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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.
| 101. |
Is the steady electric current the only source of magnetic field? Justify your answer. |
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Answer» No, the displacement current also produces magnetic field between the two plates of capacitor during charge or discharging of capacitor. |
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| 102. |
Why is the quantity `in_0(dphi_E)/(dt)` called the displacement current? Where `dphi_E//dt` is the rate of change of electric flux linked with a region or space. |
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Answer» The quntitity `in_0 (dphi_E)/(dt)` has the dimension of current and this currents exists in a region between the two plates of capacitor when displacement of charges occurs there. i.e., during charging or discharging of capacitor. |
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| 103. |
During Diwali festival, Rajender brought a new microwave oven in his house and told his wife Sarika to use the same carefully. He also told her that microwave oven is to used for slow heating of the vegetables and food articles upto moderate temperature as that will preserve the from the food. Further, the vegetables or flood items to be heated in oven are to be kept in a porecelain vessel and not in a matallic vessel. Read the above paragraph and answer and following questions: (i) What is the basic principle of working of microwave oven? (ii) Why is it advised to use porcelain vessel for heating the food items in microwave oven? (iii) What basic values do you learn from his study? |
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Answer» (i) The basic principle of working of microwave oven is to create microwave radiations of siutable frequency in the working space of oven, which may match the resonant frequency fall on food items containing water molecules (like, fruits, vegetables,cereals etc.) placed in oven, the water molecules absorb these radiations. As a result of it, the energy of water molecules increases. These molecules share their energy with neighbouring molecules. Due to it, the entire gets heated. (ii) In microwave oven, a porcelain container is used for heating the food articles in the oven. It is so becuase the size of the molecules of porcelain container is large and they vibrate and rotate with much smaller frequency than that of microwave and hence can not absorb microwaves. Due to it, the porcelain container remains at relatively lower temperature. (iii) We find that in a microwave oven, the slow and low heating of food will preserve the food values but fast and intense heating will destroy the food values. The same is true in day to day life. Sudden rises may be harmful. Instead, steady rises are appricable. |
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| 104. |
A radio can tune into any station in the `7.5 MHz` to `12 MHz` band. What is the corresponding wavelength of band? |
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Answer» Correct Answer - A::B::D `lambda=c//f` `lambda_1=(3xx10^8)/(7.5xx10^6)=40m` `lambda_2=(3xx10^8)/(12xx10^6)=25m`. |
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| 105. |
A radio can tune to any station in 7.5 MHz to 12MHz band. The corresponding wavelength band isA. 40 M to 25 mB. 30 m to 25 mC. 25 m to 10 mD. 10 m to 5 m |
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Answer» Correct Answer - A Here, `upsilon_(1)=7.5"MHz," upsilon_(2)="12 MHz"` `therefore" "lambda_(1)=(c)/(upsilon_(1))=(3xx10^(8))/(7.5xx10^(6))=40m` and `lambda_(2)=(c)/(upsilon_(2))=(3xx10^(8))/(12xx10^(6))=25m` |
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| 106. |
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is `B_0=510 nT`. What is the amplitude of the electric field part of the wave? |
| Answer» Here, `B_0=510 nT=510xx10^-9T, E_0=cB_0=3xx10^8xx510xx10^-9=153NC^-1` | |
| 107. |
A charged particle oscillates about its mean equilibrium position with a frerquency of `10^9 H_z`. The electromagnetic waves produced.A. will have frequency of `10^9 H_z`B. will have frequency of `2xx10^9 H_z`C. will have a wavelength of `0.3 m`D. fall in the region of radio waves. |
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Answer» Correct Answer - A::C::D `lambda=c/f=(3xx10^8)/(30xx10^6)=10 m` This waveslength lies in radio waves region. |
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| 108. |
The source of electromagnetic waves can be a charge.A. moving with a constant velocityB. moving in a circular orbitC. at restD. falling in an electric field |
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Answer» Correct Answer - B::D Here in option (b) charged is moving in a circular orbit. In circular motion , the direction of the motion of charge is changing continuously , thus it is an accelerated motion and this option is correct. Also we know that a charged starts accelerating when it falls in an electric field. |
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| 109. |
A charged particle oscillates about its mean equilibrium position with a frerquency of `10^9 H_z`. The electromagnetic waves produced. |
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Answer» The frequency of electromagnetic wave is the same as that of oscillating charged particle about its equilibrium position, which is `10^9Hz`. |
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| 110. |
A charged particle oscillates about its mean equilibrium position with a frerquency of `10^9 H_z`. The electromagnetic waves produced.A. will have frequency of `10^9Hz`B. will frequency of `2xx10^9Hz`C. will have a wavelength of 0.3mD. fall in the region of radiowaves |
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Answer» Correct Answer - (a,c,d) Here, `v=10^9Hz, lambda=c//v=3xx10^8//10^9=0.3m`. This wavelength radiation (=0.3m) or frequency radiation `10^9Hz` falls in the region of radio waves. Thus, options (a) , (c) and (d) are correct. |
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| 111. |
A charged particle oscillates about its mean equilibrium position with a frerquency of `10^9 H_z`. The electromagnetic waves produced.A. will have frequency of `10^(9)` HzB. will have frequency of `2xx 10^(9)` HzC. will have wavelength of 0.3 mD. fall in the region of radiowaves |
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Answer» Correct Answer - A::B::C Given , frequency by which the charged particles oscillates about its mean equilibrium position `=10^(9)Hz` So frequency of electromagnetic waves produced by the charged particle is `v=10^(9) Hz.` Wavelength `lambda =(C)/(v) =(3xx 10^(8))/(10^(9)) =0.3m` Also, frequency of `10^(9)` Hz fall in the region of rediowaves. |
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| 112. |
Suppose that the electric field of an electromagnetic wave in vacuum is `E={(3.0 N //C) cos [1.8 rad//m) y +(5.4 xx 10^6 rad // s)t]}hat i` (a) What is the direction of propagation of wave? (b) What is the wavelength lambda ? (c) What is the frequency `f` ? (d) What is the amplitude of the magnetic field of the wave (e) write an expression for the magnetic field of the wave. |
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Answer» (a) From the knowledge of wave we can see that electromagnetic wave is travelling along negative y- direction, as `omega t` and ky both are positive. (b) `k=1.8 rad//m` `k = (2pi)/(lambda)` `lambda = (2pi)/k=(2pi)/1.8=3.5 m` (c) `omega =5.4xx10^6 rads//s` `omega=2pif` `f = omega/(2pi)=(5.4xx10^6)/(2pi)` `8.6xx10^5 Hz` (d) `E_0=3.0 N//C` From the relation, `c = (E_0)/(B_0)` We have, `B_0=(E_0)/c=3.0/(3.0xx 10^8)` `=10^8 T` (e) `E` is along hati direction, wave is travelling along negative y-direction. Therefore, oscillations of `B` are along z- direction or `B = (10^-8 T) cos [(1.8 rad//m)y + (5.4 xx 10^6 rad//s)t ] hat k`. |
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| 113. |
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10 H_z` and amplitude `48V_m^-1` (a) What is the wavelength o f the wave? (b) What is the amplitude of the oscillating magnetic field. (c) Show that the average energy density of the field `E` equals the average energy density of the field `B.[c=3xx10^8 ms^-1]`. |
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Answer» Here, `v=2.0xx10^10Hz,E_0=48Vm^-1, c=3xx10^8ms^-1` (a) Wavelength of the wave, `lambda=c/v =(3xx10^8)/(2.0xx10^10)=1.5xx10^-2m` (b) Ampllitude of the oscillating magnetic field, `B_0=(E_0)/c=48/(3xx10^8)=1.6xx10^-7T` |
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| 114. |
The electric field part of an electromagnetic wave in vacuum is `E=3.1(N)/(C)cos[(1.8("rad")/("m"))y+(5.4xx10^(8)("rad")/(s))t]hati` The wavelength of this part of electromagnetic wave isA. 1.5 mB. 2 mC. 2.5 mD. 3.5 m |
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Answer» Correct Answer - D Given `E=3.1 "N C"^(-1)xxcos[(1.8"rad m"^(-1))y+(5.4xx10^(8)"rad s"^(-1)t)]hati" ….(i)"` Camparing (i) with the equation `E=E_(0)cos(ky+omegat)" …(ii)"` We get, `k=1.8"rad m"^(-1)`, `E_(0)=3.1"N C"^(-1),c=3xx10^(8)"m s"^(-1)`, `omega=5.4xx10^(8)"rad s"^(-1)` Now, `lambda=(2pi)/(k)=(2xx22)/(1.8xx7)=3.5m` |
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| 115. |
Suppose that the electric field part of an electromagnetic wave in vacuume is `E=3.1N//C cos [(1.8rad//m)y+(5.4xx10^8rad//s)t]hati` (a) Wavelength is the direction of motion? (b) What is the wavelength `lambda`? (c) What is the frequency v? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. |
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Answer» (a) From the given equation, it is clear that the direction of motion of e.m. wave is along negative y direction i.e., along `-hatj`. (b) Comparing the given equation with the equation `E=E_0cos (ky+omegat)`, we have `k=1.8rad//s, omega=5.4xx10^8rad//s, E_0=3.1 N//C` `lambda=(2pi)/k=(2xx(22//7))/1.8=3.492m~~3.5m` (c) `v=omega/(2pi) =(5.4xx10^8)/(2xx(22//7))=8.5xx10^6~~86MHz` (d) `B_0=(E_0)/c=3.1/(3xx10^8)=1.03xx10^-8T~~10.3nT` (e) `B=B_0cos (ky+omegat)hatk=(10.3nT)cos [(1.8rad//m)y+(5.4xx10^8rad//s)t]hatk` |
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| 116. |
Who discovered ultraviolet rays? Give their frequency range and mention at least two uses. |
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Answer» The ultraviolet rays were discovered by Ritter. The frequncy range of ultraviolet rays is `8xx10^(14)Hz` to `3xx10^(17)Hz.` Ultraviolet rays are used (i) to preserve the food stuff (ii) in burgular alarm. |
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| 117. |
In which of the following remote sensing technique is not used?A. Forest densityB. PollutionC. Wetland mappingD. Medical treatment |
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Answer» Correct Answer - d Remote sensing is the technique to collect information about an object in respect of its size, colour, nature, location, temperature etc. without physically touching it. There are some areas or location which are inaccessible. So to explore these areas or locations, a technique known as remote sensing is used. Remote sensing is done through a satellite. |
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| 118. |
Why are microwaves considered suitable for radar system used in aircraft nevigation? |
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Answer» Microwaves are of smaller wavelength than radiowaves. They can pass through the atmosphere without any deviation. |
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| 119. |
Find the wavelength of electromagnetic waves of frequency `4xx10^9Hz` in free space. Gives its two applications. |
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Answer» `lambda=c/v =(3xx10^8)/(4xx10^9) =0.075m` This wavelength belongs to microwave region. The microwaves are used (i) In microwave ovens (ii) in radar system for the navigation of aircraft. |
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| 120. |
Name the electromagnetic waves used for following and arrange them in increasing order of their penetrating power: (a)water purification (b) remote sensing (c) treatment of cancer. |
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Answer» (a) For water purification, ultraviolet rays are used. (b) For remote sensing, microwaves are used (c) For treatment of cancer, `gamma`-rays are used. Order of penetrating power of these radiations microwavesltultraviolet rays lt`gamma`-rays. |
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| 121. |
Identify the part of the electromagnetic specturm which is (i) Suitable for radar systems used in aircraft nevigation (ii) Adjacent to the low frequency end of the electromagnetic spectrum (iii) produced in nuclear reaction (iv) produced by bombarding a metal target by high speed electrons. |
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Answer» (i) Microwaves (ii) Radiowaves (iii) Gamma rays (iv) X-rays. |
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| 122. |
Name the parts of the electromagnetic spectrum which is (a) suitable for radar system used in aircraft navigation (b) used to treat muscular strain. (c) Use as a diagnostic tool in medicine. Write in brief, how these waves can be produced. |
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Answer» (a) Microwaves, Production:Klystron`//`magnetron (b) Infrared radiations, Production:Hot bodies`//`vibrations of atoms and molecules. (c) x-rays, Production:Hot Bombarding the high energy electons on a target of high atomic number. |
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| 123. |
The part of the spectrum of the electromagnetic radiation used to cook food isA. ultraviolet raysB. cosmic raysC. X raysD. microwaves |
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Answer» Correct Answer - D Microwaves are used to cook food. Microwave oven is a domestic application of these waves. |
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| 124. |
Name the costituent radiation of electromagnetic spectrum which (a) Is used in satellite communication (b) is used for studying crystal structure (c) is similar to the radiations emitted during decay of radioactive nuclei. (d) has its wavelength range 390 nm to 770 nm (e) is absorbed from sunlight by ozon layer (f) produces intense heating effect. |
| Answer» (a) Microwaves (b) X-rays (c) Gamma rays (d) visible light (e) Ultraviolet radiation (f) Infrared rays | |
| 125. |
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacumm is equal toA. the speed of light in vaccumB. reciprocal of speed of light in vacuumC. the ratio of magnetic permeability to the electric susceptibility of vacuumD. unity |
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Answer» Correct Answer - B The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vaccum is equal to the reciporcal of speed of light in vaccum. i.e,`(B_(0))/(E_(0)) = 1/c` |
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| 126. |
In case of linearly polarised light, the magnitude of the electric field vectorA. does not change with timeB. varies periodically with timeC. is parallel to the direction of propagationD. increases and decreases linearly with time |
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Answer» Correct Answer - B The magnitude of electric field vector varies periodically with time because it is in the form of electromagnetic wave. |
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| 127. |
The nature of electromagnetic wave is- |
| Answer» Correct Answer - transvers | |
| 128. |
Name the electromagnetic waves used for studying crystal structure of solids. What is its frequency range? |
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Answer» X-rays are used for studying crystal structure of solids. Their frequency range `3xx10^17Hz` to `3xx10^20Hz`. |
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| 129. |
A leaf which contains only green pigments, is illuminated by a laser light of wavelength `0.6328mum`. It would appear to beA. blackB. greenC. brownD. red |
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Answer» Correct Answer - A Laser light is of red colour which is not reflected by greeen, pigment of the leaf and so it appears black. |
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| 130. |
Which force in nature exits everywhere?A. Nuclear forceB. Electromagnetic forceC. Weak forceD. Gravitation |
| Answer» Correct Answer - D | |
| 131. |
What is the cause of "Green house effect"?A. Ultraviolet raysB. Infrared raysC. X-raysD. None of these |
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Answer» Correct Answer - B Infrared radiations are reflected by lower clouds and keeps the earth warm. Hence shows green house effect. |
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| 132. |
Why does galvanometer show a momentary deflection at the time of charging or discharging a capacitor? Write the necessary expression to explain this observation? |
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Answer» During charging or discharging or a capacitor, increasing or decreasing current in a circuit with time flows due to conduction current in wire and displacement current between the plates of the capacitor. When capacitor gets fully charged both conduction and displacement current becomes zero. That is why galvanometer shows a momentary deflection at the time of charging or discharging. The expression to explain this observation is : `ointvecB.vec(dl)=mu_0(I+I_D)` |
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| 133. |
In a place `e.m` wave, the electric field oscillates sinusoidally at a frequency of `2.5xx10^(10)Hz` and amplitude `480 V//m`.The amplitude of oscillating magnetic field will be,A. `1.5xx101^(-8)Wb//m^(2)`B. `1.52xx10^(-7)Wb//m^(2)`C. `1.6xx10^(-6)Wb//m^(2)`D. `1.6xx10^(-7)Wb//m^(2)` |
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Answer» Correct Answer - c `C=(E_(0))/(B_(0))` `:. B_(0)=(E_(0))/(c)=1.6xx10^(-6)Wb//m^(2)` |
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| 134. |
To which region of electromagnetic spectrum, the frequency `BHz` correspond?A. Ultraviolet waveB. radio wavesC. visibleD. `X`-rays |
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Answer» Correct Answer - b `v_(1)=BHz=10^(9)Hz` `lambda=(c)/(v)=(3xx10^(8))/(10^(9))=0.3m=30cm` |
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| 135. |
A millimetre wave has a wavelength of `2.00 mm` and the oscillating electric field associated with it has an amplitude of `20 Vm^(-1).` Determine the frequency of oscillations of the electric and magnetic fileds of this electromagnetic wave. What is the amplitude of the magnetic field oscillations of this wave? |
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Answer» Correct Answer - A |
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| 136. |
In a plane e.m. wave, the electric field varies with time having an amplitude `1Vm^-1`. The frequency of wave is `0.5xx10^(15) Hz`. The wave is propagating along Z-axis. What is the average energy density of (i) electric field (ii) magnetic field (iii) total average energy density (iv) what is the amplitude of magnetic field? |
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Answer» Correct Answer - A::B::C::D |
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| 137. |
Radio wave diffract around building although light waves do not. The reason is that radio wavesA. travel with speed larger than cB. have energy wavelength tha lightC. carry newsD. are not electromagnetic waves |
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Answer» Correct Answer - B Radio waves diffract around builiding although light waves do not. The reason is that radio waves have longer wavelength than light. |
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| 138. |
Frequency of a wave is `6 xx 10^(15)` Hz. The wave isA. radiowaveB. microwaveC. X-rayD. None of these |
| Answer» Correct Answer - D | |
| 139. |
The ferequency `1057 MHz` of radiation arising from two close energy levels in hydrogen belong to:A. Infrared raysB. X-raysC. `gamma`-raysD. Radio waves |
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Answer» Correct Answer - D This given frequency corresponds to the radio waves i.e short wavelength or high frequency. |
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| 140. |
The ferequency `1057 MHz` of radiation arising from two close energy levels in hydrogen belong to:A. radio wavesB. infrared wavesC. microwavesD. `gamma`-rays |
| Answer» Correct Answer - A | |
| 141. |
For sky wave propagation of a `10 MHz` signal, what should be the minimum electron density in ionosphere?A. `~1.2xx10^(12) m^(-3)`B. `~10^(6) m^(-3)`C. `~10^(14) m^(-3)`D. `~10^(22) m^(-3)` |
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Answer» Correct Answer - a If maximum electron density of the ionosphere is `N_(max) "per "m^(3)` then the critical frequency `f_(c)` is given by `f_(c)=9(N_(max))^(1//2)` `implies1xx10^(6)=9(N)^(1//2) implies N=1.2xx10^(12) m^(-3)` |
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| 142. |
The ferequency `1057 MHz` of radiation arising from two close energy levels in hydrogen belong to:A. radio wavesB. infrared wavesC. micro wavesD. `omega`-rays |
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Answer» Correct Answer - a `v=1057 MHz` `lambda=(c)/(v)=(3xx10^(8))/(1057xx10^(6))m` `=0.28m=28cm`(radiowaves) |
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| 143. |
An `EM` wave radiates out wards from a dipole antenna with `E_(0)` as the amplitude of its electric filed vector. The electric field `E_(0)` which transports significant energy from the source falls off asA. `(1)/(r^(3))`B. `(1)/(r^(2))`C. `(1)/(r)`D. remains constant |
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Answer» Correct Answer - C From a dipole antenna, the electromagnetic waves are radiated outwards. The amplitude of electric field vector `E_(0)` which transports significant energy from the source falls off inversely as the distance r from the antenna i.e., `E_(0)prop(1)/(r)`. |
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| 144. |
Which one of the following is the property of a monochromatic, plane electromagnetic wave in free space?A. Electric and magnetic fields have a phase difference of `pi//2`B. The energy contribution of both electric and magnetic fields are equalC. The direction of propagation is in the direction of `vecBxxvecE`D. The pressure exerted by the wave is the product of its speed energy density |
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Answer» Correct Answer - B In electromagnetic wave, electric and magnetic field are in phase. Electromagnetic wave carry energy as they travel through space and this energy is shared equally by electric and magnetic field. The direction of the propagation of electromagnetic wave is the direction of `vec(E)xxvec(B)`. The pressure exerted by the wave is equal to its energy density. |
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| 145. |
The velocity of electromagnetic waves in free space can be given by relation........A. `sqrt(mu_(0)epsi_(0))`B. `sqrt((mu_(0))/(epsi_(0)))`C. `sqrt((epsi_(0))/(mu_(0)))`D. `sqrt(1)/(sqrt(mu_(0)epsi_(0)))` |
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Answer» Correct Answer - D The velocity of electromagnetic waves in free space (vacuum) is given by `c = (1)/(sqrt(mu_(0)epsi_(0))) = 3 xx 10^(8)ms^(-1)` |
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| 146. |
Which one of the following is the property of a monochromatic, plane electromagnetic wave in free space?A. Electrica and magnetic fields have a plane difference of `pi//2`B. The energy contribution of both electric and magnetic fields are equalC. The direction of propagation is in the direction of `B xx E`D. The pressure exerted by the wave is the product of its speed and energy density |
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Answer» Correct Answer - B The energy in EM waves is divided equally between the electric and magnetic fields. |
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| 147. |
An `EM` wave radiates out wards from a dipole antenna with `E_(0)` as the amplitude of its electric filed vector. The electric field `E_(0)` which transports significant energy from the source falls off asA. `1/(r^(3))`B. `(1)/(r^(2))`C. `(1)/(r)`D. remain constant |
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Answer» Correct Answer - C From a dipole antenna, the electromagnetic waves are radiated outwards. The amplitude of electric field vector `(E_(0))` which transports singnificant energy from the source falls off inversely as the distance `(r)` from the antenna ie., `E_(0)prop (1)/(r)` |
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| 148. |
The essential distinction between `X` - rays and `gamma` - rays is thatA. `gamma`-rays have smaller wavelength than X-raysB. `gamma`-rays emanate from nucleus while X-rays emanate from outer part of the atomC. `gamma`-rays greater ionising power than X-raysD. `gamma`-rays are more penetrating than `X`-rays |
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Answer» Correct Answer - B The wavelength of the `gamma`-rays is shorter. However, the main distinguishing feature is the nature of emission. |
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| 149. |
What does an e.m. wave consists of? On what factors does its velocity in vacuum depend? |
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Answer» EM wave are those waves in which there are sinusoidal variation of electric and magnetic field vectors, at right angles to each other as well as at right angles to the direction of its propagation. These two field vectors vibrate in the same phase and with the same frequency. The velocity of e.m. wave in free space is given by `c=1//sqrt(mu_0in_0) =3xx10^8 m//s` where `mu_0` is the absolute permeability of space and `in_0` is the absolute permitivity of space. |
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| 150. |
The crystal structure can be studied by usingA. UV raysB. X-raysC. IR radiationD. Microwaves |
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Answer» Correct Answer - B Crystal structure can be studied using X-rays. |
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