# Vr's JEE Mock Test -1

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INSTRUCTIONS The Mock Test consists of 90 questions. The maximum marks are 360. The test is of 3 hours duration. There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response. 1/4 (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. There is only one correct response for each question. Read moreFilling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly.

• 1.

### A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

• A.

2

• B.

16

• C.

24

• D.

32

D. 32
Explanation
To hold 1 KV potential difference minimum four capacitors are required in series
C1= 1/4 for one series.
So for Ceq to be 2 μF, 8 parallel combinations are required.
Minimum no. of capacitors = 8 × 4 = 32

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• 2.

### A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it ?

• A.

A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

• B.

A screw gauge having 50 divisions in the circular scale and pitch as 1 mm.

• C.

A meter scale

• D.

A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.

D. A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm.
Explanation
Least count of varnier calliper is 0.01 cm .Hence it matches with the reading

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• 3.

### A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

• A.

5g/6

• B.

2g/3

• C.

G/2

• D.

G

C. G/2
Explanation
mg–T=ma
T×R= mR2 /2 * a/R
a= g /2

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• 4.

### Three conducting rods of same material and cross-section are shown in figure. Temperatures of A, D and C are maintained at 20°C, 90°C and 0°C. The ratio of lengths of BD and BC if there is no heat flow in AB is -

• A.

2/7

• B.

7/2

• C.

9/2

• D.

2/9

B. 7/2
Explanation
HAB = 0
HDB = HBC
[means TA = TB = 20]

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• 5.

### The density of water at 20°C is 998 kg/m3 and at 40°C 992 kg/m3. The coefficient of volume expansion of water is

• A.

10–4/°C

• B.

3 × 10–4/°C

• C.

2 × 10–4/°C

• D.

6× 10–4/°C

B. 3 × 10–4/°C
Explanation
Rho = m/v

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• 6.

### The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both ends is

• A.

80 cm

• B.

100 cm

• C.

120 cm

• D.

140 cm

C. 120 cm
Explanation
Given: Length of the closed organ pipe, LC = 20 cm
Let the length of the organ pipe open at both ends be LO.
The fundamental frequency of a closed organ pipe is given by: V/4Lc
The second overtone of an organ pipe open at both ends is given by: 3V/2Lo
V/4Lc. = 3V/2Lo
Lo = 3x 4 Lc / 2.
Lo = 6 Lc. = 6*20 = 120

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• 7.

### When a mass is rotating in a plane about a fixed point, its angular momentum is directed along:

• A.

The line perpendicular to the plane of rotation

• B.

The line making an angle of 45° to the plane of rotation

• C.

• D.

The tangent to the orbit

A. The line perpendicular to the plane of rotation
Explanation
The angular momentum of a mass rotating about a fixed point is L=r*p
Hence, the angular momentum will always be directed perpendicular to the radius vector and the momentum vector of the mass.
So, we can say that it is perpendicular to the plane of rotation.

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• 8.

### One way in which the operation of a n-p-n transistor differs from that of a p-n-p

• A.

The emitter junction is reversed biased in n-p-n

• B.

The emitter junction injects minority carriers into the base region of the p-n-p

• C.

The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n

• D.

The emitter injects holes into the base of n-p-n

C. The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n
Explanation
Emitter-base junction is forward biased.

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• 9.

### A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is

• A.

2 Wb

• B.

0.5 Wb

• C.

ZERO

• D.

12.5 Wb

D. 12.5 Wb
Explanation
ɸ= LI=5×2.5Wb=12.5Wb

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• 10.

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

• A.

6 V/m

• B.

9 V/m

• C.

3 V/m

• D.

12 V/m

A. 6 V/m
Explanation
Ε = CB = 3 × 108 × 20 × 10–9 = 6 V/m

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• 11.

### In a series LCR circuit R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30o. The power dissipated in the LCR circuit is

• A.

305 W

• B.

242 W

• C.

ZERO

• D.

210 W

B. 242 W
Explanation
The given circuit is under resonance as XL = XC Hence power dissipated in the circuit is
P=V2/R =242W

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• 12.

### What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

• A.

2GmM /3R

• B.

GmM /3R

• C.

5GmM /6R

• D.

GmM /2R

C. 5GmM /6R
Explanation
T. Ef = − GMm/ 6R
T. Ei = − GMm/ R
∆W = T.Ef – T.Ei = 5GMm/ 6R

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• 13.

### Transverse elastic waves can propagate –

• A.

Both in a gas and a metal

• B.

In a gas but not in a metal

• C.

In a metal but not in a gas

• D.

Neither in a gas nor in a metal

C. In a metal but not in a gas
Explanation
Transverse elastic waves can propagate in a metal but not in a gas. This is because transverse waves require a medium with sufficient elasticity to allow the particles to move perpendicular to the direction of wave propagation. Metals have a high degree of elasticity due to the arrangement of their atoms, allowing transverse waves to propagate. On the other hand, gases do not have a fixed structure and their particles are not tightly bound, resulting in low elasticity and the inability to support transverse waves.

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• 14.

### Two long parallel wires separated by a distance r have equal currents I flowing in each. Either wire experiences a magnetic force F N/m. If the distance r is increased to 3r and current in each wire is reduced to I/3, the force between them will now be -

• A.

3F N/m

• B.

9F N/m

• C.

(F/9) N/m

• D.

(F/27) N/m

D. (F/27) N/m
Explanation
F = (µ0/2π)(i1i2/r)

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• 15.

### Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]

• A.

350 Hz

• B.

361 Hz

• C.

411 Hz

• D.

448 Hz

D. 448 Hz
Explanation
fa = f [ v+v0 / v- vs]
= 400 [ 340+16.5 / 340-22]
= 448 Hz

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• 16.

### A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F, because the method involves :

• A.

Cells

• B.

• C.

A condition of no current flow through the galvanometer

• D.

A combination of cells, galvanometer and resistances

C. A condition of no current flow through the galvanometer
Explanation
Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit

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• 17.

### A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 :-

• A.

12.89 × 10–3 kg

• B.

9.89 × 10–3 kg

• C.

2.45 × 10–3 kg

• D.

6.45 × 10–3 kg

A. 12.89 × 10–3 kg
Explanation
Work done against gravity = (mgh) 1000
in lifting 1000 times
= 10 × 9.8 × 103
= 9.8 × 104 Joule
20% efficiency is to converts fat into energy. [20% of 3.8 × 107 J] × (m) = 9.8 × 104
(Where m is mass)
m = 12.89 × 10–3 kg

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• 18.

### An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :

• A.

20 times nearer

• B.

10 times taller

• C.

10 times nearer

• D.

20 times taller

D. 20 times taller
Explanation
Angular magnification is 20.

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• 19.

### Hysteresis loops for two magnetic materials A and B are given below : These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use ;

• A.

B for electromagnets and transformers.

• B.

A for electric generators and transformers.

• C.

A for electromagnets and B for electric transformers.

• D.

A for transformers and B for electric generators

A. B for electromagnets and transformers.
Explanation
For electromagnet and transformers, we require the core that can be magnitised and demagnetised quickly when subjected to alternating current. From the given graphs, graph B is suitable.

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• 20.

### The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is : (For steel Young’s modulus is 2 × 1011 N m−2 and coefficient of thermal expansion is 1.1 × 10−5 K−1)

• A.

2.2×10^7 Pa

• B.

2.2×10^6 Pa

• C.

2.2×10^8 Pa

• D.

2.2×10^9 Pa

C. 2.2×10^8 Pa
Explanation
F/A = Y(Δl/l)
(Δl/l) = αΔT
P = F/A = YαΔT
=2×10 ^11 ×1.1×10 ^-5 ×100 = 2.2×10 ^8

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• 21.

### A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is :

• A.

Io

• B.

Io/2

• C.

Io/4

• D.

Io/8

C. Io/4
Explanation
Intensity of light is halved upon passage through first polaroid.
Using Malus’ Law : I = ( Io/2 ) cos^2θ
θ = 45° (The angle between the polarization axes of the polaroids)

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• 22.

### When the current i flowing through a conductor, the drift velocity is v. If 2i current is flowed through the same metal but having double the area of cross section, the drift velocity will be

• A.

V/4

• B.

V

• C.

V/2

• D.

2V

B. V
Explanation
Vd = I/ neA

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• 23.

### A particle is executing linear simple harmonic motion amplitude A. What fraction of the total energy is kinetic when the displacement is half the amplitude.

• A.

1/2

• B.

1/4

• C.

3/4

• D.

2

C. 3/4
Explanation
When the displacement is half the amplitude in linear simple harmonic motion, the particle is at its maximum speed. At this point, the kinetic energy is at its maximum value. The total energy of the system remains constant throughout the motion. Therefore, if the kinetic energy is at its maximum, the potential energy must be at its minimum. Since the total energy is the sum of kinetic and potential energy, and the potential energy is at its minimum, the fraction of the total energy that is kinetic is the maximum value, which is 3/4.

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• 24.

### A student measures the time period of 100 oscillations of a simple pendulum four times. The datasetis90s,91s,95sand92s.Iftheminimum division in the measuring clock is 1 s, then the reported mean time should be :-

• A.

92 ± 3 s

• B.

92 ± 2 s

• C.

92 ± 5 s

• D.

92 ± 1.8 s

B. 92 ± 2 s
Explanation
TAV = 92 s
since uncertainity is 1.5 s
so digit 2 in 92 is uncertain.
so reported mean time should be
92 ± 2

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• 25.

### Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is

• A.

M1r1 :m2r2

• B.

M1 :m2

• C.

1:1

• D.

R1: r2

D. R1: r2
Explanation
a∝r

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• 26.

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

• A.

E2 =2E1

• B.

E1 >E2

• C.

E2 > E1

• D.

E1 =2E2

C. E2 > E1
Explanation
After decay, the daughter nuclei will be more stable hence binding energy per nucleon will be more than that of their parent nucleus.

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• 27.

### A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

• A.

99J

• B.

90J

• C.

100J

• D.

1J

B. 90J
Explanation
W = Q2 ( T1/T2 -1)
η = 1 - (T2/T1)

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• 28.

### Two lenses of power -15 D and + 5D are in contact with each other. The focal length of the combination is

• A.

−20 cm

• B.

+ 20 cm

• C.

−10 cm

• D.

+ 10 cm

C. −10 cm
Explanation
P=P1 +P2 =−10
f = 1/P⇒−0.1m⇒−10cm

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• 29.

### Angular momentum of the particle rotating with a central force is constant due to

• A.

Constant Force

• B.

Constant linear momentum.

• C.

Zero Torque

• D.

Constant Torque

C. Zero Torque
Explanation
The angular momentum of a particle rotating with a central force is constant due to zero torque. Torque is the rotational equivalent of force, and it is the product of the force applied and the perpendicular distance from the axis of rotation. In the case of a central force, the force always acts along the radial direction, which means the perpendicular distance is zero. As a result, the torque exerted on the particle is zero, leading to a constant angular momentum.

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• 30.

### A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

• A.

40 m/s

• B.

30 m/s

• C.

20 m/s

• D.

10 m/s

A. 40 m/s
Explanation
mgh = 1⁄2 mv^2
v= 2gh
= 2×10×80=40m/s

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• 31.

### Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point.

• A.

LiCl

• B.

NaCl

• C.

RbCl

• D.

KCl

B. NaCl
Explanation
Although lattice energy of LiCl higher than NaCl but LiCl is covalent in nature and NaCl ionic there after , the melting point decreases as we move NaCl because the lattice energy decreases as a size of alkali metal atom increases (lattice energy ∝ to melting point of alkali metal halide)

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• 32.

### Benzene and toluene form nearly ideal solutions. At 20o C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20o C for a solution containing 78 g of benzene and 46 g of toluene in torr is

• A.

50

• B.

25

• C.

37.5

• D.

53.5

A. 50
Explanation
PB=PB×B=75× 1/1.5 =50torr

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• 33.

### A reaction involving two different reactants can never be

• A.

Unimolecular reaction

• B.

First order reaction

• C.

Second order reaction

• D.

Bimolecular reaction

A. Unimolecular reaction
Explanation
A unimolecular reaction is a reaction that involves the decomposition or rearrangement of a single molecule. It does not involve the collision between two different reactant molecules. Therefore, a reaction involving two different reactants can never be a unimolecular reaction.

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• 34.

### The IUPAC name of the coordination compound K3[Fe(CN)6] is

• A.

Potassium hexacyanoferrate (II)

• B.

Potassium hexacyanoferrate (III)

• C.

Potassium hexacyanoiron (II)

• D.

Tripotassium hexcyanoiron (II)

B. Potassium hexacyanoferrate (III)
Explanation
The coordination compound K3[Fe(CN)6] contains potassium ions (K+) and a complex ion [Fe(CN)6]3-. According to the rules of IUPAC nomenclature, the name of the compound is determined by the oxidation state of the central metal ion. In this case, the iron ion (Fe) has an oxidation state of +3, which is indicated by the Roman numeral III in parentheses. Therefore, the correct IUPAC name for the compound is Potassium hexacyanoferrate (III).

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• 35.

### The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%) ; Carbon (22.9%), Hydrogen (10.0%) ; and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is

• A.

15 Kg

• B.

7.5 Kg

• C.

37.5 Kg

• D.

10 Kg

B. 7.5 Kg
Explanation
Mass in the body of a healthy human adult has :-
Oxygen = 61.4%, Carbon = 22.9%, Hydrogen = 10.0% and Nitrogen = 2.6%
Total weight of person = 75 kg
Mass due to 1H is = 75*10 /100 = 7.5kg
1H atoms are replaced by 2H atoms.
So mass gain by person =7.5 kg

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• 36.

### Which of the following species is not paramagnetic

• A.

NO

• B.

CO

• C.

O2

• D.

B2

B. CO
Explanation
NO - One unpaired electron is present in pi* molecular orbital.
CO - No unpaired electron is present
O2 - Two unpaired electrons are present in pi*molecular orbitals.
B2  Two unpaired electrons are present in pi bonding molecular orbitals.

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• 37.

### Which base is present in RNA but not in DNA?

• A.

Uracil

• B.

Thymine

• C.

Guanine

• D.

Cytosine

A. Uracil
Explanation
RNA and DNA are both nucleic acids that play a role in genetic information storage and transmission. They are made up of nucleotides, which consist of a sugar, a phosphate group, and a nitrogenous base. While DNA contains the bases adenine (A), thymine (T), guanine (G), and cytosine (C), RNA replaces thymine with the base uracil (U). This difference in bases is due to the fact that RNA is a single-stranded molecule, while DNA is double-stranded. Uracil pairs with adenine in RNA, just like thymine pairs with adenine in DNA. Therefore, the correct answer is Uracil.

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• 38.

### Which among the following is the most reactive?

• A.

Cl2

• B.

Br2

• C.

I2

• D.

ICl

D. ICl
Explanation
The interhalogen compounds are generally more reactive than halogens (except F2).

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• 39.

### The group having isoelectronic species is

• A.

O2– , F –, Na+ , Mg2+

• B.

O– , F –, Na , Mg+

• C.

O2– , F –, Na , Mg2+

• D.

O– , F –, Na + , Mg+

A. O2– , F –, Na+ , Mg2+
Explanation
O ( z=8) , F (z=9) ; Na (z=11) , Mg(z=12)
O2– (10) ,F – (10), Na+ (10), Mg2+ (10)
therefore O2– , F– , Na+ , Mg+2 are isoelectronic

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• 40.

### The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :

• A.

Distillation under reduced pressure

• B.

Simple distillation

• C.

Fractional distillation

• D.

Steam distillation

A. Distillation under reduced pressure
Explanation
Distillation under reduced pressure. Glycerol (B.P. 290°C) is separated from spent lye in the soap industry by distillation under reduced pressure, as for simple distillation very high temperature is required which might decompose the component.

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• 41.

### Which of the following atoms has the highest first ionization energy ?

• A.

Na

• B.

Sc

• C.

Rb

• D.

K

B. Sc
Explanation
Due to poor shielding of d-electrons in Sc, Zeff of Sc becomes more so that ionisation energy of Sc is more than Na, K and Rb.

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• 42.

### In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are :

• A.

Four moles of NaOH and one mole of Br2

• B.

One mole of NaOH and one mole of Br2

• C.

Four moles of NaOH and two moles of Br2

• D.

Two moles of NaOH and two moles of Br2

A. Four moles of NaOH and one mole of Br2
Explanation
4 moles of NaOH and one mole of Br2 is required during production of on mole of amine during Hoffmann's bromamide degradation reaction.

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• 43.

### Galvanization is applying a coating of

• A.

Pb

• B.

Cr

• C.

Zn

• D.

Cu

C. Zn
Explanation
Galvanization is the process of applying a protective zinc coating of steel or iron, to prevent rusting.

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• 44.

### The rate equation for the reaction 2A + B → C is found to be: rate k[A][B]. The correct statement in relation to this reaction is that the

• A.

Unit of K must be s-1

• B.

Values of k is independent of the initial concentration of A and B

• C.

Rate of formation of C is twice the rate of disappearance of A

• D.

T1/2 is a constant

B. Values of k is independent of the initial concentration of A and B
Explanation
The correct statement in relation to this reaction is that the values of k is independent of the initial concentration of A and B. This means that the rate constant, k, remains the same regardless of the initial concentrations of A and B. The rate equation, rate = k[A][B], indicates that the rate of the reaction is directly proportional to the concentrations of A and B, and the rate constant, k, represents the proportionality constant. Therefore, the values of k will not change with different initial concentrations of A and B.

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• 45.

### The synthesis of alkyl fluorides is best accomplished by :

• A.

• B.

Sandmeyer’s reaction

• C.

Finkelstein reaction

• D.

Swarts reaction

D. Swarts reaction
Explanation
R - I + AgF —> R - F + AgI (Swarts Reaction)

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• 46.

### Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is : (at. mass of Cu = 63.5 amu)

• A.

0 g

• B.

63.5g

• C.

2g

• D.

127g

B. 63.5g
Explanation
Cu+2 + 2e- —> Cu(s)
2 mol. 1 mol = 63.5

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• 47.

### Which one is classified as a condensation polymer?

• A.

Dacron

• B.

Neoprene

• C.

Teflon

• D.

Acrylonitrile

A. Dacron
Explanation
Dacron is polyester formed by condensation polymerisation of terephthalic acid and ethylene glycol.

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• 48.

• A.

1000 days

• B.

300 days

• C.

10 days

• D.

100 days

D. 100 days
• 49.

• A.

– 110.5

• B.

110.5

• C.

– 676.5

• D.

676.5

A. – 110.5
• 50.

### Identify the correct statement regarding a spontaneous process

• A.

For a spontaneous process in an isolated system, the change in entropy is positive

• B.

Endothermic processes are never spontaneous

• C.

Exothermic processes are always spontaneous

• D.

Lowering of energy in the reaction process is the only criterion for spontaneity

A. For a spontaneous process in an isolated system, the change in entropy is positive
Explanation
A spontaneous process in an isolated system is characterized by an increase in entropy. This means that the disorder or randomness of the system tends to increase over time. The statement implies that in a spontaneous process, the change in entropy is positive. It is important to note that while a positive change in entropy is a criterion for spontaneity, it is not the only criterion. Other factors such as energy changes and temperature also play a role in determining whether a process is spontaneous or not.

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• Current Version
• Sep 13, 2023
Quiz Edited by
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• Mar 31, 2018
Quiz Created by
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