents (1) The direction of electric field (2) The direction of magnetic field (3) The direction of propagation of EM wave (4) The direction opposite to the propagation of EM wave

Sol. Answer (3)    E  B S 0   E  B ⇒ direction of wave propagation 3.

A plane electromagnetic wave is incident on a plane surface of area A normally, and is perfectly reflected. If energy E strikes the surface in time t then average pressure exerted on the surface is (c = speed of light) (1) Zero

(2)

E Atc

(3)

2E Atc

(4)

E c

2I [Perfect reflection] c

  4.

2E Atc

P

2E Atc

5% of the power of 100 W bulb is converted to visible radiation. Average intensity of visible radiation at a distance of 10 m from the bulb is (1)

5 2(10 )

2

watt/m2

(2)

5 4(10)

2

watt/m2

(3)

5 (10)

2

watt/m2

(4)

5 8(10 )2

watt/m2

E P  At A

5  100 P 5   100  watt/m 2 2 2 4r 400 4 10 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

118 5.

Electromagnetic Waves

Solution of Assignment (Set-2)

Which of the following physical quantities contained in a small volume oscillates at double the frequency of passing electromagnetic wave? (1) Electric field

(2)

Magnetic field

(3)

Magnetic energy

(4)

All of these

Sol. Answer (3) The electric and magnetic energy oscillate at double the frequency as compared to electric and magnetic field. 6.

A capacitor is connected across a battery which delivers a current of 1 A at an instant in the capacitor. Displacement current through the capacitor at that instant is (1) 1 A

(2)

0A

(3)

2A

(4)

1 A 2

Sol. Answer (1) I=1A Id = 1 A 7.

The magnetic field in a plane electromagnetic wave is given by, B = 3.01 × 10–7 sin (6.28 × 102x + 2.2 × 1010t) T. [where x in cm and t in second]. The wavelength of the given wave is (1) 1 cm

(2)

628 cm

(3)

1.129 cm

(4)

314 cm

Sol. Answer (1)  B  3.01 10 –7 sin  6.28  10 2 x  2.2  10 10 t  T

k  6.28  10 2   2.2  1010

 8.

k

2 



2  10 –2 m 6.28  10 2

At a particular instant the current in the circuit given below is i. The displacement current between the plates of the capacitor shown below is

C i V (1) Zero

(2)

i

(3)

i 2

i 4

(4)

Sol. Answer (2) Displacement current id = i 9.

To establish an instantaneous displacement current of I ampere in the space between the plates of a parallel plate dV 1 farad, the value of is capacitor of dt 2 (1) 2I

(2)

I 2

(3)

1 2I

(4)

I

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electromagnetic Waves

119

A 0 dE d d dt

I

1 dE d 2 dt

I

1 dV 1  .d 2 dt d

dV  2I dt

10. A plane electromagnetic wave of frequency 28 MHz travels in free space along the positive x-direction. At a particular point in space and time, electric field is 9.3 V/m along positive y-direction. The magnetic field (in T) at that point is (1) 3.1 × 10–8 along positive z-direction

(2)

3.1 × 10–8 along negative z-direction

(3) 3.2 × 107 along positive z-direction

(4)

3.2 × 107 along negative z-direction

Sol. Answer (1) f = 28106 Hz E = 9.3 V/m   ˆj  Bc = E B

9.3 3  10 8

B = 3.1 × 10–8 along positive z-direction

SECTION - C Previous Years Questions 1.

The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong? [Re-AIPMT-2015] (1) -rays

(2)

X-rays

(3)

Infra-red rays

(4)

Ultraviolet rays

Sol. Answer (2) E = 15 keV



hc 6.6  1019  3  108  E 15  106  1.6  10–19

 = 0.8 Å  1 Å 2.

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = velocity of light) [AIPMT-2015] (1)

E C2

(2)

E C

(3)

2E C

(4)

2E C2

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

120

Electromagnetic Waves

Solution of Assignment (Set-2)

Sol. Answer (3) Surface is perfectly reflecting So change in momentum i.e., momentum transferred is p  p –  – p  p  2 p

So, p 

3.

2E C

Light with an energy flux of 25 × 104 Wm–2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm2, the average force exerted on the surface is [AIPMT-2014] (1) 1.25 × 10–6 N

(2)

2.50 × 10–6 N

(3)

1.20 × 10–6 N

(4)

3.0 × 10–6 N

Fav 

2 I A 2  25  10 4  15  10 4  N c 3  10 8

= 250 × 10–8 N = 2.5 × 10–6N 4.

The condition under which a microwave oven heats up a food item containing water molecules most efficiently is [NEET-2013] (1) The frequency of the microwaves has no relation with natural frequency of water molecules (2) Microwaves are heat waves, so always produce heating (3) Infra-red waves produce heating in a microwave oven (4) The frequency of the microwaves must match the resonant frequency of the water molecules

 The electric field associated with an e. m. wave in vacuum is given by E  iˆ 40 cos (kz – 6 × 108t), where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is [AIPMT (Prelims)-2012] (1) 6 m–1

(2)

3 m–1

(3)

2 m–1

(4)

0.5 m–1

The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to [AIPMT (Mains)-2012] (1) The speed of light in vacuum (2) Reciprocal of speed of light in vacuum (3) The ratio of magnetic permeability to the electric susceptibility of vacuum (4) Unity

E0 C B0 

B0 1  E0 C

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

7.

Electromagnetic Waves

121

The electric and the magnetic field, associated with an e.m. wave, propagating along the +z-axis, can be represented by [AIPMT (Prelims)-2011] (1)

  ⎡E  E0 jˆ, B  B0 kˆ ⎤ ⎣ ⎦

(2)

(3)

  ⎡E  E0 kˆ , B  B0 iˆ⎤ ⎣ ⎦

(4)

  ⎡E  E0 iˆ, B  B0 ˆj ⎤ ⎣ ⎦   ⎡E  E0 jˆ, B  B0 iˆ⎤ ⎣ ⎦

Sol. Answer (2)   If wave is propagating in +z direction then E and B will be in x-y plane.

  Also, E  B = direction of propagation   E  E 0 iˆ, B  B0 jˆ 8.

The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is [AIPMT (Prelims)-2011] (1) Infrared, microwave, ultraviolet, gamma rays

(2)

Microwave, infrared, ultraviolet, gamma rays

(3) Gamma rays, ultraviolet, infrared, microwaves

(4)

Microwaves, gamma rays, infrared, ultraviolet

Sol. Answer (2) Maximum wavelength = microwaves Minimum wavelength = -rays 9.

Which of the following statement is false for the properties of electromagnetic waves? [AIPMT (Prelims)-2010] (1) These waves do not require any material medium for propagation (2) Both electric and magnetic field vectors attain the maxima and minima at the same place and same time (3) The energy in electromagnetic wave is divided equally between electric and magnetic vectors (4) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave

Sol. Answer (4) 10. The electric field of an electromagnetic wave in free  space is given by E  10cos 107 t  kx jˆ V/m, where t and x are in seconds and metres respectively. It can be

inferred that (a) The wavelength  is 188.4 m (b) The wave number k is 0.33 rad/m (c) The wave amplitude is 10 V/m (d) The wave is propagating along +x direction Which one of the following pairs of statements is correct ? (1) (c) & (d)

(2)

(a) & (b)

(3)

[AIPMT (Mains)-2010] (b) & (c)

(4)

(a) & (c)

Sol. Answer (4) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

122

Electromagnetic Waves

Solution of Assignment (Set-2)

11. The electric field part of an electromagnetic wave in a medium is represented by ⎡⎛ N ⎛ 6 rad ⎞ 2 rad ⎞ ⎤ Ex = 0; E y =2.5 C cos ⎢ ⎜⎝ 2  10 s ⎟⎠ t  ⎜⎝   10 m ⎟⎠ x ⎥ ; ⎣ ⎦

Ez = 0. The wave is

[AIPMT (Prelims)-2009]

(1) Moving along x-direction with frequency 106 Hz and wavelength 100 m (2) Moving along x-direction with frequency 106 Hz and wavelength 200 m (3) Moving along –x-direction with frequency 106 Hz and wavelength 200 m (4) Moving along y-direction with frequency 2 × 106 Hz and wavelength 200 m Sol. Answer (2) Ex = 0 E y  2.5

N as  2  10 6 t –   10 –2 x  C

Ez = 0 f 

2  10 6 w   10 6 s –1 2 2



2 2   200 m .    10 –2

12. The velocity of electromagnetic radiation in a medium of permittivity 0 and permeability 0 is given by [AIPMT (Prelims)-2008]

(1)

0 0

(2)

0 0

(3)

0 0

(4)

1  0 0

C

1  0 0

13. The electric and magnetic field of an electromagnetic wave are

[AIPMT (Prelims)-2007]

(1) In opposite phase and perpendicular to each other (2)

In opposite phase and parallel to each other

(3) In phase and perpendicular to each other

In phase and parallel to each other

(4)

Sol. Answer (3) 14. If v, x and m represent the wavelengths of visible light, X-rays and microwaves respectively, then : [AIPMT (Prelims)-2005] (1) m > x > v

(2)

v > m > x

(3)

m > v > x

(4)

v > x > m

Sol. Answer (3) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electromagnetic Waves

123

15. For a medium with permittivity and permeability , the velocity of light is given by (1)

 

(2)



1 

(3)

 

(4)

Sol. Answer (3) Speed of light in vacuum C 

in any other medium, V 

1  0 0

1 

16. Which of the following electromagnetic radiations have the smallest wavelength? (1) X-rays

(2)

-rays

(3)

UV waves

(4)

Microwaves

Sol. Answer (2) -rays  minimum wavelength. 17. If 0 and 0 are the electric permittivity and magnetic permeability in a free space,  and  are the corresponding quantities in medium, the index of refraction of the medium is

(1)

 0 0 

(2)

  0 0

(3)

0   0

(4)

(3)

X-rays

(4)

 0

Sol. Answer (2) Refractive index =  V 

C V

1 

C

1  0 0



  0 0

18. What is the cause of “Greenhouse effect”? (1) Infra-red rays

(2)

Ultraviolet rays

Sol. Answer (1) Green house effect is mainly due to the infrared radiation, methane gas, SO2 etc. 19. The velocity of electromagnetic wave is parallel to   (1) B  E

(2)

  E B

(3)

 E

(4)

 B

  E  B  Direction of wave propagation   perpendicular to the plane of E and B . Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

124

Electromagnetic Waves

Solution of Assignment (Set-2)

SECTION - D Assertion-Reason Type Questions 1.

A : Different electromagnetic waves differ considerably in their mode of interaction with matter. R : Different electromagnetic waves have different wavelength or frequency.

A : All electromagnetic waves travel through vacuum with same speed but they have different wavelength or frequency. R : The wavelength of the electromagnetic waves is often correlated with the characteristic size of the system that produces and radiates them.

A : High frequency electromagnetic waves are detected by some means based on the physical effects they produce on interacting with matter. R : The oscillating fields of an electromagnetic wave can accelerate charges and can produce oscillating currents therefore, an apparatus designed to detect EM waves is based on this fact.

A : Infrared waves are often called heat waves. R : Infrared waves vibrate not only the electrons, but entire atoms or molecules of a substance which increases the internal energy and temperature of the substance.

A : The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. R : Humans have evolved with visions most sensitive to the strongest wavelength from the sun.

A : Long distance radio broadcasts use short-wave bands. R : Ionosphere reflectes waves in these bands.

A : It is necessary to use satellites for long distance TV transmission. R : Television signals are not properly relfected by the ionosphere therefore, relfection is effected by satellites.

A : Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. R : Atmosphere absorbs X-rays, while visible and radiowaves can penetrate it.

A : If the earth did not have an atmosphere, its average surface temperature would have been lower. R : In the absence of atmosphere, the green house effect will be absent.

Sol. Answer (1) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

Electromagnetic Waves

125

126

Electromagnetic Waves

Solution of Assignment (Set-2)

21. A : Light waves can be polarised. R : All electromagnetic waves move with same speed in vacuum. Sol. Answer (2) 22. A : In an electromagnetic wave the energy density in electric field is equal to energy density in magnetic field. R : Electromagnetic waves are transverse in nature. Sol. Answer (2)

   E B represents the instantaneous intensity at a point. 23. A : The Poynting vector given as S  0   R : The velocity of an electromagnetic wave is in the direction of the vector E  B .

Sol. Answer (2) 24. A : The radiation pressure due to light waves is maximum when the surface is a perfect reflector. R : The momentum transfer by the photons to a perfectly reflecting surface is maximum. Sol. Answer (1) 25. A : In a material medium the speed of a particle can be more than the speed of light in that medium. R : In the phenomenon of green house effect, low wavelength radiation is allowed to pass but high wavelength radiation is not allowed to pass. Sol. Answer (2)







Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456