MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics


113 Questions MCQ Test Class 12 Physics 35 Years JEE Mains &Advance Past year Paper | MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics


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This mock test of MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics for JEE helps you for every JEE entrance exam. This contains 113 Multiple Choice Questions for JEE MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics (mcq) to study with solutions a complete question bank. The solved questions answers in this MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics quiz give you a good mix of easy questions and tough questions. JEE students definitely take this MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics exercise for a better result in the exam. You can find other MCQs Of Modern Physics, Past Year Questions JEE Main, Class 12, Physics extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is

Solution:

KEY CONCEPT : The energy of  nth orbit of hydrogen is given by

Therefore the energy required to remove electron from n = 2 is + 3.4 eV.

QUESTION: 2

At absolute zero, Si acts as

Solution:

Pure silicon, at absolute zer o, will contain al l the electrons in bounded state. The conduction band will be empty. So there will be no free electrons (in conduction band) and holes (in valence band) due to thermal agitation. Pure silicon will act as insulator.

QUESTION: 3

At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit

(i) electrons
(ii) protons
(iii) He2+
(iv) neutrons

The emission at instant can be

Solution:

Charged particles are deflected in magnetic field.

QUESTION: 4

Sodium and copper have work functions 2.3 eV and 4.5 eV respectively. Then the ratio of the wavelengths is nearest to 

Solution:

We know that work function is the energy required and energy E = hu

QUESTION: 5

Formation of covalent bonds in compounds exhibits

Solution:

For mation of covalent bond is best explain ed by molecular orbital theory.

QUESTION: 6

If N0 is the original mass of the substance of half-life period t1/2 = 5 years, then the amount of substance left after 15 years is

Solution:

After every half-life, the mass of the substance reduces to half its initial value.

QUESTION: 7

By increasing the temperature, the specific resistance of a conductor and a semiconductor

Solution:

Specific resistance is resistivity which is given by

where n  = no. of free electrons per unit volume and τ  = average relaxation time For a conductor with  rise in temperature n increases and τ decreases. But decrease in t is more dominant than increase in n resulting an increase in the value of ρ.
For a semiconductor with rise in temperature, n increases and τ decreases. But the increase in n is more dominant than decrease in τ resulting in a decrease in the value of ρ.

QUESTION: 8

The energy band gap is maximum in

Solution:

The energy band gap is maximum in insulators.

QUESTION: 9

The part of a transistor which is most heavily doped to produce large number of majority carriers is

Solution:

Emitter sends the majority charge carrriers towards the collector. Therefore emitter is most heavily doped.

QUESTION: 10

Which of the following are not electromagnetic waves?

Solution:

β -rays are fast moving beam of electrons.

QUESTION: 11

A strip of copper and another of germanium are cooled from room temperature to 80K. The resistance of 

Solution:

The resist ance of metal (like Cu) decr eases with decrease in temperature whereas the resistance of a semi-conductor (like Ge) increases with decrease in temperature.

QUESTION: 12

Which of the following radiations has the least wavelength ?

Solution:

The electromagnetic spectrum is as follows

∴ γ-rays has least wavelength

QUESTION: 13

When a U238 nucleus originally at rest, decays by emitting an alpha particle having a speed ‘u’, the recoil speed of the residual nucleus is

Solution:

(c) : Linear momentum is conserved

QUESTION: 14

The difference in the variation of resistance with temeperature in a metal and a semiconductor arises essentially due to the difference in the

Solution:

When the temperature increases, certain boun ded electrons become free which tend to promote conductivity. Simultaneously  number of collisions between electrons and positive kernels increases

QUESTION: 15

A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is

Solution:

QUESTION: 16

A nucleus with Z = 92 emits the following in a sequence:

Then Z of the resulting nucleus is

Solution:

The number of  α - particles released = 8 Therefore the atomic number should decrease by 16
The number of β- -particles released = 4
Therefore the atomic number should increase by 4.
Also the number of β+ particles released is 2, which should decrease the atomic number by 2.
Therefore the final atomic number is
92 –16 + 4 – 2 = 78

QUESTION: 17

Two identical photocathodes receive light of frequencies f1 and f2. If the velocites of the photo electrons (of mass m) coming out are respectively v1 and v2, then

Solution:

For one photocathode

For another photo cathode

Subtracting (ii) from (i) we get

QUESTION: 18

Which of the following cannot be emitted by radioactive substances during their decay ?

Solution:

The radioactive substances emit α -particles (Helium nucleus), β – particles (electrons) and neutrinoes.

QUESTION: 19

In the nuclear fusion reaction 

given that the repulsive potential energy between the two nuclei is ~ 7.7 × 10 -14 J , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s Constant k = 1.38 × 10 -23 J/K]

Solution:

The average kinetic energy per molecule 

This kinetic energy should be able to provide the repulsive potential energy

QUESTION: 20

Which of the following atoms has the lowest ionization potential ?

Solution:

The ionisation potential increases from left to right in a period and decreases from top to bottom in a group.
Therefore ceasium will have the lowest ionisation potential.

QUESTION: 21

The wavelengths involved in the spectrum of deuterium  are slightly different from that of  hydrogen spectrum, because

Solution:

The wavelength of spectrum is given by

where m = mass of electron
M = mass of nucleus.
For different M, R is different and therefore λ is different

QUESTION: 22

In the middle of the depletion layer of a reverse- biased p-n junction, the

Solution:

As in reverse bias, the current through the 0000 is zero through the electric field is also zero.

QUESTION: 23

If the binding energy of  the electron in a hydrogen atom is 13.6eV,  the energy required to remove the electron from the first excited state of Li ++ is

Solution:

For lithium ion  Z = 3  ; n = 2  (for first excited state)

QUESTION: 24

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Solution:

Momentum of photon = E/c

Change in momentum = 2E/c

= momentum transferred to the surface(the photon will reflect with same magnitude of momentum in opposite direction)

QUESTION: 25

According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal Vs the frequency, of the incident radiation gives as straight the whose slope 

Solution:

From E quation K .E = hv-φ
slope of graph of  K.E  & v is h (Plank's constant) which is same for all metals

QUESTION: 26

The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately.

Solution:

For the longest wavelength to emit photo electron

QUESTION: 27

A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of 2 : 1. The ratio of their nuclear sizes will be

Solution:

From conservation of momentum m1v1 = m2v2

QUESTION: 28

The binding energy per nucleon of deuteron  and helium nucleus  is 1.1 MeV and 7 MeV respectively.. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

Solution:

The nuclear reaction of  process is  He Energy released  = 4 × (7) – 4(1.1) = 23.6 MeV

QUESTION: 29

An α-particle of energy 5 MeV is scattered through 180º by a fixed uranium nucleus. The distance of closest approach is of the order of

Solution:

KEY CONCEPT :

Distance of closest approach

QUESTION: 30

When npn transistor is used as an amplifier

Solution:

Electrons move from base to emmitter.

QUESTION: 31

For a transistor amplifier in common emitter configuration for load impedance of 1k Ω (h fe = 50 and hoe = 25) the current gain is

Solution:

In common emitter configuration current gain

QUESTION: 32

A piece of copper and another of germanium are cooled from room temperature to 77K, the resistance of

Solution:

Copper is a conductor, so its resistance decreases on decreasing temperature as thermal agitation decreases,; whereas germanium is semiconductor therefore  on decreasing temperature resistance increases.

QUESTION: 33

The manifestation of band structure in solids is due to

Solution:

Pauli’s exclusion principle.

QUESTION: 34

When p-n junction diode is forward biased then

Solution:

Both the depletion region and barrier height is reduced.

QUESTION: 35

If radius of the  nucleus is estimated to be 3.6 fermi then the radius of  nucleus be nearly

Solution:

KEY CONCEPT : R = R0(A)1/3

QUESTION: 36

Starting with a sample of pure  of it decays into Zn in 15 minutes. The corresponding half life is

Solution:

7/8 of Cu decays in 15 minutes.

⇒ T = half life period = 15/3 = 5 minutes

QUESTION: 37

A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 1/2 m away,, the number of electrons emitted by photocathode would

Solution:

When intensity becomes 4 times, no. of photoelectrons emitted would increase by 4 times, since number of electrons emitted per second is directly proportional to intensity.

QUESTION: 38

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor is

Solution:

Band gap = energy of photon of wavelength 2480 nm.
So,

QUESTION: 39

The intensity of gamma radiation from a given source is I.On passing through 36 mm of lead, it is reduced to 1/8. The thickness of lead which will reduce the intensity to I/2 will be

Solution:

KEY CONCEPT :  Intensity I = I0 .e -μd ,
Applying logarithm on both sides,

Dividing (i) by (ii),

QUESTION: 40

In a common base amplifier, the phase difference between the input signal voltage and output voltage is

Solution:

Zero; In common base amplifier circuit, input and output voltage are in the same phase.

QUESTION: 41

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

Solution:

E will be maximum for the transition for which  is maximum. Here n2 is the higher energy
level.

Clearly,  is maximum for the thirdtran sition, i.e. 2 → 1 . I transition represents the absorption of energy.

QUESTION: 42

If the kinetic energy of a free electron doubles, it’s deBroglie wavelength changes by the factor

Solution:

de-Broglie wavelength,

 

If K.E is doubled, wavelength becomes 

QUESTION: 43

A nuclear transformation is denoted by X (n, α)  Which of the following is the nucleus of element X ?

Solution:

On comparison,

A = 7 + 4 – 1 = 10, z = 3 + 2 – 0 = 5

It is boron 5B10

QUESTION: 44

In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be

Solution:

Input frequency, f = 50 Hz  

For full wave rectifier, 

QUESTION: 45

In a common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β) will be

Solution:

QUESTION: 46

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

Solution:

This range lies in ultra violet range.

QUESTION: 47

An alph a nucleus of energy  bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Solution:

Work done to stop the a particle is equal to K.E.

QUESTION: 48

The time taken by a photoelectron to come out after the photon strikes is approximately

Solution:

The order of time is 10–105.

QUESTION: 49

When 3Li7 nuclei are bombar ded by pr oton s, and th e resultant nuclei are 4Be8, the emitted particles will be

Solution:

QUESTION: 50

The energy spectrum of β-particles [number N(E) as a function of β-energy E] emitted from a radioactive source is

Solution:

The range of energy of β-particles is from zero to some maximum value.

QUESTION: 51

A solid which is not transparent to visible light and whose conductivity increases with temperature is formed by

Solution:

Van der Waal's bonding is attributed to the attractive forces between molecules of a liquid. The conductivity of semiconductors (covalent bonding) and insulators (ionic bonding) increases with increase in temperature while that of metals (metallic bonding) decreases.

QUESTION: 52

If the ratio of the concentration of electrons to that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4, then what is the ratio of their drift velocities?

Solution:

QUESTION: 53

The circuit has two oppositively connected ideal diodes in parallel. What is the current flowing in the circuit?

Solution:

D2 is forward biased whereas D1 is reversed biased.
So effective resistance of the circuit

R = 4 + 2 = 6Ω

QUESTION: 54

In the following, which one of the diodes reverse biased?

Solution:

p-side con nected to low potential and n-side is connected to high potential.

QUESTION: 55

The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows

Solution:

As λ decreases, n increases and hence the speed of photoelectron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.

QUESTION: 56

If the binding energy per nucleon in  and  nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction  energy of proton must be

Solution:

Let E be the energy of proton, then
E + 7 × 5.6 = 2 × [4 × 7.06]
⇒ E = 56.48 - 39.2= 17.28MeV

QUESTION: 57

The 'rad' is the correct unit used to report the measurement of

Solution:

The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue.
Absorbed dose is expressed in rad. A rad is equal to 100 ergs of energy absorbed by 1 gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist L. H. Gray); one gray equals 100 rad.

QUESTION: 58

If the lattice constant of this semiconductor is decreased, then which of the following is correct?

Solution:

A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cells in various directions is called its lattice parameters or constants. Increasing these lattice constants will increase or widen the band-gap (Eg), which means more energy would be required by electrons to reach the conduction band from the valence band. Automatically Ec and Ev decreases.

QUESTION: 59

The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is

Solution:

Erms = 720

The average total energy density

QUESTION: 60

If MO is the mass of an oxygen isotope 8O17 ,MP and MN are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

Solution:

Binding energy

= [ZMP + (A – Z)MN – M]c2
= [8MP + (17 – 8)MN – M]c2
= [8MP + 9MN – M]c2
= [8MP + 9MN – Mo]c2

QUESTION: 61

In gamma ray emission from a nucleus

Solution:

There is no change in the proton number and the neutron number as the g-emission takes place as a result of excitation or de-excitation of nuclei. γ-rays have no charge or mass.

QUESTION: 62

If in a p-n junction diode, a square input signal of 10 V is applied as shown

Then the output signal across RL will be

Solution:

The current will flow through RL when the diode is forward biased.

QUESTION: 63

Photon of frequency v has a momentum associated with it. If c is the velocity of light, the momentum is

Solution:

Energy of a photon of frequency n is given by E = hv .

QUESTION: 64

The half-life period of a radio-active element X is same as the mean life time of another radio-active element Y. Initially they have the same number of atoms. Then

Solution:

According to question,


Now, the rate of decay is given by

∴ Y will decay faster than X. [∵N is some]

QUESTION: 65

Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate ?

Solution:

Si and Ge are semiconductors but C is an insulator.
Also, the conductivity of Si and Ge is more than C because the valence electrons of Si, Ge and C lie in third, fourth and second orbit respectively.

QUESTION: 66

Which of the following transitions in hydrogen atoms emit photons of highest frequency?

Solution:

We have to find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options (c) and (d).
Frequency is given by

For transition from n = 6 to n = 2,

QUESTION: 67

Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

Q. Electrons accelerated by potential V are diffracted from a crystal. If d = 1Å and i = 30°, V should be about (h = 6.6 × 10 – 34 Js, me = 9.1 × 10–31 kg, e = 1.6 × 10 – 19 C)

Solution:

Using Bragg's equation 2d sinθ = nλ

Here n = 1, θ = 90 – i = 90 – 30 = 60°
∴ 2d sin θ = λ      ......(i)

QUESTION: 68

Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

Q. If a strong diffraction peak is observed when electrons are incident at an angle ‘i’ from the normal to the crystal planes with distance ‘d’ between them (see figure), de Broglie wavelength λdB of electrons can be calculated by the relationship (n is an integer)

Solution:

2d cos i = nλdB

QUESTION: 69

This question contains Statement-1 and statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. 

Statement-1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and

Statement-2 : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z.

Solution:

We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Therefore statement (1) is correct.
The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing Z and for light nuclei, B.E/nucleon increases with increasing Z.

QUESTION: 70

A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor?

Solution:

It is a n-p-n transistor with R as base.

QUESTION: 71

Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be ‘rn’ and the kinetic energy of the electron to be ‘Tn’.
Then which of the following is true

Solution:

When  
⇒ mv2 = constat ⇒ kinetic energy is constant
⇒ T is  independent of n.

QUESTION: 72

In the circuit below, A and B represent two inputs and C represents the output.

The circuit represents

Solution:

The truth table for the above logic gate is : This truth table follows the boolean algebra  C = A + B which is for OR gate

QUESTION: 73

The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :

Solution:

It is given that transition from the state n = 4 to n = 3 in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is 5 → 4 .

QUESTION: 74

The surface of a metal is illuminted with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : 

(hc = 1240 eV.nm)

Solution:

QUESTION: 75

The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions :

(i) A + B → C + ε
(ii) C → A + B  + ε
(iii) D + E → F + ε and
(iv) F→ D + E + ε,

where ε is the energy released? In which reactions is ε positive?

Solution:

For A + B → C + ε,ε is positive. This is because Eb for C is greater than the Eb for A and B.
Ag ai n for F → D +E + ε,ε is positive. Th is is because Eb for D and E is greater than Eb for F .

QUESTION: 76

The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform.

Output is

Solution:

Here  Thus it is an AND gate for which truth table is

QUESTION: 77

A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.

The current (I) in the resistor (R) can be shown by :

Solution:

We know that a single p-n junction diode connected to an a-c source acts as a half wave rectifier [Forward biased in one half cycle and reverse biased in the other half cycle].

QUESTION: 78

Statement -1 : When ultraviolet light is inciden t on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax .When the ultraviolet light is replaced by X-rays, both V0 and Kmax increase.

Statement -2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

Solution:

We know that

where, φ is the work function .
Hence, as v increases (note that frequency of X-rays is greater than that of U.V. rays), both V0 and Kmax increase. So statement - 1 is correct

QUESTION: 79

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

Q. The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then

Solution:

In nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleus.

QUESTION: 80

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

Q. The speed of daughter nuclei is

Solution:

By conservation of energy,

where v is the speed of the daughter nuclei

QUESTION: 81

A radioactive nucleus (initial mass number A and atomic number Z emits 3 a - particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Solution:

QUESTION: 82

The combination of gates shown below yields

Solution:

The final boolean expression is,

QUESTION: 83

If a source of power 4kW produces 1020  photons/second , the radiation belongs to a part of the spectrum called

Solution:

QUESTION: 84

This question has Statement – 1 and Statement – 2. Of the four choices given after the statements, choose the one that best describes the two statements.             

Statement – 1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.
Statement – 2 : The state of ionosphere varies from hour to hour, day to day and season to season.

Solution:

For long distance communication, sky wave signals are used.
Also, the state of ionosphere varies every time. So, both statements are correct.

QUESTION: 85

Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is :

Solution:

Energy of excitation,

QUESTION: 86

The half life of a radioactive substance is 20 minutes. The approximate time interval (t2 – t1) between the time t2 when 2/3 of it had decayed and time t1 when 1/3 of it had decayed is :

Solution:

Number of undecayed atom after time t2 ;

Number of undecayed atom after time t1;

Solving (iii) and (iv), we get t2 – t1 = 20 min

QUESTION: 87

This question has Statement – 1 and Statement – 2. Of the four choices given after the statements, choose the one that best describes the two statements. 

Statement – 1: A metallic surface is irradiated by a monochromatic light of frequency v > v0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency incident on the surface is doubled, both the Kmax and V0 are also doubled.

Statement – 2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surfac

Solution:

By Einstein photoelectric equation,
Kmax = eV0 = hv – hv0
When v is doubled, Kmax and V0 become more than double.

QUESTION: 88

Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be :

Solution:

The possible number of the spectral lines is given

QUESTION: 89

Truth table for system of four NAND gates as shown in figure is :

Solution:

By expanding this Boolen expression

Thus the truth table for this expression should be (1).

QUESTION: 90

A radar has a power of 1kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m.
The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 × 106m) is :

Solution:

Let d is the maximum distance, upto it the objects From ΔAOC




QUESTION: 91

Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron= 1.6725 × 10–27 kg, mass of proton = 1.6725 × 10–27 kg,mass of electron = 9 × 10–31 kg).

Solution:

The mass defect during the process

= 1.6725 × 10–27 – (1.6725 × 10–27+ 9 × 10–31kg) =  – 9 × 10–31 kg

The energy released during the process

E = Δmc2

E =  9 × 10–31× 9 × 1016 = 81 × 10–15 Joules

 

QUESTION: 92

A diatomic molecule is made of two masses m1 and m2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by : (n is an integer)

Solution:

The energy of the system of two atoms of diatomic molecule 

where I = moment of inertia

ω = Angular velocity = L/I,

L = Angular momentum

QUESTION: 93

A diode detector is used to detect an amplitudemodulated wave of 60% modulation by using a condenser of capacity 250 picofarad in parallel with a load resistance 100 kilo ohm.
Find the maximum modulated frequency which could be detected by it.

Solution:

Given : Resistance R = 100 kilo ohm = 100 × 10 3 ω Capacitance C = 250 picofarad = 250 × 10–12F

= 2.5 × 10–5 sec

The higher frequency whcih can be detected with tolerable distortion is

This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated singnal voltage for proper detection of mdoulated signal.

QUESTION: 94

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

Solution:

From question,

B0 = 20 nT = 20 × 10–9T

(∴ velocity of light in vacuum C = 3 × 108 ms–1)

   = 6 V/m.

QUESTION: 95

The anode voltage of a photocell is kept fixed. Th e wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

Solution:

As λ is increased, there will be a value of λ above which photoelectrons will be cease to come out so photocurrent will become zero. Hence (d) is correct answer.

QUESTION: 96

The I-V characteristic of an LED is

Solution:

For same value of current higher value of voltage is required for higher frequency hence (1) is correct answer.

QUESTION: 97

In a hydrogen like atom electron make transition from an energy level with quantum number n to another with quantum number (n – 1). If n>>1, the frequency of radiation emitted is proportional to :

Solution:

QUESTION: 98

Th e cur r en t voltage r elation of a diode is given by , where the applied voltage V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ±0.01V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?

Solution:

The current voltage relation of diode is

(By exponential function)

QUESTION: 99

During the propagation of electromagnetic waves in a medium:

Solution:

Electric energy density

Magnetic energy density

Energy is equally divided between electric and magnetic field

QUESTION: 100

The radiation corresponding to 3→2 transition of hydrogenatom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4  T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

Solution:

Radius of circular path followed by electron is given by,

For transition between 3 to 2.

Work function = 1.88 eV – 0.8 eV = 1.08 eV ≈ 1.1eV

QUESTION: 101

Hydrogen  Deuterium singly ionised Helium  and doubly ionised lithium  all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are respectively then approximately which one of the following is correct?

Solution:

Wave number 

By question n = 1 and  n1 = 2

QUESTION: 102

The forward biased diode connection is:

 

Solution:

For forward bias, p-side must be at higher potential than n-side. ΔV = (+)Ve

QUESTION: 103

Match List - I (Electromagnetic wave type) with List - II (Its association/application) and select the correct option from the choices given below the lists:

List 1                                                                        List 2

1. Infrared waves                                                        (i) To treat muscular strain

2. Radio waves                                                           (ii) For  broadcasting

3. X-rays                                                                     (iii) To detect fracture of bones

4. Ultraviolet rays                                                      (iv) Absorbed by the ozone layer of the atmosphere

1  2  3 4

Solution:

(1) Infrared rays are used to treat muscular strain because these are heat rays. (2) Radio waves are used for broadcasting because these waves have very long wavelength ranging from few centimeters to few hundred kilometers (3) X-rays are used to detect fracture of bones because they have high penetrating power but they can't penetrate through denser medium like bones. (4) Ult r aviol et r a ys ar e absorbed by ozone of the atmosphere.

QUESTION: 104

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is :

 

Solution:

∴E0 = √6 = 2.45V/m

QUESTION: 105

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are :

Solution:

Amplitude modulated wave consists of three frequencies are

i.e.  2005 KHz, 2000KHz, 1995 KHz

 

QUESTION: 106

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion : 

Solution:

QUESTION: 107

Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct option from the choices given below the list:

Solution:

Frank-Hertz experiment - Discrete energy levels of atom Photoelectric effect - Particle nature of light Davison - Germer experiment - wave nature of electron.

*Multiple options can be correct
QUESTION: 108

For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β is :

Solution:

We know that 

Also Ie = Ib + Ic

Option (b) and (d) are therefore correct.

 

*Multiple options can be correct
QUESTION: 109

If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is :

Solution:

In case of an 'OR' gate the input is zero when all inputs are zero. If any one input is ' 1', then the output is '1'.

*Multiple options can be correct
QUESTION: 110

 Choose the correct statement : 

Solution:

In amplitude modulation, the amplitude of the high frequency carrier wave made to vary in proportional to the amplitude of audio signal.

*Multiple options can be correct
QUESTION: 111

Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to  the speed of the fastest emitted electronwill be:

Solution:

*Multiple options can be correct
QUESTION: 112

Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed number of A and B nuclei will be :

Solution:

For A = 20 min, t = 80 min, number of half lifes n = 4

∴ Nuclei remaining  Therefore nuclei decayed 

For B = 40 min., t = 80 min, number of half lifes n = 2

∴ Nuclei remaining  =N0/2.Therefore nuclei decayed 

∴ Required ratio 

*Multiple options can be correct
QUESTION: 113

Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d) :

Solution:

Graph (a) is for a simple diode.
Graph (b) is showing the V Break down used for zener diode.
Graph (c) is for solar cell which shows cut-off voltage and open circuit current.
Graph (d) shows the variation of resistance h and hence current with intensity of light.