InterviewSolution
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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.
| 3051. |
The ionization potential of hydrogenic ions P and Q are V_(P) and V_(Q) respectively. IfV_(Q) lt V_(P) then radii ...... |
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Answer» `r_(P) GT r_(Q)` `:. V prop (1)/(r) :. (V_(P))/(V_(Q))=(r_(Q))/(r_(P))` `:. V_(Q) lt V_(P) rArr r_(Q) gt r_(P) so, r_(P) lt r_(Q)` |
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| 3052. |
Three capacitors of capacitance 2pF,3pF and 4pF are connected in parallel. Determine the cahrge on each capacitor, if the combination is connected to a 100 V supply? |
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Answer» Solution :EQUIVALENT capacitance, `C=C_(1)+C_(2)+C_(3)=2pF+3pF+4pF=9pF` or `C=9xx10^(-12)F` Charge on the parallel combination of the capacitor and HENCE the charge on each capacitor, `q=CV=9xx10^(-12)xx100=9xx10^(-10)C`. |
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| 3053. |
Read the following passage and then answer questions (a) to (e) on the basis of your under- standing of the passage and the related studied concepts. In accordance with Faraday.s law of electromagnetic induction a magnetic field, changing with time, gives rise to an electric field. Again as per Ampere-Maxwell.s law an electric field, changing with time, gives rise to a magnetic field. It means that if we consider a charge oscillating with some frequency, it produces an oscillating electric field in space, which produces an oscillating mag- netic field, which in turn, is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other and an electromagnetic wave propagates through space. The energy associated with the propagating wave comes at the expense of the energy of the oscillating charge. Electromagnetic wave propagates through free space as a transverse wave in which vecE and vecB are perpendicular to each other as well as perpendicular to the direction of wave propagation. Like other waves electromagnetic waves carry energy and momentum. As electromagnetic waves contains both electric and magnetic fields, it has an electrical energy density mu_(E)=(1)/(2)in_(0)E^(2) as well as a magnetic energy density mu_(B)=(B^(2))/(2mu_(0)). Both of these vary with time. The magnetic field in a plane electromagnetic wave is given by : B_(y)=2xx10^(-7)sin[1.5xx10^(11)t-0.5xx10^(3)x]T What is the speed of electromagnetic wave? |
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Answer» Solution :(a) Comparing the GIVEN equation of MAGNETIC field with the standard equation `B_(y)=B_(0)sin(omegat-kx)T, ` We have `B_(0)=2xx10^(-7)T, omega=1.5xx10^(11)s^(-1) and k=0.5xx10^(3)m^(-1)` `therefore"SPEED of electromagnetic wave " =(omega)/(k)=(1.5xx10^(11))/(0.5xx10^(3))=3.0xx10^(8)ms^(-1)` |
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| 3054. |
For the electromagnetic waves propagating in the X - direction, oscillation of vec(E ) takes places along y - axis. What from the following is right ? |
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Answer» `E_(X)=E_(y)=0, E_(z)NE 0` |
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| 3055. |
Read the following passage and then answer questions (a) to (e) on the basis of your under- standing of the passage and the related studied concepts. In accordance with Faraday.s law of electromagnetic induction a magnetic field, changing with time, gives rise to an electric field. Again as per Ampere-Maxwell.s law an electric field, changing with time, gives rise to a magnetic field. It means that if we consider a charge oscillating with some frequency, it produces an oscillating electric field in space, which produces an oscillating mag- netic field, which in turn, is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other and an electromagnetic wave propagates through space. The energy associated with the propagating wave comes at the expense of the energy of the oscillating charge. Electromagnetic wave propagates through free space as a transverse wave in which vecE and vecB are perpendicular to each other as well as perpendicular to the direction of wave propagation. Like other waves electromagnetic waves carry energy and momentum. As electromagnetic waves contains both electric and magnetic fields, it has an electrical energy density mu_(E)=(1)/(2)in_(0)E^(2) as well as a magnetic energy density mu_(B)=(B^(2))/(2mu_(0)). Both of these vary with time. What is the frequency and wavelength of the electromagnetic wave? |
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Answer» Solution :(b) Frequency of ELECTROMAGNETIC wave = Frequency of `VECB` VECTOR `=(OMEGA)/(2pi)` `=(1.5xx10^(11))/(2xx3.14)=2.39xx10^(10)Hz=23.9GHz` and wavelength of electromagnetic wave `=(2pi)/(K)=(2xx3.14)/(0.5xx10^(3))=1.26xx10^(-2)m=1.26cm` |
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| 3056. |
Read the following passage and then answer questions (a) to (e) on the basis of your under- standing of the passage and the related studied concepts. In accordance with Faraday.s law of electromagnetic induction a magnetic field, changing with time, gives rise to an electric field. Again as per Ampere-Maxwell.s law an electric field, changing with time, gives rise to a magnetic field. It means that if we consider a charge oscillating with some frequency, it produces an oscillating electric field in space, which produces an oscillating mag- netic field, which in turn, is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other and an electromagnetic wave propagates through space. The energy associated with the propagating wave comes at the expense of the energy of the oscillating charge. Electromagnetic wave propagates through free space as a transverse wave in which vecE and vecB are perpendicular to each other as well as perpendicular to the direction of wave propagation. Like other waves electromagnetic waves carry energy and momentum. As electromagnetic waves contains both electric and magnetic fields, it has an electrical energy density mu_(E)=(1)/(2)in_(0)E^(2) as well as a magnetic energy density mu_(B)=(B^(2))/(2mu_(0)). Both of these vary with time. In which direction does the electric field oscillates? |
| Answer» SOLUTION :(c ) As ELECTROMAGNETIC wave is propagating along X - AXIS and `vecB` is along Z - axis, hence `vecE` must be along Z - axis. | |
| 3057. |
Which is more dangerous, A.C or D.C? Why? |
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Answer» Solution :AC is more DANGEROUS than D.C of the same voltage. Because the peak value of A.C is MORETHAN the INDICATED value. e.g. Peak value of D.C = 220 V Then value of A.C `= 220 sqrt2 = 311 V` |
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| 3058. |
The electric flux due to an electric field barEthrough a surface DeltabarSis given by barE DeltabarS.The SI unit of electric flux is |
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Answer» NEWTON /Coulomb |
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| 3059. |
Read the following passage and then answer questions (a) to (e) on the basis of your under- standing of the passage and the related studied concepts. In accordance with Faraday.s law of electromagnetic induction a magnetic field, changing with time, gives rise to an electric field. Again as per Ampere-Maxwell.s law an electric field, changing with time, gives rise to a magnetic field. It means that if we consider a charge oscillating with some frequency, it produces an oscillating electric field in space, which produces an oscillating mag- netic field, which in turn, is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other and an electromagnetic wave propagates through space. The energy associated with the propagating wave comes at the expense of the energy of the oscillating charge. Electromagnetic wave propagates through free space as a transverse wave in which vecE and vecB are perpendicular to each other as well as perpendicular to the direction of wave propagation. Like other waves electromagnetic waves carry energy and momentum. As electromagnetic waves contains both electric and magnetic fields, it has an electrical energy density mu_(E)=(1)/(2)in_(0)E^(2) as well as a magnetic energy density mu_(B)=(B^(2))/(2mu_(0)). Both of these vary with time. What is the amplitude of electric field vector? |
| Answer» Solution :(d) AMPLITUDE of electric FIELD `E_(0)=B_(0)xxc=(2xx10^(-7))XX(3.0xx10^(8))=60Vm^(-1)` | |
| 3060. |
A:For microscope particles or sub microscopic particles wavelength of de-Broglie is important. R:If velocity is constant then de-Broglie wavelength is inversely proportional to mass of particle. |
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Answer» <P>Both ASSERTION and reason are true and the reason is correct explanation of the assertion. de-Broglie wavelength, I `LAMBDA=(h)/(p)=(h)/(MV)`,if v is CONSTANT then , `lambda prop(1)/(m)[becauseh="constant"]` |
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| 3061. |
If P represents radiation pressure, C speed of light,and Q radiation energy striking unit area per second and x,y,z are non zero integers, then p^x , Q^y , C^zis dimensionless. The values of x,y and z are respectively |
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Answer» 1,1-1 |
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| 3062. |
An electric dipole whose modulus is constant and whose moment is equal to p rotates with constant angular velocity omega about the axis draws at right angles to the axis of the dipole and passing through its midpoint. Find the power radiated by such a dipole. |
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Answer» Solution :The rotating dispole has moments `p_(x) = pcos omegat, p_(y) = p SIN omegat` Thus `P = (1)/(4PI epsilon_(0))(2)/(3c^(3))omega^(4)p^(2) (p^(2)omega^(4))/(6pi epsilon_(0)c^(3))`. |
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| 3063. |
The kinetic energies of the photoelectrons are E_(1) and E_(2) with wavelength of incident light lamda_(1) and lamda_(2). The work function of the metal is : |
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Answer» `(E_(1)lamda_(1)-E_(2)lamda_(2))/(lamda_(2)-lamda_(1))` `:.(E_(1)+w)/(E_(2))+w=(lambda_(2))/(lambda_(1)) or w=(E_(1) lambda_(1)-E_(2)lambda_(2))/(lambda_(2)-lambda_(1))` |
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| 3064. |
A 2A current is flowing through a circular coil of radius 10 cm containing 100 turns. Find the magnetic flux density at the centre of the coil. |
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Answer» Solution :`B=N(mu_(0)i)/(2R)=100xx(2pixx10^(-7)xx2)/(10xx10^(-2))` `=1.26xx10^(-3)Wb//m^(2)` |
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| 3065. |
Assertion: Two prisms of different materials can be used to attain a condition of angular dispersion without mean deviation. Reason: When two prisms of appropriate materials and refracting angles are oppositely directed then mean deviation produced by one can be cancelled by the other due to opposite sense. |
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Answer» If both ASSERTION and REASON are CORRECT and rason is a correct explanation of the assertion. |
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| 3066. |
Read the following passage and then answer questions (a) to (e) on the basis of your under- standing of the passage and the related studied concepts. In accordance with Faraday.s law of electromagnetic induction a magnetic field, changing with time, gives rise to an electric field. Again as per Ampere-Maxwell.s law an electric field, changing with time, gives rise to a magnetic field. It means that if we consider a charge oscillating with some frequency, it produces an oscillating electric field in space, which produces an oscillating mag- netic field, which in turn, is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus regenerate each other and an electromagnetic wave propagates through space. The energy associated with the propagating wave comes at the expense of the energy of the oscillating charge. Electromagnetic wave propagates through free space as a transverse wave in which vecE and vecB are perpendicular to each other as well as perpendicular to the direction of wave propagation. Like other waves electromagnetic waves carry energy and momentum. As electromagnetic waves contains both electric and magnetic fields, it has an electrical energy density mu_(E)=(1)/(2)in_(0)E^(2) as well as a magnetic energy density mu_(B)=(B^(2))/(2mu_(0)). Both of these vary with time. Write an expression for the electric field. |
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Answer» SOLUTION :(e ) The expression for the ELECTRIC will be : `E_(2)=60sin[1.5xx10^(11)t-0.5xx10^(3)x]Vm^(-1)` |
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| 3067. |
What are example of electromagnetic waves ? |
| Answer» SOLUTION :Infrared-rays,MICROWAVES,RADIO WAVES,X-rays, UV-rays ETC. | |
| 3068. |
A body at rest may have : |
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Answer» Velocity |
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| 3069. |
It is desired to make a 20Omegacoil of wire with zero thermal coefficient of resistance. To do this a carbon resistor of resistance R_1 is placed in series with an iron resistor of resistance R_2 . The proportions of iron and carbon are so chosen that R_1+ R_2 = 20 Omega at all temperatures near 20^@C. How large are R_1 and R_2? [alpha_c = - 7.5xx10^(2)""^@ C^(-1) and alpha_(Fe) = 5.0xx10^3 ""^@C^(-1) ? |
| Answer» SOLUTION :`R_1 = 1.25 OMEGA and R_2 = 18.75 Omega` | |
| 3070. |
An electron and a proton move in a uniform magnetic field with same speed directed perpendicular to the magnetic field. They experience force F_e and F_p respectively in mutually opposite directions, where F_p = 1840 F_e. |
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Answer» |
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| 3071. |
A radioactive nucleus of mass M emits a photon of frequency nu and the nucleus recoils. The recoil energy will be |
| Answer» Answer :C | |
| 3072. |
A coil of inductance 1H and neligible resistance is connected to a source of supply, whose voltage is given by V=4VOLT. If the voltage is applied at t=0, find the energy stored in the coil in 4s |
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Answer» 512J |
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| 3074. |
Which ones are the main characters? |
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Answer» TOMMY and his mother |
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| 3075. |
The figure shows electronic wave function for a hydrogen atom. |
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Answer» The quantum number of this STATE is 6 (B) `2pir=nlambda Rightarrow 2pi3^(2)r_(0)=nlambda Rightarrow lambda=6pir_(0)` (C) No `=^(3)C_(2)=3` (D) all lines in Balmer or lyman SERIES. |
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| 3076. |
Calculate the energy of the neutron , if it's wavelength is 1 A^@ ? |
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Answer» SOLUTION :`lambda=h/sqrt(2mE)` where E is the ENERGY of the NEUTRON. `E=h^2/sqrt(2mlambda^2) = ((6.62xx10^-34)^2)/(2xx1.67xx10^-27xx(10^-10)^2)=1.301xx10^-20` `E=(1.301xx10^-20)/(1.6xx10^-19)=0.081eV` |
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| 3077. |
Draw energy band diagram for a n-type extrinsic semiconductor. |
Answer» SOLUTION :
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| 3078. |
In the double-slit experiment, the viewing screen is at distance D = 4.00 m, point P lies at distance y = 20.5 cm from the center of the pattern, the slit separation d is 4.50 mu m, and the wavelength lambda is 650 nm. (a) Determine where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. (b) What is the ratio of the intensity I_(P) at point P to the intensity I_("cen") at the center of the pattern? |
| Answer» SOLUTION :(a) between the CENTRAL maximum (zero WAVELENGTH difference) and the FIRST minimum (`1 // 2` wavelength difference), (b) 0.196 | |
| 3079. |
Calculate the critical angle for a pair of media, water and glass. Given refractive index of glass and water is 1.45 and 1.33 respectively. |
| Answer» SOLUTION :`66.49^(@)` | |
| 3080. |
A system of 2 capacitors of capacitance 2muF and 4muF is connected in series across potential difference of 6 V. The electric charge and energy stored in the system are |
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Answer» `10 mu C` and `30 mu J`  The capacitor are connected in series combination, So, `(1)/(C_(EQ))=(1)/(C_(1))+(1)/(C_(2))` `(1)/(C_(eq))=(1)/(2)+(1)/(4)` `rArr C_(eq)=(4)/(3)mu F` We KNOW that, `Q = CV = (4)/(3)xx 6 = 8 mu C` `Q=CV=(4)/(3)xx6=8 mu C` Energy, `E=(1)/(2)CV^(2)=(1)/(2)xx(4)/(3)xx(6)^(2)` `= 24 mu J` |
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| 3081. |
For a transistor, beta = 100. The value of a is |
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Answer» 1.01 As `ALPHA=(beta)/(1+beta)=(100)/(1+100)=(100)/(101)=0.99` |
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| 3082. |
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for the of sight communication ? A TV transmitting antenna is 81m tall. How much service area can it cover, if the receiving antenna is at the ground level ? |
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Answer» 3800 `KM^(2)` |
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| 3083. |
Three resistor 4 Omega, 6 Omega and 8 Omega, are combined in parallel.If the combination is connected to a battery of emf 25 V and negligible Internal resistance, then determine the current through each resistor and total current drawn from the battery. |
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Answer» SOLUTION :Given : `R_(1)=4 Omega, R_(2) = 6 Omega,R_(3)= 8OmegaV= EPSILON= 25 V, r=0` a. the total resistance of the PARALLEL COMBINATION `(1)/(R_(rho))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))` `(1)/(R_(rho))=(1)/(4)+(1)/(6)+(1)/(8)` `(1)/(R_(rho))=(6+4+3)/(24)=(13)/(24)` `R_(rho)=(24)/(13)=1.846 Omega` B. The current through `4 Omega` is `I_(1)=(V)/(R_(1))=(25)/(4)` `I_(1)= 6.25 A` The current through `6 Omega`is `I_(2) = (V)/(R_(2))=(25)/(6)=4.16` A The current through `8 Omega` is`I_(3)=(V)/(R_(3))=(25)/(8) ` `I_(3)= 3.125` Total current drawn from the batter is `1=1_(1)+1_(2)+1_(3)=6.25+4.16+3.125` 1=13.535 A |
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| 3084. |
When light falls on two polaroid sheets having their axies matually perpendicular, then it is |
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Answer» completly extriguished |
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| 3085. |
A body starting from rest moves with constant acceleration 'a'. This acceleration is a pr where p is a constant. What is the displacement of particle in a time interval=0 to t = t_1 ? |
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Answer» `(1)/(3)pt_(1)^(3)` dv=pt DT `:. v=int_(0)^(H) pt dt=1/2pt_(1)^(2)` But v=dx/dt `:. Dx=int_(0)^(t_1) 1/2pt_(1)^(2) dt` `=1/2p(t_(1)^(3))/(3)` `=(1)/(6)pt_(1)^(3)` |
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| 3086. |
Which of the following phenomenon doesn.t depend on total internal reflection ? |
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Answer» Work of optical FIBRE. (C) Phenomenon DEPENDS on refraction of light. |
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| 3087. |
The critical angle of a transparent crystal is 45^(0). Then its polarizing angle is |
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Answer» `THETA= tan^(-1)""(sqrt(2))` |
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| 3088. |
Two blocks A and B of masses 2kg and 3kg are connected by a light string as shown in the figure and placed on a horizontal surface. mu between all surfaces is 0.1 and g=10 ms^(-2). The acceleration of the system is, when the force applied F = 45 N |
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Answer» SOLUTION :`F-mu(m_(1)+m_(2))G=(m_(1)+m_(2))` `45-0.1(5)10=5a` `45-5=5a` `a=8 ms^(-2)` 
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| 3089. |
There are two spherical balls A and B of the same material with same surface finish but the radius of A is half than of B. If A and B are heated to the same temperature and allowed to cool then : |
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Answer» RATE of cooling of both is same `RARR""` rate of cooling `prop"(radius)"^(2)`. `:.` since radius of A is `1/2` od radius of sphere B. `:.` rate of cooling of A is `1/4` of rate of cooling of sphere B. Thus CORRECT choice is (d). |
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| 3090. |
यदि फलन fA से B में परिभाषित है तथा फलन g C से D में परिभाषित है तो gof परिभाषित होगा यदि- |
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Answer» A=B |
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| 3091. |
The wavelength of an electron of energy 100 eV will be |
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Answer» `1.2A^(@)` |
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| 3092. |
Match the List I with the List II and select the correct answer using the code given below the lists. |
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Answer» P-2, Q-3, R-1, S-4 |
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| 3093. |
Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two — the parent or the daughter nucleus — would have higher binding energy per nucleon ? |
| Answer» Solution :Since energy is RELEASED during FUSION PROCESS, hence the binding energy PER nucleon of the daughter nucleus is more. | |
| 3094. |
There are atomic (Calcium) clocks capable of measuring time with an accuracy of 1 part in 10^11. If two such clocks are operated to precision, then after running for 5000 years, these will record a difference of |
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Answer» 1 day |
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| 3095. |
(A) : In a cavity within a conductor, the electric field is zero. (R ): Charges in a conductor reside only at its surface. |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
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| 3096. |
A convex lens, of focal length 20 cm, has a point object placed on its principal axis at distance of 40 cm from it. A plane mirror is placed 30 cm behind the convex lens. Locate the position of image formed by this combination |
Answer» SOLUTION : We first CONSIDER the effect of the lens. For the lens, we have u=-40 cm and f= + 20 cm Using the lens formula, we get ` (1)/(v_1) - (1)/(-40) = 1/20 "" thereforev_1= + 40cm` Had there been the lens only the image WOULD have have been formed at `Q_1` . The plane mirror M is at a distance of 30 cm from the lens L. We can,therefore, think of `Q_1` as a VIRTUAL object, located at a distance of 10 cm, behind the plane mirror M. The plane mirror therefore forms a real image (of this virtual object `Q_1`) at Q, 10 cm in front of it. |
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| 3097. |
Resistance of a resistor wire is 5 Omega " at " 50" "^(@)C and 6 Omega " at " 100 " "^(@)C , then its resistance at 0" "^(@) Cwill be ..... |
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Answer» `2 Omega` `R_(t) = R_(0) (1 + alpha (t - t_(0))` `therefore 5 = R_(0) (1 + alpha (50- 0)) ` `therefore 5 = R_(0) (1 + 50 alpha) "" `.... (1) and 6 = `R_(0) (1 + alpha (100- 0)) ` ` therefore 6 = R_(0) ( 1+ 100 alpha)` .... (2) `therefore(5)/(6) = (1 + 50 alpha)/(1 + 100 alpha) ` `therefore5 + 500 alpha = 6 + 300 alpha` `therefore200 alpha = 1 ` `therefore alpha = (1)/(200 )` Now , by USING `alpha = (1)/(200)` in equation (1), `5 = R_(0)[ 1 + 50 xx (1)/(200) ]` `therefore 5 = R_(0) [ 1 + (1)/(4) ]` `therefore 5 = R_(0) xx (5)/(4)` `therefore R_(0) = 4 Omega` |
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| 3098. |
Unpolarized light of intensity 32Wm^(-2) passes through three polarizers such that transmission axes of the first and second polarizer make an angle 30^@ with each other and the transmission axis of the last polarizer is crossed with that of the first . The intensity of the final emerging light will be |
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Answer» `32 WM^(-2)` |
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| 3099. |
Each of two concentric spheres of radii 5 cm and 10 cm are given a charge of 10 mu C. (ii) The electric potential at a point situated at a distance of 8 cm from the centre is |
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Answer» `10.52 XX 10^(4) V` |
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| 3100. |
What is the effect of dielectric medium on the electric potential of a capacitor ? |
| Answer» Solution :The ELECTRIC potential DECREASES with the introduction of DIELECTRIC medium. | |