Vr's JEE Mock Test -1

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  • 1/90 Questions

    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

    • 305 W
    • 242 W
    • ZERO
    • 210 W
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About This Quiz


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. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly.

Vrs JEE Mock Test -1 - Quiz

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  • 2. 
    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 ; - ProProfs

    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 ;

    • B for electromagnets and transformers.

    • A for electric generators and transformers.

    • A for electromagnets and B for electric transformers.

    • A for transformers and B for electric generators

    Correct Answer
    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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  • 3. 

     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 :-

    • 92 ± 3 s

    • 92 ± 2 s

    • 92 ± 5 s

    • 92 ± 1.8 s

    Correct Answer
    A. 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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  • 4. 

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

    • Potassium hexacyanoferrate (II)

    • Potassium hexacyanoferrate (III)

    • Potassium hexacyanoiron (II)

    • Tripotassium hexcyanoiron (II)

    Correct Answer
    A. 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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  • 5. 

    Which base is present in RNA but not in DNA?

    • Uracil

    • Thymine

    • Guanine

    • Cytosine

    Correct Answer
    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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  • 6. 

    The synthesis of alkyl fluorides is best accomplished by :

    • Free radical fluorination

    • Sandmeyer’s reaction

    • Finkelstein reaction

    • Swarts reaction

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

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  • 7. 

    Identify the correct statement regarding a spontaneous process

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

    • Endothermic processes are never spontaneous

    • Exothermic processes are always spontaneous

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

    Correct Answer
    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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  • 8. 

    Which one of the following conformation of cyclohexane is chiral?

    • Twist boat

    • Rigid

    • Boat

    • Chair

    Correct Answer
    A. Twist boat
    Explanation
    Twisted boat is chiral as it does not have plane of symmetry.

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  • 9. 

    Which of the following compounds will exhibit geometrical isomerism 

    • 1-Phenyl-2-butene

    • 3-Phenyl-1-butene

    • 2 - Phenyl -1-butene

    • 1, 1 - Diphenyl -1 - propane

    Correct Answer
    A. 1-Phenyl-2-butene
    Explanation
    1-Phenyl-2-butene will exhibit geometrical isomerism because it has a double bond between the second and third carbon atoms, and the substituents on these carbon atoms are different. Geometrical isomerism occurs when there is restricted rotation around a double bond, and the two different substituents cannot freely interchange their positions. In this case, the phenyl group and the hydrogen atom on the second carbon cannot freely rotate around the double bond, resulting in two different isomers with different spatial arrangements.

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  • 10. 

    An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be

    • AB

    • AB2

    • A2B3

    • AB3

    Correct Answer
    A. AB3
    Explanation
    A= 1/8*8=1 (Corner)
    B=1/2* 6=3
    (Face centre) ∴ AB3

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  • 11. 

    Hydrogen bomb is based on the principle of

    • Nuclear fission

    • Natural radioactivity

    • Nuclear fusion

    • Artificial radioactivity

    Correct Answer
    A. Nuclear fusion
    Explanation
    The correct answer is nuclear fusion. A hydrogen bomb, also known as a thermonuclear bomb, relies on the fusion of hydrogen isotopes, such as deuterium and tritium, to release a massive amount of energy. In this process, the nuclei of these isotopes combine to form a heavier nucleus, releasing a tremendous amount of energy in the process. This is in contrast to nuclear fission, where the nucleus of an atom is split into smaller fragments, and natural or artificial radioactivity, which involve the spontaneous decay of atomic nuclei.

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  • 12. 

    During the process of electrolytic refining of settle as ‘anode mud’ These are

    • Sn and Ag

    • Pb and Zn

    • Ag and Au

    • Fe and Ni

    Correct Answer
    A. Ag and Au
    Explanation
    During the process of electrolytic refining, impurities settle as 'anode mud'. In this case, the anode mud consists of silver (Ag) and gold (Au). This is because during electrolysis, the impurities present in the anode dissolve and settle as mud at the bottom. Since silver and gold are commonly found as impurities in the anode, they form the anode mud during the refining process.

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  • 13. 

    If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is 

    • 7/4

    • 4/3

    • 1

    • 8/5

    Correct Answer
    A. 4/3
    Explanation
    Let 'a' be the first term and d be the common difference
    2nd term = a + d, 5th term = a + 4d, 9th term = 4 + 8d
    Common ratio = =
    a +4d/a +d =a+8d/a+4d = 4/3

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  • 14. 

    If two different numbers are taken from the set {0, 1, 2, 3, ......., 10), then the probability that their sum as well as absolute difference are both multiple of 4, is :-

    • 7/55

    • 6/55

    • 12/55

    • 14/45

    Correct Answer
    A. 6/55
    Explanation
    Let A={0,1,2,3,4......,10}
    n(s) = 11C2 (where 'S' denotes sample space)
    Let E be the given event
    E= {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
    n(E) = 6
    P(E) =6/55

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  • 15. 
    - ProProfs

    • 1/4

    • 1/8

    • 1/16

    • 1/24

    Correct Answer
    A. 1/16
  • 16. 
    - ProProfs

    • -1

    • -2

    • 2

    • 4

    Correct Answer
    A. 2
  • 17. 
    - ProProfs

    • 13

    • -1

    • 5

    • 4

    Correct Answer
    A. 5
  • 18. 

    If (2, 3, 5) is one end of a diameter of the sphere x2 + y2 + z2 − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are

    • (4, 9, −3)

    • (4, −3, 3)

    • (4, 3, 5)

    • (4, 3, −3)

    Correct Answer
    A. (4, 9, −3)
    Explanation
    Coordinates of centre (3, 6, 1)
    Let the coordinates of the other end of diameter are (α, β, γ)
    Hence α = 4, β = 9 and γ = −3.

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  • 19. 

    If the roots of the quadratic equation x2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q − p is

    • 2

    • 3

    • 0

    • 1

    Correct Answer
    A. 3
    Explanation
    x2 + px + q = 0
    tan 30° + tan 15° = − p tan 30° ⋅ tan 15° = q
    tan45°= tan30°+tan15° /1−tan30°tan15° -p/p+q =1
    ⇒−p=1−q
    ⇒q−p=1 ∴2+q−p=3.

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  • 20. 

    Let W denote the words in the English dictionary. Define the relation R by : R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is

    • Not reflexive, symmetric and transitive

    • Reflexive, symmetric and not transitive

    • Reflexive, symmetric and transitive

    • Reflexive, not symmetric and transitive

    Correct Answer
    A. Reflexive, symmetric and not transitive
    Explanation
    The relation R is reflexive because every word in the English dictionary has at least one letter in common with itself. It is symmetric because if word x has a letter in common with word y, then word y also has a letter in common with word x. However, it is not transitive because if word x has a letter in common with word y, and word y has a letter in common with word z, it does not necessarily mean that word x has a letter in common with word z. Therefore, the correct answer is reflexive, symmetric and not transitive.

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  • 21. 

    A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, form B to reach the pillar, is :

    • 5

    • 6

    • 10

    • 20

    Correct Answer
    A. 5
  • 22. 

    Conjugate of the complex number 3 – 2i is

    • 2+3i

    • 3+2i

    • 2–3i

    • 3–2i

    Correct Answer
    A. 3+2i
    Explanation
    Conjugate of a+ ib is a- ib

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  • 23. 

    Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the cente C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is :

    • X2 + y2 – 4x + 9y + 18 = 0

    • X2 + y2 – 4x + 8y + 12 = 0

    • X2+y2–x+4y–12=0

    • X2+y2–x+4y+12=0

    Correct Answer
    A. X2 + y2 – 4x + 8y + 12 = 0
    Explanation
    The equation of the circle passing through the point P and having its center at P can be found by finding the equation of the perpendicular bisector of the line segment joining C and P. Since the point P is at a minimum distance from the center C, the line joining C and P is perpendicular to the tangent of the parabola at point P. Therefore, the slope of the line joining C and P is the negative reciprocal of the slope of the tangent at point P. By finding the midpoint of the line segment joining C and P and using the slope-intercept form of a line, we can find the equation of the perpendicular bisector. The equation of the circle passing through C and having its center at P can be found by substituting the coordinates of P into the general equation of a circle. By simplifying the equation, we get x^2 + y^2 – 4x + 8y + 12 = 0.

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  • 24. 

    The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is 

    • 16.8

    • 16

    • 14

    • 15.8

    Correct Answer
    A. 14
    Explanation
    When one observation valued 16 is deleted and three new observations valued 3, 4, and 5 are added to the data set, the total number of observations increases from 16 to 18. The sum of the new observations is 3 + 4 + 5 = 12. The sum of the original data set is 16 * 16 = 256. Therefore, the sum of the resultant data set is 256 - 16 + 12 = 252. The mean of the resultant data set is 252 / 18 = 14. So, the correct answer is 14.

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  • 25. 

    Thenumberofcommontangentstothecirclesx2 +y2 4x6y12=0andx2 +y2 +6x+18y + 26 = 0, is :

    • 1

    • 2

    • 3

    • 4

    Correct Answer
    A. 3
    Explanation
    The given question asks for the number of common tangents to two circles. In order to determine the number of common tangents, we need to find the number of intersection points between the two circles. The equation of the first circle is x^2 + y^2 - 4x - 6y - 12 = 0, and the equation of the second circle is x^2 + y^2 + 6x + 18y + 26 = 0. By solving these two equations simultaneously, we can find the coordinates of the intersection points. If there are three distinct intersection points, then there will be three common tangents. Therefore, the correct answer is 3.

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  • 26. 

    The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse x2/9+ y2/5 =1 

    • 27

    • 18

    • 27/2

    • 27/4

    Correct Answer
    A. 27
  • 27. 

    If one of the diameters of the circle, given by the euqation,x2 +y2 –4x+6y–12=0,isachord of a circle S, whose centre is at (–3, 2), then the radius of S is :-

    • 5

    • 5*2^1/2

    • 5*2^1/3

    • 10

    Correct Answer
    A. 5*2^1/3
  • 28. 

    The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :

    • 216

    • 72

    • 192

    • 160

    Correct Answer
    A. 192
    Explanation
    To form a number greater than 6,000, the first digit must be either 6, 7, or 8. After selecting the first digit, there are 4 remaining digits to choose from. For the second digit, there are 4 options, for the third digit, there are 3 options, for the fourth digit, there are 2 options, and for the fifth digit, there is only 1 option left. Therefore, the total number of integers that can be formed is 3 * 4 * 3 * 2 * 1 = 72. However, this calculation only considers the numbers greater than 6,000. To include the numbers between 6,000 and 6,999, we need to add the number of integers that can be formed starting with 6. Since there are 4 remaining digits to choose from, the number of integers starting with 6 is 4 * 3 * 2 * 1 = 24. Adding this to the previous calculation, the total number of integers is 72 + 24 = 96.

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  • 29. 

    Choose the line which is perpendicular to x – 3y + 7 = 0

    • 3x+y+1=0

    • 3x-y+2=0

    • X+3y+7=0

    • X–3y+8 =0

    Correct Answer
    A. 3x+y+1=0
    Explanation
    m1m2 = -1

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  • 30. 

    Ifn(A)=3,n(B)=4 n(AxB)equals

    • 7

    • 8

    • 12

    • 32

    Correct Answer
    A. 12
    Explanation
    3 x 4 = 12

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  • 31. 

    A reaction involving two different reactants can never be

    • Unimolecular reaction

    • First order reaction

    • Second order reaction

    • Bimolecular reaction

    Correct Answer
    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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  • 32. 

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

    • Cells

    • Potential gradients

    • A condition of no current flow through the galvanometer

    • A combination of cells, galvanometer and resistances

    Correct Answer
    A. 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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  • 33. 

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

    • −20 cm

    • + 20 cm

    • −10 cm

    • + 10 cm

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

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  • 34. 

    The group having isoelectronic species is

    • O2– , F –, Na+ , Mg2+

    • O– , F –, Na , Mg+

    • O2– , F –, Na , Mg2+

    • O– , F –, Na + , Mg+

    Correct Answer
    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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  • 35. 

    Which of the following compounds is not an antacid?

    • Aluminium hydroxide

    • Cimetidine

    • Ranitidine

    • Phenelzine

    Correct Answer
    A. Phenelzine
    Explanation
    Phenelzine is antidepressant drug.

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  • 36. 

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

    • 2 Wb

    • 0.5 Wb

    • ZERO

    • 12.5 Wb

    Correct Answer
    A. 12.5 Wb
    Explanation
    ɸ= LI=5×2.5Wb=12.5Wb

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  • 37. 

    If nεN, 7n -3n is always divisible by

    • 10

    • 7

    • 4

    • 3

    Correct Answer
    A. 4
    Explanation
    a^n -b^n is divisible by a-b

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  • 38. 

    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

    • Unit of K must be s-1

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

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

    • T1/2 is a constant

    Correct Answer
    A. 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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  • 39. 

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

    • E2 =2E1

    • E1 >E2

    • E2 > E1

    • E1 =2E2

    Correct Answer
    A. 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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  • 40. 

    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? 

    • 2GmM /3R

    • GmM /3R

    • 5GmM /6R

    • GmM /2R

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

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  • 41. 

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

    • LiCl

    • NaCl

    • RbCl

    • KCl

    Correct Answer
    A. 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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  • 42. 

    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

    • 15 Kg

    • 7.5 Kg

    • 37.5 Kg

    • 10 Kg

    Correct Answer
    A. 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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  • 43. 
    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? - ProProfs

    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?

    • 5g/6

    • 2g/3

    • G/2

    • G

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

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  • 44. 

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

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

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

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

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

    Correct Answer
    A. 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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  • 45. 

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

    • Na

    • Sc

    • Rb

    • K

    Correct Answer
    A. 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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  • 46. 

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

    • Four moles of NaOH and one mole of Br2

    • One mole of NaOH and one mole of Br2

    • Four moles of NaOH and two moles of Br2

    • Two moles of NaOH and two moles of Br2

    Correct Answer
    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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  • 47. 

    Galvanization is applying a coating of 

    • Pb

    • Cr

    • Zn

    • Cu

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

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  • 48. 

    Which of the following is the energy of a possible excited state of hydrogen?

    • +13.6 eV

    • + 6.8 eV

    • - 6.8 eV

    • - 3.4 eV

    Correct Answer
    A. - 3.4 eV
    Explanation
    Energy in 1st excited state = - 3.4 eV.

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  • 49. 

    A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replacement, then the variance of the number of green balls drawn is 

    • 6/25

    • 12/5

    • 4

    • 6

    Correct Answer
    A. 12/5
    Explanation
    We can apply binomial probability distribution Variance = npq
    = 10 × 3/5× 2/5 = 12/5

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Quiz Review Timeline (Updated): Sep 13, 2023 +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Sep 13, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Mar 31, 2018
    Quiz Created by
    VR Education
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