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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Give a simple possible argument to suggest that an accelerated charge must emit electromagnetic radiation. |
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Answer» An electric charge can produce electric as well as magnetic field which very with time and are perpendicular to each other. These time varying fields, mutually perpendicular to each other will produce a disturbance which can travel through a space in the form of e.m. wave. |
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| 152. |
An electromagnetic radiation has an energy of 13.2 keV. Then the radiation belongs to region ofA. visible lightB. ultravioletC. infraredD. X-ray |
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Answer» Correct Answer - D Given: `E = 13.2 keV` `lambda("in" Å) = (hc)/(E("in"eV)) = (12400)/(13.2 xx 10^(3)) = 0.939 Å ~~ 1Å` X-rays cover wavelengths ranging from about `10^(-8)m (10 nm)` to `10^(-13)m (10^(-4)nm)`. An electromagnetic radiation of energy 13.2 keV belongs to X-ray region of electromagnetic spectrum. |
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| 153. |
Which of the following statement about e.m. wave is /are / correctA. 1 and 2B. 2 and 3C. 1,2 and 3D. 2 only |
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Answer» Correct Answer - d The energy of a photon, `E=hc//lambda`, since `lambda` for visible light is more than that of x-rays, hence energy of x-rays photon is more than that of light. |
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| 154. |
The energy of the electromagetic wave is of the order of `15` keV. To which part of the spectrum dose it belong?A. `gamma`-raysB. x-raysC. Infrared raysD. Ultra-violet rays |
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Answer» Correct Answer - b `E=15KeV =15000eV=15000xx1.6xx10^(-19)J` Wavelength `lambda=(hc)/E=(6.64xx10^(-34)xx3xx10^8)/(1500xx1.6xx10^(-19)J)` `=0.083xx10^-9m~~10^-1nm` Since,` lambda` for x-rays is in between 1nm to `10^-3nm`, so the given radiation is x-rays. |
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| 155. |
The energy of the electromagetic wave is of the order of `15` keV. To which part of the spectrum dose it belong?A. `gamma`-raysB. `X`-raysC. Infrared raysD. Ultraviolet rays |
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Answer» Correct Answer - b `E=15` ke V `lambda =(hc)/(E)=(6.6xx10^(-19)xx3xx10^(8))/(15xx10^(6)xx1.6xx10^(-19))` `lambda=0.8Å ~~1Å` |
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| 156. |
Assertion : Electromagnetic waves exert radiation pressure Reason : Electromagnetic waves carry energy.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - A Electromagnetic waves have linear momentum as well as energy. This concludes that they can exert radiation pressure by falling beam of electromagnetic radiation on an object. |
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| 157. |
The waves used by artificial satellites for communication isA. microwavesB. infrared wavesC. radio wavesD. X-rays |
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Answer» Correct Answer - A In artificial satellite microwaves are used for communication. |
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| 158. |
Match Column I (electromagnetic wave type) with Column II (its application) and select the correct option from the choices given below the columns. |
| Answer» Correct Answer - `A rarr p, B rarr q, C rarr r, D rarr s` | |
| 159. |
The electric field of an electromagnetic wave travelling through vaccum is given by the equation ` E = E_(0) "sin"(kx - omegat)` The quantity that is independent of wavelength isA. `komega`B. `(k)/(omega)`C. `k^(2)omega`D. `omega` |
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Answer» Correct Answer - B Here, `k=(2pi)/(lambda),omega=2piupsilon` `therefore" "(k)/(omega)=(2pi//lambda)/(2piupsilon)=(1)/(upsilon lambda)=(1)/(c)" "(thereforec=ulambda)` where c is the speed of electromagnetic wave in vacuum. It is a constant whose value is `3xx10^(8)"m s"^(-1)`. |
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| 160. |
In a plane electromagnetic wave the ainusoidal electrical oscillations have a lre-quency of `5xx10^(10)`Hz and amplitude `48Vm^(-1).` Calculate (a) its wavelength (b) the amplitude of the oscillating magnetic field. |
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Answer» Correct Answer - A::C |
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| 161. |
When a capacitor of capacitance C after charging with a charge Q is connected to inductor of self inductance L, the oscillation of charge takes place with time between the two plates of capacitor. If one plate of capacitor is connected to antenna and other plate is earthed, then em wave are produced, which are sinusoidal variation of electric and magnetic field vectors, perpendicular to each other as well as perpendicular to the direction of propagation of wave. The velocity of these waves depends upon the electric and magnetic properties of the medium. the em wave were produced experimentally by Hertz in 1888 using Hertz Oscillator, which were of wavelength 6m. Jagdish chander bose in 1895 produced these waves which were of wave length 5mm to 25mm and in 1896, G. Marconi established a wireless communication between two stations 50km apart using em waves. In an em wave, the amplitude of electric field is `10Vm^-1`. The frequency of wave is `5xx10^(14)Hz .` the wave is propagating along z-axis. In em wave, the average energy density due to magnetic field isA. `8.85xx10^(-10)Jm^-3`B. `4.42xx10^(-10)Jm^-3`C. `2.21xx10^(-10)Jm^-3`D. `6.63xx10^(-10)Jm^-3` |
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Answer» Correct Answer - c Average energy density due to magnetic field in em wave is `u_B=1/2 (B_(rms)^2)/(mu_0) =1/(2mu_0) ((B_0)/(sqrt2))^2=1/4 (B_0^2)/(mu_0)` `=1/4 ((E_0//c^2)/(1//mu_0epsilon_0))=1/4 in_0 E_0^2` `=1/4xx(8.85xx10^(-12))xx10^2=2.21xx10^(-10)Jm^-3` |
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| 162. |
When a capacitor of capacitance C after charging with a charge Q is connected to inductor of self inductance L, the oscillation of charge takes place with time between the two plates of capacitor. If one plate of capacitor is connected to antenna and other plate is earthed, then em wave are produced, which are sinusoidal variation of electric and magnetic field vectors, perpendicular to each other as well as perpendicular to the direction of propagation of wave. The velocity of these waves depends upon the electric and magnetic properties of the medium. the em wave were produced experimentally by Hertz in 1888 using Hertz Oscillator, which were of wavelength 6m. Jagdish chander bose in 1895 produced these waves which were of wave length 5mm to 25mm and in 1896, G. Marconi established a wireless communication between two stations 50km apart using em waves. In an em wave, the amplitude of electric field is `10Vm^-1`. The frequency of wave is `5xx10^(14)Hz .` the wave is propagating along z-axis. Sun also sends em wave to earth. which one of em wave out of the visible portion, from sun will be reaching the surface of earth earlier than others:A. violet wavesB. green wavesC. yellow wavesD. red waves |
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Answer» Correct Answer - d `mu=c/v or v=c/mu=and mu=A+B/(lambda^2)+C/(lambda^4)` As `lambda_r gt lambda_v` so `mu_r lt mu_v`, therefore `v_r gt v_v` |
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| 163. |
When a capacitor of capacitance C after charging with a charge Q is connected to inductor of self inductance L, the oscillation of charge takes place with time between the two plates of capacitor. If one plate of capacitor is connected to antenna and other plate is earthed, then em wave are produced, which are sinusoidal variation of electric and magnetic field vectors, perpendicular to each other as well as perpendicular to the direction of propagation of wave. The velocity of these waves depends upon the electric and magnetic properties of the medium. the em wave were produced experimentally by Hertz in 1888 using Hertz Oscillator, which were of wavelength 6m. Jagdish chander bose in 1895 produced these waves which were of wave length 5mm to 25mm and in 1896, G. Marconi established a wireless communication between two stations 50km apart using em waves. In an em wave, the amplitude of electric field is `10Vm^-1`. The frequency of wave is `5xx10^(14)Hz .` the wave is propagating along z-axis. If `mu_0,mu_r, in_0` and `in_r` as the absolute permeabilty, relative permeability , absolute permitivity and relative permitivity of the medium, then the velocity of em wave in a medium isA. `1/(sqrt(mu_0in_0))`B. `1/(sqrt(mu_rin_r))`C. `1/(sqrt(mu_0in_rmu_rin_r))`D. `sqrt((mu_rin_r)/(mu_0in_0))` |
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Answer» Correct Answer - c Velocity of em wave, `v=1/(sqrt(mu in))=1/(sqrt(mu_0 in_0 mu_0 in_r))` |
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| 164. |
The force exerted by electromagnetic wave on unit area of the surface is called......... |
| Answer» Correct Answer - radiation pressure | |
| 165. |
If a portion of elctromagnetic wave of energy U is propaging with speed c, then linear momentum of the electromagnetic wave........... |
| Answer» Correct Answer - `U//c` | |
| 166. |
The cross product `vecExxvecB` (where `vecE` =electric field vector and `vecB` is the magnetic field vector ) always gives the.............. of electromagnetic wave. |
| Answer» Correct Answer - direction of propagation | |
| 167. |
Show that the average energy density of the electric field `vecE` equals the average energy density of the magnetic field `vecB`, in electromagnetic waves. |
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Answer» Average energy density of electric field is given by, `u_E=1/2in_0E_(rms)^2=1/2in_0((E_0)/(sqrt2))^2=1/4in_0E_0^2....(i)` Average energy density of magnetic field is given by, `u_B=1/(2mu_0) B_(rms)^2 =1/(2mu_0)((B_0)/(sqrt2))^2=1/4 (B_0^2)/(mu_0).....(ii)` We know `E_0=cB_0 or B_0 =E_0//c` Also `c=1/sqrt(mu_0in_0)` Putting values in (ii) , we get `u_B= 1/(4mu_0) (E_0^2)/(c^2) =1/(4mu_0) (E_0^2)xxmu_0 in_0=1/4 in_0E_0^2=u_E` |
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| 168. |
Light can travel in vacuum whereas sound can not do so. Why? |
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Answer» Light being electromagnetic wave do not require any material medium for its propagation. Hence light can travel in vacuum. On the other hand sound is a mechanical wave. It requires a material medium for its propagation. Hence sound can not travel in vacuum. |
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| 169. |
In an apparatus the electric field was found to oscillate with an amplitude of `18` `V//m`. The magnitude of the oscillating magnrtic field will beA. `4xx10^(-6) T`B. `6xx10^(-8) T`C. `9xx10^(-9) T`D. `11xx10^(-11) T` |
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Answer» Correct Answer - b Here, `E_(0)=18 V//m, B_(0)=?` `:. B_(0)=(E_(0))/(c)=(18)/(3xx10^(8))=6xx10^(-8) T` |
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| 170. |
In an apparatus the electric field was found to oscillate with an amplitude of `18` `V//M`. The magnitude of the oscillating magnrtic field will beA. `4xx10^(-6)T`B. `6xx10^(-8)T`C. `9xx10^(-9)T`D. `11xx10^(-11)T` |
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Answer» Correct Answer - b `c=(E)/(B)rArr B=(E)/(C)=(18)/(3xx10^(8))=6xx10^(-8)T` |
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| 171. |
The ozone layer absorbs |
| Answer» Correct Answer - Ultraviolet rays | |
| 172. |
Which part of electromagnetic spectrum has largest penterating power. |
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Answer» Gamma rays (frequency range `5xx10^23 Hz` to `3xx10^18Hz`) has largest penetrating power. |
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| 173. |
Why does microwave oven heats up a food item containing water molecules most efficiently? |
| Answer» Microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules. | |
| 174. |
Why dose microwave oven heats up a food item containing water molecules most efficiently? |
| Answer» Because the frequency of the micro-wave matches the resonant frequency of water molecules. | |
| 175. |
The condition under which a microwave oven heats up a food item containing water molecules most efficiently is:A. Infra-red wave produce heating in a microwave ovenB. The frequency of the microwave must match the resonant frequency of the water moleculesC. The frequency of the microwave has no relation with natural frequency of water moleculesD. Microwave are heat waves, so they always produce heating |
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Answer» Correct Answer - b For efficient working of microwave oven, the frequency of microwave must match the resonant frequency of the water molecules, then maximum heating is produced. |
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| 176. |
The charge on a parallel plate capacitor varies as `=q_0 cos 2pi ft`. The plates are very large and close together (area=a,separation=d). Neglecting the edge effects, find the displacement current through the capacitor. |
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Answer» Displacement current, `I_D`=conduction current `I_C`. `:. I_D=I_C=(dq)/(dt)=d/(dt)[q_0 cos 2pivt]=-q2piv sin 2pivt.` |
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| 177. |
The charge on a parallel plate capacitor varies as `q = q_(0) "cos" 2 pit`. The plates are very large and close together (area = A, separation = d). The displacement current through the capacitor isA. `q_(0)2pi upsilon sin pi upsilont`B. `-q_(0)2pi upsilon sin 2pi upsilont`C. `q_(0)2pi sin pi upsilont`D. `q_(0)pi upsilon sin2pi upsilont` |
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Answer» Correct Answer - B Displacement current, `I_(D)` : conduction current, `I_(C)` `therefore" "(dq)/(dt)=(d)/(dt)[q_(0)cos 2piupsilont]=-q_(0)2piupsilon sin 2pi upsilont` |
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| 178. |
A parallel plate capacitor is made out of two rectangular metal pates of sides `30 cmxx15 cm` and separated by a distance of 2.0 mm. The capcitor is charged in such a way that the charging current has a constant value of 100mA. What must be the rate of change of potential of the charging source to ensure this and what will be the displacement current in the region between the capacitor plates? |
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Answer» Correct Answer - `5xx10^8Vs^-1 0.1A` `I=(dQ)/(dt)=(d(CV))/(dt)=(CdV)/(dt)=((in_0A)/d)(dV)/(dt)` or `(dV)/(dt)=(Id)/(in_0A)` or `(dV)/(dt)=(0.1xx(2xx10^-3))/((8.85xx10^(-12))xx(30xx15xx10^-4))` `=5xx10^8Vs^-1`, `I_d=I=0.1A` |
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| 179. |
Find the value of magnetic field between pates capacitor at distance `1m` from center, where electric field varies by `10^(10) V//m` per second.A. `5.56xx10^(-8)T`B. `5.56xx10^(-3)T`C. `5.56 muT`D. `5.56T` |
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Answer» Correct Answer - a `B=(mu_(0)epsilon_(0)^(r))/(2) (dE)/(dt) =(1)/(9xx10^(16)xx2)xx10^(10)` `=5.56xx10^(-8)T` |
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| 180. |
An antenna is a deviceA. That converts electromegnetic energy into radio frequency signalB. That covert radio frequency signal into electromagnetic energyC. That convert guided electromagnetic wave into free space electromagnetic wave and vice-vesaD. None of these |
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Answer» Correct Answer - c An antenna is a metallic structure used to radiate or recxevie `EM` wave. |
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| 181. |
The electromagnetic waves which are used in the working of solar water heaters and cookers are called......... |
| Answer» Correct Answer - infrared waves | |
| 182. |
The electromagnetic waves detected using a thermopile and used in physical therapy areA. gamma radiationsB. X-raysC. ultraviolet radiationsD. infrared radiations |
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Answer» Correct Answer - D The electromagnetic waves detected using a thermopile and used in physical therapy are infrared radiation for treating muscular strain. |
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| 183. |
A parallel capacitor made of circular plates radius 10.0 cm has a capacitance of 200 pF. The capacitor is connected to a 200 a.c. supply with an angular frequency of `200 "rad"" "s^(-1).` (i) What is the r.m.s. value of conduction current (ii) Is the conduction current equal to displacement current? (iii) Find peak value of displacement current ? (iv) Determine the amplitude of magnetic field at a point 2.0cm from the axis between the plates. |
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Answer» Correct Answer - A::B::C::D |
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| 184. |
A small metallic ball is charged positively and negatively in a sinusoidally manner at a frequency of `10^6cps.` The maximum charge on the ball is `10^-6C`. What is the displacement current due to the alternating current?A. `6.28A`B. `3.8A`C. `3.75xx10^-4A`D. `122.56A` |
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Answer» Correct Answer - a Here, `v=10^6 cps, q_0=10^-6C, I_D=?` As `I_D=(dq)/(dt)=d/(dt)(q_0sin omegat)=q_0omegat cos omegat` for `I_D` to be maximum, `cos omegat=1` So, maximum value `I_D=q_0omega=q_0(2piv)` `=10^6xx2xx3.14xx10^6=6.28A` |
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| 185. |
Assertion The frequency of the electromagnetic wave naturally equals equals the frequency of oscillations of the charge. Reason The energy associated with the propagating wave comes at the expense of the energy of the source.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B The oscillating electric and magnetic fields, thus regenerate each othe, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. The energy associated with the propagating wave comes at the expense of the energy of the source the accelerated charge. |
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| 186. |
An electromagnetic wave can be produced, when charge isA. moving with a constant velocityB. moving in a circular orbitC. falling in an electric fieldD. both (b) and (c) |
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Answer» Correct Answer - D An electromagnetic wave can be produced by accelerated or oscillating charge. In options (b) and (c), the charge is in accelerated state, hence it will be a source of electromagnetic waves. |
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| 187. |
A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency? |
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Answer» Current through capacitor, `I_(~)=E_(~)/(X_C)=(E_(~))/(1//omegaC)=omegaCE_(~) or I_(~)propv`. therefore, decrease in frequency v of a.c. source decrease the conduction current. As displacement current =conduction current, hence decrease in v decrease displacement current in circuit. |
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| 188. |
A variable frequency a.c. source is connected to a capacitor. Will the displacement current increase of decrease with increase in frequency? |
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Answer» The increase in frequency of a.c., will decrease the reactance of the capacitor `[X_c=1//(2pivC)]` and hence will increase the conduction current. Since the displacement current is equal to the conduction current, therefore, the displacement current will increase with the increase in frequency of a.c. |
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| 189. |
If a variable frequency ac source is connected to a capacitor then with decrease in frequency the displacement current willA. increaseB. decreaseC. remains constantD. first decrease then increase |
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Answer» Correct Answer - B Current through capacitor, `I=(E)/(X_(C))=(E)/((1)/(omegaC))=omegaCE=2piupsilonCEor Ipropupsilon.` `therefore` decrease in frequency `upsilon` of ac source decreases the conduction current. As displacement current is equal to conduction current, decrease in `upsilon` decreases displacement current in circuit. |
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| 190. |
Wave which cannot travel in vacumm isA. X-raysB. InfrasonicC. UltraviloetD. Radiowaves |
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Answer» Correct Answer - b Infrasoic waves are mechanical waves. |
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| 191. |
Find the photon energy in (i) calories (ii) watt hour (iii) electron volt, for e.m. wave of wavelength `300 mu m`. Give, `h=6.6xx10^-34Js.` |
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Answer» Correct Answer - (i) `1.6xx10^(-22) cal. (ii) 1.83xx10^(-25)Wh (iii) 4.125xx10^-3eV` `lambda=300 mum=300xx10^-6m=3xx10^-4m`, `E=(hc)/lambda=(6.6xx10^(-34)xx3xx10^8)/(300xx10^-6)` `=6.6xx10^(-22)J` `=(6.6xx10^(-22))/4.2 cal=(6.6xx10^(-22))/(60xx60)` watt hour `=(6.6xx10^(-22))/(1.6xx10^(-19))` electron-volt. |
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| 192. |
Consider telecommunication through optical fibres. Which of the following statements is not true?A. Optical fibre may have homogeneous core with a suitable claddingB. Optical fibre can be of graded refractive indexC. Optical fibre are subject to electromagnetic interference from outsideD. Optical fibres have extremely low transmission loss |
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Answer» Correct Answer - c Optical fibres are not subjected to electromagnetic interefrence from outised. |
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| 193. |
Assertion: Ultraviolet radiation are of higher frequency waves are dangerous to huhman beaing. Reasion: Ultraviolet radiation are absorbed by the atmosphereA. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B The mavelength of these wave ranges between `4000Å` to `100Å` that is smaller wavelength and higher frequency. They are absorbed by atmosphere and covert oxygen into ozone. They cause skin diseases and they are harmful to eye and cause permanent blindess. |
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| 194. |
Obtain the temperature ragnes for ultraviolet part of radiation of e.m.waves. Use the formulae `lambda_m T=2.9xx10^-3mK`. Take frequency of ultraviolet part of radiations as `8xx10^14Hz` to `5xx10^17Hz`. |
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Answer» The corresponding wavelength to the frequency `8xx10^14Hz` is `lambda_1=c/(v_1)=(3xx10^8)/(8xx10^14)=3.75xx10^-7m` The corresponding wavelength to the frequency `5xx10^17` is `lambda_2=c/(v_2)=(3xx10^8)/(5xx10^17)=6xx10^-10m` As `lambda_mT=2.9xx10^-3 or T=(2.9xx10^-3)/(lambda_m)` For, `lambda_1=3.75xx10^-7m,` `T_1=(2.9xx10^-3)/(3.75xx10^-7)=7.73xx10^3K` For, `lambda_2=6xx10^-10m,` `T_2=(2.9xx10^-3)/(6xx10^-10)=4.83xx10^6K` Temperature range is `7.73xx10^3K to 4.83xx10^6K` |
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| 195. |
In an amplitude modulated wave for audio frequency of `500 "cycle"//second`,the appropriate carrier frequency will beA. 50 cycles/secB. 100 cycles/secC. 500 cycle/secD. 50,000 cycle/sec |
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Answer» Correct Answer - d Carrier frequency `gt` audio frequency |
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| 196. |
Assertion: Light can travel in vacuum but sound cannot do so. Reason: Light is an em wave and sound is a mechanical wave.A. Both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. Both Assertion and Reason are true, but Reason is not true explanation of the Assertion.C. Assertion is true, but the Reason is false.D. both Assertion and Reason is false. |
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Answer» Correct Answer - a Both Assertion and Reason are correct and Reason is the correct explanation of Assertion. |
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| 197. |
Out of the following options which one can be used produce a propagating electromagnetic wave?A. A charged moving at constant velocityB. A stationary chargeC. A chargeless particleD. An accelerating charge |
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Answer» Correct Answer - d Accelerating charge produce electromagnetic wave. |
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| 198. |
Out of the following options which one can be used produce a propagating electromagnetic wave?A. A stationary chargeB. A chargeless particleC. An accelerating chargeD. A charge moving at constant velocity |
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Answer» Correct Answer - C Electric charge at rest has electric field in the region around it, but no magnetic field. Whereas,a moving charge produces both the electric and magnetic fields. If a charge is moving with a constant velocity, then the electric and magnetic fields will not change with time, hence no EM wave will be produced. But, if the change is moving with a non-zero acceleration, then both electric and magnetic filed will change with space and time, it then produces EM wave. |
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| 199. |
The electric vector vibration of an electromagnetic wave is given by `E = (50NC^(-1)) sin omega(t-(x)/(c))`, The intensity of the wave isA. `2.3 Wm^(-2)`B. `4.3 Wm^(-2)`C. `3.3 Wm^(-2)`D. `1.8 Wm^(-2)` |
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Answer» Correct Answer - C The intensity is, `I = 1/1 epsi_(0)E_(0)^(2)c` `= 1/2 (8.85 xx 10^(-12)N-m^(2)C^(-2))xx (50 NC^(-1))^(2) xx 3 xx 10^(8)ms^(-1)` `=3.3 wm^(-2)` |
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| 200. |
The frequencies of `X`-rays, `gamma`-rays and ultraviolet rays are respectively `a, b and c` .ThenA. `p gt q, gt r`B. `p lt q, q gt r`C. `p lt q, q lt r`D. `p gt q, g lt r` |
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Answer» Correct Answer - B Increasing order of wavelength as given below `gamma`-rays `lt "X-rays" lt "ultraviolet rays"` Frequency order, `gamma-"ray" gt "X-rays" gt "ultravilot rays"` `rArr q gt p gt r` Thus, `q gt p` `q gt r` ` p gt r` |
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