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CBSE Questions for Class 12 Medical Physics Electromagnetic Waves Quiz 5 - MCQExams.com
CBSE
Class 12 Medical Physics
Electromagnetic Waves
Quiz 5
Two waves having same velocity enter electric and magnetic fields respectively. If $$\lambda _{1}$$and$$\lambda _{2}$$ are their wavelengths as they move in the fields, then
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$$\lambda _{1},\lambda _{2}$$ are constants
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$$\lambda _{1}$$ and $$\lambda _{2}$$ are variable
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$$\lambda _{1}$$ is variable $$\lambda _{2}$$ is constant
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$$\lambda _{2}$$ is variable, $$\lambda _{1}$$ is constant
Explanation
Velocity of a wave is given by:
$$v = \cfrac{E}{B}$$
Hence wave velocity change in both the cases.
Frequency of the wave remains the same.
Using $$v = f\lambda$$, it can be concluded that both $$\lambda_1$$ and $$\lambda_2$$ are variable.
We consider the radiation emitted by the human body. Which of the following statements is true :
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The radiation emitted is in the infrared region
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The radiation is emitted only during the day
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The radiation is emitted during the summers and absorbed during the winters
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The radiation emitted lies in the ultraviolet region and hence is not visible
Explanation
Every body at all time, at all temperatures emit radiation (except at T = 0), which fall in the infrared region.
PAR range of wavelength is
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$$400-700 nm$$
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$$400-800 nm$$
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$$450-950 nm$$
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$$500-600 nm$$
Explanation
Photosynthetically Active Radiation $$(PAR)$$ Is the amount of light available for photosynthesis, which is light in the $$400$$ to $$700$$ nanometer wavelength range.
PAR changes seasonally and varies depending on the latitude and time of day.
Wave length of X-rays is
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1 $$\overset{\circ}A$$ to 10 $$\overset{\circ}A$$
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0.01 $$\overset{\circ}A$$ to 100 $$\overset{\circ}A$$
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0.001 $$\overset{\circ}A$$ to 1 $$\overset{\circ}A$$
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0.1 $$\overset{\circ}A$$ to 1 $$\overset{\circ}A$$
Which of the following relations is correct for electromagnetic waves:
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$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =\cfrac { 1 }{ { c }^{ 2 } } \cfrac { { \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$
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$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } ={ \left( \cfrac { \partial E }{ \partial t } \right) }^{ 2 }$$
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$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =\cfrac { { c }^{ 2 }{ \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$
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$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =c\cfrac { { \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$
Explanation
The equation of electromagnetic wave propagating along x direction in vaccum is: $$\dfrac{ \partial ^2 E}{ \partial x^2}=\dfrac{1}{c^2}\dfrac{ \partial ^2 E}{ \partial t^2}$$
In which colour an yellow coloured dress would appear when viewed through a blue filter?
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yellow
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blue
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black
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green
Explanation
Yellow light is a combination of red and green light. A dress appears yellow if it reflects red and green light to our eyes. In order to reflect red and green light, these two primary colors of light must be present in the incident light.
So, blue colored filter allows only blue color and absorbs all the remaining colors.
Therefore, yellow colored dress appears black.
The solar spectrum in general is:
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radiation spectrum
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absorption spectrum
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emission spectrum
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continuous spectrum
Explanation
Solar spectrum is not a continuous spectrum, it is an absorption spectrum. It shows a lot of dark lines. These dark lines are because of absorption of that particular frequency by some atoms or molecules on the way of the path of light. Absorption spectrum is the spectrum of light which has passed through a medium capable of absorbing a portion of the rays. It is characterized by dark spaces, bands, or lines. The diagram shows a merged solar spectrum.
Radio isotope useful in treating blood cancer is
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Cobalt 40
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Carbon 60
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Cobalt 60
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Carbon 14
Explanation
Cobalt therapy or cobalt $$60$$ therapy is the medical use of gamma rays from the radio isotope cobalt $$-60$$ to treat conditions such as cancer. Thus, Cobalt $$-60$$ is the radio isotope useful in treating the blood cancer.
Wave length of visible spectrum extends from
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0.4 $$\mu$$ m- 0.7 $$\mu$$m
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0.5 nm - 6 nm
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5 $$\mu$$ m - 60 $$\mu$$m
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50 nm - 60 nm
Which of the following are produced by the high-frequency electromagnetic oscillators in electric circuits.
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Microwaves
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Longitudinal waves
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Magnetic waves
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None of these
Consider the radiations emitted by the human body. Which of the following statements is true?
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radiations lie in ultraviolet region
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radiations lie in infrared region
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radiations are emitted only during the day
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radiations are emitted during summer and absorbed during winter
Explanation
As we know that human body release there temperature in form of heat. So, h
eat radiations lie in infrared and microwave regions.
The method of using radio waves to cure some diseases are called
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Physiotherapy
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Netratherapy
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Radiotherapy
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Radiometry
If we denote microwaves, X rays, infrared, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum by M, X, I, G, U, R and V, the following is the arrangement in ascending order of wavelength :
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R, M, I, V, U, X and G
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M, R, V, X, U, G and I
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G, X, U, V, I, M and R
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I, M, R, U, V, X and G
If $${ \lambda }_{ v },{ \lambda }_{ x },{ \lambda }_{ m }$$ represents the wavelengths of vision, X-ray and microwaves respectively then
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$${ \lambda }_{ m }>{ \lambda }_{ x }>{ \lambda }_{ v }$$
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$${ \lambda }_{ m }>{ \lambda }_{ v }>{ \lambda }_{ x }$$
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$${ \lambda }_{ v }>{ \lambda }_{ x }>{ \lambda }_{ m }$$
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$${ \lambda }_{ v }>{ \lambda }_{ m }>{ \lambda }_{ x }$$
Explanation
The above diagram shows the increasing order of wavelength for different waves from gamma rays to radio waves.
According to the diagram, among X-rays, micro waves and visible rays, the micro waves have the largest wavelength and X-ray has the smallest.
$$\implies$$ $$\lambda_m > \lambda_v >\lambda_x$$
A parallel plate capacitor consists of two circular plates each of radius $$12cm$$ and separated by $$5.0mm$$. The capacitor is being charged by an external source. The charging current is constant and is equal to $$0.15A$$.
The displacement current is:
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$$15A$$
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$$1.5A$$
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$$0.15A$$
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$$0.015A$$
Explanation
According to Maxwell's hypothesis, a displacement current will flow through a capacitor when the potential difference across its plates is varying. Thus a varying electric field will exist between the plates and this displacement current is same in magnitude to the current flowing in outer circuit.
Here, the current in the outer circuit is 0.15 A. Thus $$0.15 A$$ will be the displacement current.
The AM range of radiowaves has frequency:
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less than $$30$$MHz
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more than $$30$$MHz
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less than $$20000$$Hz
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more than $$20000$$Hz
Explanation
AM range of radio-waves has frequency between
535-1605 kHz i.e. less that 30MHz.
The charge on a parallel plate capacitor is varying as $$q={ q }_{ 0 }\sin { \left( 2\pi nt \right) } $$. The plates are very large and close together. neglecting the edge effects, the displacement current through the capacitor is:
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$$\cfrac { q }{ { \varepsilon }_{ 0 }A } $$
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$$\cfrac { { q }_{ 0 } }{ { \varepsilon }_{ 0 } } \sin { 2\pi } nt$$
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$$2\pi n{ q }_{ 0 }\cos { 2\pi } nt $$
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$$\cfrac { 2\pi n{ q }_{ 0 } }{ { \varepsilon }_{ 0 } } \cos { 2\pi } nt $$
Explanation
Displacement current through a surface is $$\epsilon_0$$ times the rate of change of flux through the surface.
Since $$E=\dfrac{q}{A\epsilon_0}$$.
Therefore electric flux through the surface is $$\phi = EA=\dfrac{q}{\epsilon_0}$$
So real current though the capacitor $$i = \dfrac{d\phi}{dt} = \epsilon_0 \dfrac{d}{dt}(\dfrac{q}{\epsilon_0})=\dfrac{dq}{dt}=i$$
Thus displacement current is $$\dfrac{dq}{dt}=2\pi n q_0\cos2\pi ft$$
The $$rms$$ value of conduction current will be:
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$$5.7\mu A$$
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$$6.3\mu A$$
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$$9.6\mu A$$
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$$6.9\mu A$$
Explanation
Rms value of conduction current, $$I=V/X_c$$
Where,
$$X_c$$
= Capacitive reactance
$$=1/ωC$$
∴ $$ I = V × ωC $$
= $$230\times 300\times 100\times 10^{-12}$$
= $$6.9\times 10^{-6} $$ A
= $$6.9$$ μA
If the frequencies of an ultrasonic wave and an electromagnetic wave are same, then:
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their wavelength will be same
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wavelength of electromagnetic wave will be less than that of ultrasonic wave
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wavelength of electromagnetic wave will be more than that of ultrasonic wave
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the wavelengths of both will be nearly equal
Explanation
Speed of a wave=$$v=\lambda\nu$$
Thus $$v\propto \lambda$$
Since the speed of electromagnetic wave or the speed of light is more than the speed of ultrasonic wave(sound), the wavelength of electromagnetic wave is more than that of ultrasonic wave.
Maxwell's four equations are written as
(i) $$\oint { \overrightarrow { E } .\overrightarrow { d } s } =\cfrac { { q }_{ 0 } }{ { \varepsilon }_{ 0 } } \quad $$
(ii)$$\oint { \overrightarrow { B } .\overrightarrow { d } s } =0$$
(iii)$$\oint { \overrightarrow { E } .\overrightarrow { d } l } =\cfrac { d }{ dt } \quad \oint { \overrightarrow { B } .\overrightarrow { d } s } $$
(iv)$$\oint { \overrightarrow { B } .\overrightarrow { d } s } ={ \mu }_{ 0 }{ \varepsilon }_{ 0 }\cfrac { d }{ dt } \quad \oint { \overrightarrow { E } .\overrightarrow { d } s } $$
Which of the above Maxwell's equations shows that electric field lines do not form closed loops?
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(i) only
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(ii) only
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(iii) only
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(iv) only
Explanation
$$\oint \vec{E}.\vec{d}s $$ represents electric flux which is equal to the number of electric field lines crossing through an area.
Now for $$q_o \neq 0$$, implies
$$\oint \vec{E}.\vec{d}s \neq 0 $$ which indicates Electric field lines do not form closed loops.
The Sun delivers $${{10}^{3}}W/{{m}^{2}}$$ of electromagnetic flux to the Earth's surface.
The radiation force on the roof
of dimensions $$8m\times 20m$$
will be:
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$$3.33\times { 10 }^{ -5 }N$$
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$$5.33\times { 10 }^{ -4 }N$$
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$$7.33\times { 10 }^{ -3 }N$$
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$$9.33\times { 10 }^{ -2 }N$$
Explanation
Poynting vector, $$S={10}^3 W/m^2$$
Pressure, $$P=\dfrac{S}{c}=$$$$\dfrac{1000}{3\times 10^8}$$
$$=3.33\times {10}^{-6} N/m^2$$
Radiation force on the roof is given by:
$$F=P\times 8\times 20 = 5.33 \times {10}^{-4}N$$
For treating arthritis
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radio waves are used
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direct current is used
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mircowaves are used
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AC of high frequency is used
Explanation
AC sources with high frequencies($$10^6Hz$$) are used for treatment of arthritis.
Waves with same frequencies can also be used.
Wavelength of such waves$$\approx 3km$$.Thus radio waves have such a frequency.
Maxwell's four equations are written as
(i) $$\oint { \overrightarrow { E } .\overrightarrow { d } s } =\cfrac { { q }_{ 0 } }{ { \varepsilon }_{ 0 } } \quad $$
(ii)$$\oint { \overrightarrow { B } .\overrightarrow { d } s } =0$$
(iii)$$\oint { \overrightarrow { E } .\overrightarrow { d } l } =\cfrac { d }{ dt } \quad \oint { \overrightarrow { B } .\overrightarrow { d } s } $$
(iv)$$\oint { \overrightarrow { B } .\overrightarrow { d } s } ={ \mu }_{ 0 }{ \varepsilon }_{ 0 }\cfrac { d }{ dt } \quad \oint { \overrightarrow { E } .\overrightarrow { d } s } $$
Out of the four Maxwell's equation above, which one shows the non-existence of monopoles?
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(i) and (iv)
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(ii) only
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(iii) only
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none of these
Explanation
Had there been a monopole, the integral of $$\vec{B}.\vec{ds}$$ on a closed surface enclosing the monopole would have been non-zero.
Thus the second equation shows the non-existence of monopole.
Instantaneous displacement current $$1A$$ in the space between the parallel plates of $$1\mu F$$ capacitor can be established by changing the potential difference at the rate of:
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$$0.1V/s$$
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$$1V/s$$
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$${10}^{6}V/s$$
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$${10}^{-6}V/s$$
Explanation
In a capacitor of capacitance $$C$$,
$$V=\dfrac{q}{C}$$
$$\implies \dfrac{dV}{dt}=\dfrac{i}{C}=\dfrac{1A}{1\mu F}=10^{6}V/s$$
Assume you are sitting in sun for 2.5 hours. The area of your body exposed normally to sun rays $$ { 1.3m }^{ 2 }$$. The intensity of sun rays is 1.1 Kilowatt/$${ m }^{ 2 }$$. If your body completely absorbs the sun rays then the momentum transferred to your body will be (in Kg-m/s):
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$$0.043$$
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$$0.037$$
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$$0.61$$
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$$-0.91$$
Explanation
$$Power=intensity\times area$$
Intensity and area are given ,
$$Power=1.3\times 1.1KW$$
$$Energy = power\times time$$
Time is 2.5hr
$$Energy= 1300\times 1.1\times 2.5\times 3600J$$
$$=12870000J$$
$$Momentum=\dfrac{energy}{c}$$
$$=\dfrac{12870000}{3\times 10^{8}}$$
$$=0.043$$
The wavelengths for the light of red and blue colours are roughly $$7.8\times 10^{-7}m$$ and $$4.8\times 10^{-7}m$$ respectively. Which colour has the greater speed in vacuum?
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Red
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Violet
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Yellow
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All colors have same speed
Explanation
All electromagnetic waves travels at the same speed in vacuum i.e $$3\times10^8$$ m/s. Due to the absence of matter, there is no change in speed corresponding to wavelength. There is no absorption and re-emission of light in the absence of matter.
For the propagation of electromagnetic waves
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Medium is required
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no medium is required
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E and B are in mutually opposite phase
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E and b do not contriute
Explanation
Electromagnetic waves are a result of changing electric field creating magnetic field and the magnetic field does the same for electric field. Unlike mechanical waves, electromagnetic waves does not require material medium to transfer energy from one place to another.
Name the radiations of wavelength just shorter than $$4\times 10^{-7}m$$.
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Infraviolet
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Ultraviolet
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Infrared
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Ultrared
Explanation
The range of wavelength for visible light lies between, $${ 400 \ nm }<\lambda <{ 700 \ nm }$$.
$$\lambda $$ just less than 400 nm is Ultraviolet.
$$\lambda $$ just greater than 700 nm is Infrared.
Arrange the following radiations in the order of their increasing wavelength:
$$X$$-rays, infra red rays, radio waves, gamma rays and micro waves.
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Gamma rays, $$X$$-rays, infra red rays, radio waves, micro waves.
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Gamma rays, micro waves, $$X$$-rays, infra red rays, radio waves.
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Gamma rays, infra red rays, $$X$$-rays, micro waves, radio waves.
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Gamma rays, $$X$$-rays, infra red rays, micro waves, radio waves.
Explanation
Electromagnetic radiation is classified into types according to the frequency of the wave. These types include, in order of increasing frequency, radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.
The frequency is the inverse of wavelength. Hence, in the order of increasing wavelength, the waves are Gamma rays ($$<1nm$$), X-rays (
1-10 nm
), infra red rays ($$700-{ 10 }^{ 5 }nm$$), micro waves ($${ 10 }^{ 5 }-{ 10 }^{ 8 }nm$$), radio waves ($${ >10 }^{ 8 }nm$$).
Name the radiation which can be detected by a thermophile.
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infrared radiation
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ultraviolet radiation
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X-rays
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All of the above
Explanation
Infrared radiations can be detected by thermopile. Thermopiles are designed to measure temperature from a distance by detecting an object's infrared (IR) energy. Hence option 'A' is correct.
Give the range of wavelength of the electromagnetic waves visible to us.
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$$1000A^o$$ to $$4000A^o$$
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$$4000A^o$$ to $$7000A^o$$
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$$8000A^o$$ to $$9000A^o$$
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$$400A^o$$ to $$800A^o$$
Explanation
$$\Large \underline{ \textbf{Correct answer: B}}$$
$$\Large \underline{ \textbf{Explanation}}$$
Light is a form of electromagnetic wave. The wavelength range of light which is detectable by the human eyes is; $$4000-7000 \overset{0}{\text{A}}$$ w
here violet has the lowest wavelength$$(\sim4000\overset{0}{\text{A}})$$ and Red has the maximum wavelength
$$(\sim7000\overset{0}{\text{A}})$$
A wave has wavelength $$50A^o$$. Which of the following is a use of this wave?
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in detection of fracture in bones.
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in detection of planets
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in detection of water
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in detection of air
Explanation
X-rays has a wavelength of
$$50 \mathring{A}$$.
X-rays are used by doctors to see inside people. The machines are managed by a trained x-ray technician. They pass easily through soft tissues, but not so easily through bones.
Hence, they help in detecting fractures in bones.
A wave has a wavelength of $$0.01A^o$$. Name the wave:
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X rays
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Micro waves
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Gamma waves
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Radio waves
Explanation
$$\large \underline {\textbf{Concept applied:}}$$
Electromagnetic spectrum
$$\large \underline {\textbf{Correct answer:}}$$ Gamma waves
$$\large \underline {\textbf{Explanation:}}$$
$$\bullet$$
Electromagnetic radiations are classified according to the frequency of the wave. Their types include: Radio waves, Microwaves, Infrared radiation, Visible light, Ultraviolet radiation, X-rays and Gamma rays,
in order of increasing frequency,
.
$$\bullet$$ Since W
avelength is inversely proportional to the frequency. Hence, the order of increasing wavelength will be just opposite, as shown in the figure below:
$$\bullet$$ Range of wavelength of
Gamma rays is $$< 10^{-2}nm$$
$$\bullet$$ Wave length of given wave is $$0.01 A^0 $$$$=0.01 \times 10^{-10}m$$
$$=1 \times 10^{-3}nm$$
$$\bullet$$ Therefore the wave will lie in the region of Gamma rays.
A wave has wavelength $$50A^o$$. State the speed of wave in vacuum.
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$$3\times 10^8ms^{-1}$$
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>$$3\times 10^8ms^{-1}$$
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<$$3\times 10^8ms^{-1}$$
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Insufficient information
Explanation
X-rays has a wavelength of
$$50\ A^o$$.
All electromagnetic radiation, of which radio waves and X-rays are examples, travels at the velocity of light, in a vacuum. The only difference between the two is that the frequency of X-rays is very much higher than radio waves.
Hence, X-rays travels at a speed of
$${ 3\times 10 }^{ 8 }m/s$$.
When light propagates in vaccum there is an electric field and a magnetic field. Which of the following is not true about these field?
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They are constant in time
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They have zero average value
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They are perpendicular to the direction of propagation of light
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They are mutually perpendicular
Explanation
$$Answer:-$$ A
They vary with time following a wave function (sinuosoidal) and average value of these function is zero and also we can see in figure they are mutually perpendicular and also perpendicular to direction of propagation.
Name the rays or waves used for taking photographs in dark.
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Infrared rays
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Ultraviolet rays
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X-rays
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Gamma rays
Explanation
Photographers use film that is sensitive to infrared rays to take pictures in places where there is no visible light.
IR light is longer in wavelength than visible light and focuses differently. Regular photographic lenses are made for visible light photography without any regard to what happens when photographing out of band light like IR. Therefore each lens design and focal length, even each focal length within a zoom lens will focus IR light differently.
Hence, infrared light is used for taking pictures in the dark.
A radiation $$A$$ is focused by a proper device on the bulb of a thermometer. Mercury in the thermometer shows a rapid increase. The radiation $$A$$ is:
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infrared radiation
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visible light
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ultraviolet radiation
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$$X$$-rays
Explanation
If radiation A is focused by a proper device on the bulb of a thermometer then thermometers measure the heat cause in rapid increase,and the radiation A is infrared radiation.because we know Infrared radiation is as heat.
Name the rays or waves produced by the changes in the nucleus of an atom.
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X-rays
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UV rays
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Gamma rays
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IR rays
Explanation
Gamma rays are classically produced by the decay of atomic nuclei as they transition from a high energy state to a lower state known as gamma decay, but may also be produced by other processes. Paul Villard, a French chemist and physicist, discovered gamma radiation in $$1900$$, while studying radiation emitted from radium. Villard's radiation was named "gamma rays" by Ernest Rutherford in $$1903$$.
Hence,
the rays or waves produced by the changes in the nucleus of an atom is Gamma rays.
An electromagnetic wave has a frequency of $$500MHz$$ and a wavelength of $$60cm$$. Name the medium through which it is traveling.
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air
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water
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salt
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cardboard
Explanation
The formula for velocity, when given wavelength and the frequency of the wave, is written as: $$v=f\lambda $$.
In this formula, $$f$$ represents frequency, $$V$$ represents the velocity of the wave, and $$\lambda $$ represents the wavelength of the wave.
Here, the frequency of the wave is $$500 MHz = 500000000 Hz$$ and the wavelength is $$60 cm = 0.6 m$$
Therefore, the velocity of the wave is $$3\times { 10 }^{ 8 }m/s$$.
This determined velocity is the velocity of the light in air.
Hence, the medium through which the wave is travelling is air.
In the propagation of light waves, the angle between the plane of vibration & plane of polarisation is:
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0$$^0$$
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90$$^0$$
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45$$^0$$
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80$$^0$$
Explanation
The plane of polarisations is that plane in which there is no vibration. While a plane including the direction of light propagation and the direction of electric field is called the plane of vibration. The angle between them is $$ { 90 }^{ o }$$.
Read the following statements.
a. Ozone in found mostly in the stratosphere.
b. Ozone layer lies 55-75 kms above the surface of the earth.
c. Ozone absorbs ultraviolet radiation from the Sun.
d. Ozone layer has no significance for life on the earth.
Which set of the statements is correct ?
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a and c
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b and d
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b and c
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a and d
Explanation
Ozone layer also called ozonosphere contains relatively high concentrations of ozone ($$O_{3}$$) molecules. Approximately 90% of atmospheric ozone occurs in stratosphere, the region extending from 10 - 18 km to 50 km above the earth's surface. The ozone layer effectively blocks certain types of ultraviolet and other forms of radiations that could injure or kill most living things.
High energetic radiations among the following:
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$$\lambda=0.01 A $$
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$$\lambda=0.10 A $$
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$$\lambda=1.00 A $$
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$$\lambda=40.0 A $$
Explanation
We know that, $$E = hv = \dfrac{hc}{\lambda}$$
Hence highest energetic rediation is whose wavelength is atleast i.e $$ \lambda = 0.01 A^{\circ}$$
On the basis of the following features identify correct option:
I. They require no material medium.
II. They are always transverse.
II. They are produced by accelerating charged particles
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Mechanical waves
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Electro-magnetic waves
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Both (1) and (2)
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Neither (1) and (2)
Explanation
Electromagnetic waves do not need any material medium for their propagation and they are always transverse in nature e.g. light waves.
Two waves $$A$$ and $$B$$ have wavelength $$0.01A^o$$ and $$9000A^o$$ respectively. Compare the speeds of these waves when they travel in vacuum.
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$$1:9$$
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$$9:1$$
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$$1:1$$
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None of the above
Explanation
The electromagnetic spectrum consists of the full range of known electromagnetic waves, including gamma rays, X-rays, ultraviolet light, visible light, infrared light, microwaves, and radio waves. All of these are produced by various forms of oscillating charges or currents. Electromagnetic waves are transverse waves which travel at the speed of light and involve oscillating electric and magnetic fields at right angles to each other.
Hence, the speeds of these waves when they travel in vacuum is same as the speed of visible light, that is, $$3\times { 10 }^{ 8 }m/s$$.
Therefore, the ratio of their speeds is $$1:1$$.
Which of the following is the infrared wavelength?
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$${10}^{-4}\:cm$$
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$${10}^{-5}\:cm$$
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$${10}^{-6}\:cm$$
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$${10}^{-7}\:cm$$
Explanation
The wavelength of infrared region is $$8\times{10}^{-5}\:cm$$ to $$3\times{10}^{-3}\:cm$$. So maximum wavelength of infrared region $$=8\times{10}^{-5}\:cm\approx{10}^{-4}\:cm$$.
Maxwell's modified form of Ampere's circuital law is
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$$\displaystyle\oint_{\displaystyle s}{\vec{B}.\vec{ds}}=0$$
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$$\displaystyle\oint_{\displaystyle s}{\vec{B}.\vec{dl}}=\mu_0I$$
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$$\displaystyle\oint_{\displaystyle s}{\vec{B}.\vec{dl}}=\mu_0I+\frac{I}{\varepsilon_o}\frac{dq}{dt}$$
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$$\displaystyle\oint_{\displaystyle s}{\vec{B}.\vec{dl}}=\mu_0I+\mu_0\varepsilon_o\frac{d\phi_E}{dt}$$
Explanation
$$Answer:-$$ D
Treating all charges on the same footing (disregarding whether they are bound or free charges), the generalized Ampère's equation, also called the Maxwell–Ampère equation, is in integral form.
All components of the electromagnetic spectrum in vacuum have the same
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energy
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velocity
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wavelength
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frequency
Explanation
$$Answer:-$$ B
speed of electromagnetic wave in vacuum is given by:-
c(speed of light)=frequency$$\times$$ wavelength =$$\dfrac{1}{\mu_0 \epsilon_0}$$=constant
as we go from gamma rays to radio waves frequency decreases and wavelength increases thereby maintaining the product constant.
Energy is given by:- $$E=h\nu$$ since $$\nu$$ is different so energy is also different.
Also wavelength and frequency varies.
The range of wavelength of visible light is:
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$$10\mathring{A}$$ to $$100\mathring{A}$$
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$$4000\mathring{A}$$ to $$7000\mathring{A}$$
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$$8000\mathring{A}$$ to $$10,000\mathring{A}$$
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$$10,000\mathring{A}$$ to $$15,000\mathring{A}$$
Explanation
The displacement current is
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$$\displaystyle\varepsilon_o\frac{d\phi_E}{dt}$$
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$$\displaystyle\frac{\varepsilon_o}{R}\frac{d\phi_E}{dt}$$
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$$\displaystyle\varepsilon_o\frac{E}{R}$$
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$$\displaystyle\varepsilon_oq\frac{C}{R}$$
If a source is transmitting an electromagnetic wave of frequency $$8.2\times{10}^6\:Hz$$, then the wavelength of the electromagnetic waves transmitted from the source will be:
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$$36.6\ m$$
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$$40.5\ m$$
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$$42.3\ m$$
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$$50.9\ m$$
Explanation
Here, $$\displaystyle\lambda=\frac{c}{v}=\frac{3\times{10}^8}{8.2\times{10}^6}=36.6\:m$$
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Practice Class 12 Medical Physics Quiz Questions and Answers
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