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Test: MCQs (One or More Correct Option): Modern Physics | JEE Advanced

38 Questions MCQ Test Class 12 Physics 35 Years JEE Mains &Advance Past year Paper | Test: MCQs (One or More Correct Option): Modern Physics | JEE Advanced

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Attempt Test: MCQs (One or More Correct Option): Modern Physics | JEE Advanced | 38 questions in 80 minutes | Mock test for JEE preparation | Free important questions MCQ to study Class 12 Physics 35 Years JEE Mains &Advance Past year Paper for JEE Exam | Download free PDF with solutions

*Multiple options can be correct

QUESTION: 1

The shortest wavelength of X-rays emitted from an X-ray tube depends on

A.
the current in the tube
B.
the voltage applied to the tube
C.
the nature of the gas in tube
D.
the atomic number of the target material

Solution:

Note : Shortest wavelength means highest frequency.
This means highest energy.
The energy of X-rays depends on the accelerating voltage provided in the X-ray tube.
Also, accordin g to Moseley’s law √n= a(Z –b) .
Thus the frequency also depends on the atomic number.

*Multiple options can be correct

QUESTION: 2

The threshold wavelength for photoelectric emission from a material is 5200 Å. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a

A.
50 watt infrared lamp
B.
1-watt infra-red lamp
C.
50 watt ultraviolet lamp
D.
1-watt ultraviolet lamp

Solution:

The threshold wavelength is 5200Å. For ejection of electrons, the wavelength of the light should be less than 5200 Å, so that frequency increases and hence the energy of incident photon increases. U.V light has less wavelength than 5200 Å.

*Multiple options can be correct

QUESTION: 3

From the following equations pick out the possible nuclear fusion reactions

A.
₆C¹³+ ₁H¹ → ₆C¹⁴ + 4.3MeV
B.
₆C¹²+ ₁H¹ → ₇N¹³ + 2MeV
C.
₇N¹⁴ + ₁H¹ → ₈O¹⁵ + 7.3MeV
D.
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴+ ₀n¹
+ ₀n¹ + γ+ 200MeV

Solution:

Nuclear fusion occurs when two or more lighter nuclei combine to form a heavier nucleus with release of a huge amount of energy.

*Multiple options can be correct

QUESTION: 4

In Bohr ’s model of the hydrogen atom

A.
the radius of the nth orbit is proportional to n²
B.
the total energy of the electron in nth orbit is inversely proportional to n
C.
the angular momentum of the electron in an orbit is an integral multiple of h/2π
D.
the magnitude of potential energy of the electron in any orbit is greater than its K.E.

Solution:

We know, r_n ∝ n²

| P.E. | = 2 × | K.E. |

*Multiple options can be correct

QUESTION: 5

Select the correct statement from the following

A.
A diode can be used as a rectifier
B.
A triode cannot be used as a rectifier
C.
The current in a diode is always proportional to the applied voltage
D.
The linear portion of the I–V characteristic

Solution:

A diode can be used as a rectifier.

*Multiple options can be correct

QUESTION: 6

For a given plate voltage, the plate current in a triode valve is maximum when the potential of

A.
the grid is positive and plate is negative
B.
the grid is zero and plate is positive
C.
the grid is negative and plate is positive
D.
the grid is positive and plate is positive

Solution:

is the correct option. The electrons emitted by emitter are collected to the maximum by the plate in this case.

*Multiple options can be correct

QUESTION: 7

The X-ray beam coming from an X-ray tube will be

A.
monochromatic
B.
having all wavelengths smaller than a certain maximum wavelength
C.
having all wavelengths larger than a certain minimum wavelength
D.
having all wavelengths lying between a minimum and a maximum wavelength

Solution:

is the correct option.

*Multiple options can be correct

QUESTION: 8

The mass number of a nucleus is

A.
always less than its atomic number
B.
always more than its atomic number
C.
sometimes equal to its atomic number
D.
sometimes more than and sometimes equal to its atomic number

Solution:

In the case of hydrogen, atomic number = mass number In the other atoms, atomic number < mass number.

*Multiple options can be correct

QUESTION: 9

Four physical quantities are listed in Column I. Their values are listed in Column II in a random order:

The correct matching of Columns I and II is given by

A.
a – e, b – h, c – g, d – f
B.
a – e, b – g, c – f, d – h
C.
a – f, b –e, c – g, d – h
D.
a – f, b – h, c – e, d – g.

Solution:

*Multiple options can be correct

QUESTION: 10

Photoelectric effect suppor ts quantum nature of light because

A.
there is a minimum frequency of light below which no photoelectrons are emitted
B.
the maximum kinetic energy of photo electrons depends only on the frequency of light and not on its intensity
C.
even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately
D.
electric charge of the photoelectrons is quantized

Solution:

*Multiple options can be correct

QUESTION: 11

During a negative beta decay

A.
an atomic electron is ejected
B.
an electron which is already present within the nucleus is ejected
C.
a neutron in the nucleus decays emitting an electron
D.
a part of the binding energy of the nucleus is converted into an electron

Solution:

*Multiple options can be correct

QUESTION: 12

During a nuclear fusion reaction

A.
a heavy nucleus breaks into two fragments by itself
B.
a light nucleus bombarded by thermal neutrons breaks up
C.
a h eavy nucleus bom barded by ther mal n eutr on s breaks up
D.
two light nuclei combine to give a heavier nucleus and possibly other products

Solution:

*Multiple options can be correct

QUESTION: 13

The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation

A.
the intensity increases
B.
the minimum wavelength increases
C.
the intensity remain unchanged
D.
the minimum wavelength decreases

Solution:

*Multiple options can be correct

QUESTION: 14

A freshly prepared radioactive source of half life 2 hr emits radiation of intensity which is 64 times the permissibe safe level. The minimum time after which it would be possible to work safely with this source is

A.
6 hr
B.
12 hr
C.
24 hr
D.
128 hr

Solution:

Note: The intensity of radiation emitted is proportional to the rate of decay which in turn is proportional to number of atoms left (radioactive).

*Multiple options can be correct

QUESTION: 15

The impurity atoms with which pure silicon should be doped to make a p-type semiconductor are those of

A.
phosphorus
B.
boron
C.
antimony
D.
aluminium

Solution:

Boron and Aluminium are trivalent impurities.

*Multiple options can be correct

QUESTION: 16

Two identical p-n junctions may be connected in series with a battery in three ways, fig. The potential drops across the two p – n junctions are equal in

A.
circuit 1 and circuit 2
B.
circuit 2 and circuit 3
C.
circuit 3 and circuit 1
D.
circuit 1 only

Solution:

Since the p-n junction arrangement are in series, therefore the potential drop across a p-n junction will be proportional to their resistances. When the resistances will be equal, the potential drops will be equal. In circuit I, the two p-n junctions are attached such that one is forward biased (low resistance) and other is reverse biased (high resistance). Whereas in the other two circuits both are either forward biased or reversed biased.

*Multiple options can be correct

QUESTION: 17

The decay constant of a radioactive sample is λ . The halflife and mean-life of the sample are respectively given by

A.
1/ λ and (ln 2)/ λ
B.
(ln 2)/ λ and 1/ λ
C.
λ (ln 2) and 1/ λ
D.
λ /(ln 2) and 1/ λ

Solution:

*Multiple options can be correct

QUESTION: 18

When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then

A.
the stopping potential will be 0.2 volt
B.
the stopping potential will be 0.6 volt
C.
the saturation current will be 6.0 mA
D.
the saturation current will be 2.0 mA

Solution:

Since the stopping potential depends on the frequency and not on the intensity and the source is same, the stopping potential remains unaffected. The saturation current depends on the intensity of incident light on the cathode of the photocell which in turn depends on the distance of the source from cathode. The intensity (I) of light is inversely proportional to the square of the distance between the light source and photocell.

*Multiple options can be correct

QUESTION: 19

In an n-p-n transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector,

A.
the emitter current will be 9 mA
B.
the base current will be 1 mA
C.
the emitter current will be 11 mA
D.
the base current will be – 1 mA

Solution:

I_c = 10 mA

90% of electrons emitted pr produce a collector current of 10 mA. The base current

Now, I_e = I_b + I_c = 1 + 10 = 11mA

*Multiple options can be correct

QUESTION: 20

A star initially has 10⁴⁰ deuterons. It produces energy via the processes and If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of

The masses of the nuclei are as follows :

M (H²) = 2.014 amu;
M (p) = 1.007 amu; M(n) = 1.008 amu; M (He⁴) = 4.001amu.

A.
10⁶s.
B.
10⁸s.
C.
10¹² s.
D.
10¹⁶ s.

Solution:

₁H² + ₁H² → ₁H³ + p
₁H² + ₁H³ → ₂He⁴ + n
Net Reaction 3₁H² → ₂He⁴ + p + n
Δm = 3 (2.014) – [4.001 + 1.007 + 1.008] = 0.026
3 deuterons release 3.87 × 10^–12 J

= 1.29 × 10²⁸J

*Multiple options can be correct

QUESTION: 21

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, T_A eV and de Broglie wavelength λ_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is T_B = (T_A- 1.50) eV. If the de Broglie wavelength of these photoelectrons is λ_B = 2λ_A, then

A.
The work function of A is 2.25 eV
B.
The work function of B is 4.20 eV
C.
T_A = 2.00 eV
D.
T_B = 2.75 eV

Solution:

For metal A

4.25 = W_A + T_A ...(i)

For metal B

4.7 = (T_A – 1.5) + W_B ...(iii)

Dividing equation (iv) by (ii),

[∵ λ B = 2λA given]

⇒ 4TA – 6 =TA ⇒ TA = 2 eV
From (i), WA = 2.25 eV
From (iii), WB = 4.2 eV
Also T_B = T_A – 1.5 ⇒ T_B = 0.5eV

*Multiple options can be correct

QUESTION: 22

Which of the following statement(s) is (are) correct?

A.
The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons
B.
The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons
C.
In nuclear fission, energy is released by fusing two nuclei of medium mass (approximately 100 amu)
D.
In nuclear fission, energy is released by fragmentation of a very heavy nucleus

Solution:

*Multiple options can be correct

QUESTION: 23

Holes are charge carriers in

A.
intrinsic semiconductors
B.
ionic solids
C.
p-type semiconductors
D.
metals

Solution:

Holes are electron vacancies which participate in electrical conductivity. These are produced in semiconductors.

*Multiple options can be correct

QUESTION: 24

A transistor is used in the common emitter mode as an amplifier. Then

A.
the base-emitter junction is forward-biased
B.
the base-emitter junction is reverse-biased
C.
the input signal is connected in series with the voltage applied to bias the base-emitter junction
D.
the input signal is connected in series with the voltage applied to bias the base-collector junction

Solution:

The circuit for a p-n-p transistor used in the common emitter mode as an amplifier is shown in figure. The base emitter junction is forward-biased and the input signal is connected in series with the voltage applied to bias the base emitter junction.

*Multiple options can be correct

QUESTION: 25

Let m_p be the mass of a proton, m_n the mass of a neutron, M₁ the mass of a nucleus and M₂ the mass of a nucleus. Then

A.
M₂ = 2M₁
B.
M₂ > 2M₁
C.
M₂ < 2M₁
D.
M₁ < 10 (m_n+m_p)

Solution:

KEY CONCEPT : Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles.

is made up of 10 protons plus 10 neutrons.
Therefore, mass of nucleus
M₁ < 10 (m_p + m_n)
Note :
Heavier the nucleus, more is the mass defect.
20 (m_n + m_p) – M₂ > 10 (m_p + m_n) – M₁
Thus, 10 (m_n + m_p) > M₂ – M₁
or M₂ < M₁ + 10 (m_p + m_n)
Now, since M₁ < 10 (m_p + m_n)
∴ M₂ < 2M₁

*Multiple options can be correct

QUESTION: 26

The electron in a hydrogen atom makes a transition n₁→ n₂ where n₁ and n₂ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of n₁ and n₂ are

A.
n₁ = 4,n₂=2
B.
n₁ = 8,n₂ = 2
C.
n₁ = 8,n₂ = 1
D.
n₁ = 6,n₂ = 3

Solution:

The time period of the electron in a Bohr orbit is given

Since for the nth Bohr orbit, mvr = n (h/2π), the time period becomes

Since the radius of the orbit r depends on n, we replace r. Bohr radius of a hydrogen atom is

It is given that T₁ / T₂ = 8, hence, n₁/n₂ = 2.

*Multiple options can be correct

QUESTION: 27

The half-life of ¹³¹I is 8 days. Given a sample of ¹³¹I at time t = 0, we can assert that

A.
no nucleus will decay before t = 4 days
B.
no nucleus will decay before t = 8 days
C.
all nuclei will decay before t = 16 days
D.
a given nucleus may decay at any time after t = 0

Solution:

The result follows from the formula based on laws of radioactive decay N = N₀e^–λt The nucleus start decaying after time t = 0

*Multiple options can be correct

QUESTION: 28

In a p-n junction diode not connected to any circuit,

A.
the potential is the same everywhere
B.
the p-type side is at a higher potential than the n-type side
C.
there is an electric field at the junction directed from the n-type side to the p-type side
D.
there is an electric field at the junction directed from the p-type side to the n-type side

Solution:

At junction a potential barrier/depletion layer is formed as shown, with n-side at higher potential and p-side at lower potential. Therefore, there is an electric field at the junction directed from the n-side to p-side

*Multiple options can be correct

QUESTION: 29

X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from

A.
0 to ∞
B.
λ_min to ∞ where λ_min > 0
C.
0 to λ_max where λ_max < ∞
D.
λ_min to λ_max where 0 < λ_min < λ_max < ∞

Solution:

The continuous X-ray spectrum is shown in figure.

All wavelengths > λ_min are found.

Here V is the applied voltage.

*Multiple options can be correct

QUESTION: 30

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

A.
540 nm
B.
400 nm
C.
310 nm
D.
220 nm

Solution:

*Multiple options can be correct

QUESTION: 31

The half-life period of a radioactive element X is same as the mean-life time of another radioactive element Y. Initially both of them have the same number of atoms. Then

A.
X and Y have the same decay rate initially
B.
X and Y decay at the same rate always
C.
Y will decay at a faster rate than X
D.
X will decay at a faster rate than Y

Solution:

∴ λ_x = 0.693 λ_Y
λ_x < λ_Y. Now, rate of decay = λN
Initially, number of atoms (N) of both are equal but since λ_Y < λ_x, therefore Y will decay at a faster rate than x.

*Multiple options can be correct

QUESTION: 32

The graph between the stopping potential (V₀) and is shown in the figure. φ₁ , φ₂ and φ₃ are work functions, which of the following is/are correct

A.
φ₁ : φ₂ : φ₃ = 1 : 2 : 4
B.
φ₁ : φ₂ : φ₃ = 4 : 2 : 1
C.
D.
ultravioletlight can be used to emit photoelectrons from metal 2 and metal 3 only

Solution:

Therefore option (a) is correct

By Einstein’s photoelectric equation,

Comparing equation (i) by y = mx + c, we get the slope of the line

⇒ Option (c) is correct.
From the graph it is clear that,

Note : Violet colour light will have wavelength less than 400 nm.
Therefore, this light will be unable to show photoelectric effect on plate 3 ⇒Þ Option (d) is wrong.

*Multiple options can be correct

QUESTION: 33

Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below.

A.
Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy.
B.
Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy
C.
Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments
D.
Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments

Solution:

Note : When binding energy per nucleon increases for a nuclear process, energy is released.
When two nuclei of mass numbers between 51 to 100 fuse, the mass number of the resulting nuclei will come out to be between 100 to 200. The graph shows that in this process the binding energy per nucleon increases and therefore energy is released.
When nucleus of mass number 200 to 260 breaks; it will produce nuclei of mass numbers lying between 100 to 200 if we assume that the two daughter nuclei are of nearly same mass. This in fact happens practically that when a heavy nucleus splits into two parts during nuclear fission, two moderate size nuclei are formed in general. The graph shows that in this process also the binding energy per nucleon increases. Therefore energy is released.

*Multiple options can be correct

QUESTION: 34

The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a₀, where a₀ is the Bohr radius. Its orbital angular momentum is 3h/2π It is given that h is Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

A.
9/32R
B.
9/16R
C.
9/5R
D.
4/3R

Solution:

Angular momentum

we know that

(a), (c) are correct options

*Multiple options can be correct

QUESTION: 35

For photo-electric effect with incident photon wavelength λ, the stopping potential is V₀. Identify the correct variation(s) of V₀ with λ and 1/λ.

Solution:

we should get a straight line with negativeslope and positive intercept.
For V₀vesus λ, we will get a hyperbola. As λ decreases V₀ increases. (a) and (c) are the correct options

*Multiple options can be correct

QUESTION: 36

A fission reaction is given by where x and y are two particles. Considering to be at rest, the kinetic energies of the products are denoted by K_Xe, K_Sr, K_x(2 MeV) and K_y(2 MeV), respectively. Let the binding energies per nucleon of be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is(are)

A.
x = n, y = n, K_Sr = 129 MeV, K_Xe = 86 MeV
B.
x = p, y = e^–, K_Sr = 129 MeV, K_Xe = 86 MeV
C.
x = p, y = n, K_Sr = 129 MeV, K_Xe = 86 MeV
D.
x = n, y = n, K_Sr = 86 MeV, K_Xe = 129 MeV

Solution:

The number of proton in reactants is equal to the products (leaving x and y) and mass number of product (leaving x and y) is two less than reactants
∴ x = p, y = e^– is ruled out [B] is incorrect and x = p, y = n is ruled out [C] is incorrect

Total energy loss = (236 × 7.5) – [140 × 8.5 + 94 × 8.5] = 219 MeV
The energies of kx and ky together is 4MeV The energy remain is distributed by Sr and Xe which is equal to 219 – 4 = 215 MeV
∴ A is the correct option Also momentum is conserved

The energies of kx and ky together is 4MeV The energy remain is distributed by Sr and Xe which is equal to 219 – 4 = 215 MeV

∴ A is the correct option Also momentum is conserved .
Therefore K.E_sr > K.E_xe

*Multiple options can be correct

QUESTION: 37

Highly excited states for hydrogen–like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n>>1. Which of the following statement(s) is(are) true?

A.
Relative change in the radii of two consecutive orbitals does not depend on Z
B.
Relative change in the radii of two consecutive orbitals varies as 1/n
C.
Relative change in the energy of two consecutive orbitals varies as 1/ n³
D.
Relative change in th e angular momen ta of two consecutive orbitals varies as 1/n

Solution:

Relative change in the radii of two consecutive orbitals

does not depend on Z

Relative change in the energy of two consecutive orbitals

*Multiple options can be correct

QUESTION: 38

Light of wavelength λ_ph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes.
If the minimum de Broglie wavelength of the electrons passing through the anode is λ_e, which of the following statement(s) is (are) true?

A.
λ_e decreases with in crease in φ and λ_ph
B.
λ_e is approximately halved, if d is doubled
C.
For large potential difference (V >> φ/e), λ_e is approximately halved if V is made four times
D.
λ_e increases at the same rate as λ_ph for λ_ph < hc/φ

Solution:

The wavelength of emitted photoelectron as per de Broglie is

When φ increases, K.E. decreases and therefore λ_e increases
When λ_ph increases, N_ph decreases , K.E decreases and therefore λ_e increases. le is independent of the distance d.