VR's JEE MOCK TEST -1

1. Which base is present in RNA but not in DNA?

Uracil

Thymine

Guanine

Cytosine

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.

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.

2. 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

Explanation

3. 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

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.

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.

4. 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

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.

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.

5. 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

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

Explanation

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

6. 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

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.

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.

7.

13

-1

5

4

Explanation

8. 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

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.

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.

9. Hydrogen bomb is based on the principle of

Nuclear fission

Natural radioactivity

Nuclear fusion

Artificial radioactivity

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.

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.

10. 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)

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.

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.

11. 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

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.

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.

12. Which one of the following conformation of cyclohexane is chiral?

Twist boat

Rigid

Boat

Chair

Twisted boat is chiral as it does not have plane of symmetry.

Explanation

Twisted boat is chiral as it does not have plane of symmetry.

13. 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

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

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

14. 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

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.

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.

15. Let W denote the words in the English dictionary. Define the relation R by : R = {(x, y) ∈ W m 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

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.

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.

16. 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

Explanation

17.

1/4

1/8

1/16

1/24

Explanation

18. 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

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

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

19.

-1

-2

2

4

Explanation

20. 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

m1m2 = -1

Explanation

m1m2 = -1

21. Conjugate of the complex number 3 – 2i is

2+3i

3+2i

2–3i

3–2i

Conjugate of a+ ib is a- ib

Explanation

Conjugate of a+ ib is a- ib

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

Potassium hexacyanoferrate (II)

Potassium hexacyanoferrate (III)

Potassium hexacyanoiron (II)

Tripotassium hexcyanoiron (II)

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).

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).

23. The synthesis of alkyl fluorides is best accomplished by :

Free radical fluorination

Sandmeyer’s reaction

Finkelstein reaction

Swarts reaction

R - I + AgF —> R - F + AgI (Swarts Reaction)

Explanation

R - I + AgF —> R - F + AgI (Swarts Reaction)

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 :-

92 ± 3 s

92 ± 2 s

92 ± 5 s

92 ± 1.8 s

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

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

25. 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

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.

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.

26. 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

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

Explanation

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

27. 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

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.

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.

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

1

2

3

4

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.

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.

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

7

8

12

32

3 x 4 = 12

Explanation

3 x 4 = 12

30. 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

Explanation

31. A reaction involving two different reactants can never be

Unimolecular reaction

First order reaction

Second order reaction

Bimolecular reaction

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.

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.

32. 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

P=P1 +P2 =−10
f = 1/P⇒−0.1m⇒−10cm

Explanation

P=P1 +P2 =−10
f = 1/P⇒−0.1m⇒−10cm

33. 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

Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit

Explanation

Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit

34. The group having isoelectronic species is

O2– , F –, Na+ , Mg2+

O– , F –, Na , Mg+

O2– , F –, Na , Mg2+

O– , F –, Na + , Mg+

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

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

35. Which of the following compounds is not an antacid?

Aluminium hydroxide

Cimetidine

Ranitidine

Phenelzine

Phenelzine is antidepressant drug.

Explanation

Phenelzine is antidepressant drug.

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

ɸ= LI=5×2.5Wb=12.5Wb

Explanation

ɸ= LI=5×2.5Wb=12.5Wb

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

10

7

4

3

a^n -b^n is divisible by a-b

Explanation

a^n -b^n is divisible by a-b

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

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.

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.

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

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

Explanation

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

40. 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

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)

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)

41. 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

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

Explanation

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

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

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

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

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?

5g/6

2g/3

G/2

G

mg–T=ma
T×R= mR2 /2 * a/R
a= g /2

Explanation

mg–T=ma
T×R= mR2 /2 * a/R
a= g /2

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

Emitter-base junction is forward biased.

Explanation

Emitter-base junction is forward biased.

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

Na

Sc

Rb

K

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.

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.

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

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

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.

47. Galvanization is applying a coating of

Pb

Cr

Zn

Cu

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

Explanation

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

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

Energy in 1st excited state = - 3.4 eV.

Explanation

Energy in 1st excited state = - 3.4 eV.

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

We can apply binomial probability distribution Variance = npq
= 10 × 3/5× 2/5 = 12/5

Explanation

We can apply binomial probability distribution Variance = npq
= 10 × 3/5× 2/5 = 12/5

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

Constant Force

Constant linear momentum.

Zero Torque

Constant Torque

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.

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.

51. The3 termofanAPis4.Whatisthesumoffirst5terms?

12

15

20

24

The terms are 4-2d,4-d,4,4+d,4+2d Sum = 4x5 = 20

Explanation

The terms are 4-2d,4-d,4,4+d,4+2d Sum = 4x5 = 20

52. The eccentricity of the ellipse 5x²+ 9y² = 1 is

2/3

3/2

4/5

1/2

Eccentricity = c/a where c = √a2- b2

Explanation

Eccentricity = c/a where c = √a2- b2

53. Which of the following compounds is metallic and ferromagnetic ?

MnO2

TiO2

CrO2

VO2

CrO2 is metallic as well as ferromagnetic

Explanation

CrO2 is metallic as well as ferromagnetic

54. 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)

0 g

63.5g

2g

127g

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

Explanation

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

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

Distillation under reduced pressure

Simple distillation

Fractional distillation

Steam distillation

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.

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.

56. The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

40

20

60

80

52x + 42y = 50 (x + y)
2x = 8y
X/y = 4/1 ,x/x+y =4/5
∴ % of boys = 80.

Explanation

52x + 42y = 50 (x + y)
2x = 8y
X/y = 4/1 ,x/x+y =4/5
∴ % of boys = 80.

57. 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 m 1011 N m−2 and coefficient of thermal expansion is 1.1 m 10−5 K−1)

2.2×10^7 Pa

2.2×10^6 Pa

2.2×10^8 Pa

2.2×10^9 Pa

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

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

58. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then :

2x = r

2x = (pi + 4)r

(4 – pi)x = pi*r

X = 2r

The given equation x = 2r indicates that the side length of the square is twice the radius of the circle. This means that the square will have a larger area compared to the circle, as the area of a square is determined by multiplying the length of one side by itself, while the area of a circle is determined by multiplying pi by the square of the radius. Therefore, by making x equal to 2r, the sum of the areas of the square and the circle will be minimized.

Explanation

The given equation x = 2r indicates that the side length of the square is twice the radius of the circle. This means that the square will have a larger area compared to the circle, as the area of a square is determined by multiplying the length of one side by itself, while the area of a circle is determined by multiplying pi by the square of the radius. Therefore, by making x equal to 2r, the sum of the areas of the square and the circle will be minimized.

59. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is :

464

485

468

465

Explanation

60. Which one is classified as a condensation polymer?

Dacron

Neoprene

Teflon

Acrylonitrile

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

Explanation

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

61. The heats of combustion of carbon and carbon monoxide are – 393.5 and – 285.5 kJ mol–1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :-

– 110.5

110.5

– 676.5

676.5

Explanation

62. 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

V/4

V

V/2

2V

Vd = I/ neA

Explanation

Vd = I/ neA

63. 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 :

20 times nearer

10 times taller

10 times nearer

20 times taller

Angular magnification is 20.

Explanation

Angular magnification is 20.

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

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

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

A meter scale

Least count of varnier calliper is 0.01 cm .Hence it matches with the reading

Explanation

Least count of varnier calliper is 0.01 cm .Hence it matches with the reading

65. 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 -

3F N/m

9F N/m

(F/9) N/m

(F/27) N/m

F = (µ0/2π)(i1i2/r)

Explanation

F = (µ0/2π)(i1i2/r)

66. Which of the following species is not paramagnetic

NO

CO

O2

B2

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.

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.

67. 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

10–4/°C

3 × 10–4/°C

2 × 10–4/°C

6× 10–4/°C

Rho = m/v

Explanation

Rho = m/v

68. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?

E1, E2 and E3 are independent.

E1 and E2 are independent.

E2 and E3 are independent.

E1 and E3 are independent.

The statement "E1, E2 and E3 are independent" is NOT true. The events E1, E2, and E3 are not independent because the occurrence of one event affects the probability of the other events. If die A shows up four (event E1), it reduces the probability of the sum of numbers on both dice being odd (event E3), as the only way for the sum to be odd is if die B shows up an odd number. Therefore, E1 and E3 are dependent events.

Explanation

The statement "E1, E2 and E3 are independent" is NOT true. The events E1, E2, and E3 are not independent because the occurrence of one event affects the probability of the other events. If die A shows up four (event E1), it reduces the probability of the sum of numbers on both dice being odd (event E3), as the only way for the sum to be odd is if die B shows up an odd number. Therefore, E1 and E3 are dependent events.

69. Which among the following is the most reactive?

Cl2

Br2

I2

ICl

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

Explanation

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

70. 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]

350 Hz

361 Hz

411 Hz

448 Hz

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

Explanation

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

71. 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

80 cm

100 cm

120 cm

140 cm

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

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

72. Transverse elastic waves can propagate –

Both in a gas and a metal

In a gas but not in a metal

In a metal but not in a gas

Neither in a gas nor in a metal

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.

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.

73. 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

40 m/s

30 m/s

20 m/s

10 m/s

mgh = 1⁄2 mv^2
v= 2gh
= 2×10×80=40m/s

Explanation

mgh = 1⁄2 mv^2
v= 2gh
= 2×10×80=40m/s

74. A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

1000 days

300 days

10 days

100 days

Explanation

75. A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be

700

300

350

360

Let the vapour pressure of pure ethyl alcohol be P,
According to Raoult’s law
290 = 200 × 0.4 + P × 0.6
P =350mmHg

Explanation

Let the vapour pressure of pure ethyl alcohol be P,
According to Raoult’s law
290 = 200 × 0.4 + P × 0.6
P =350mmHg

76. 1 gram of a carbonate (M₂CO₃) on treatment with excess HCl produces 0.01186 mole of CO2. the molar mass of M₂CO₃ in g mol–1 is :-

1186

118.6

11.86

84.3

1/M = 0.01186
M = 84.3 gm/mol

Explanation

1/M = 0.01186
M = 84.3 gm/mol

77. 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 m 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2 :-

12.89 × 10–3 kg

9.89 × 10–3 kg

2.45 × 10–3 kg

6.45 × 10–3 kg

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

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

78. 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 -

2/7

7/2

9/2

2/9

HAB = 0
HDB = HBC
[means TA = TB = 20]

Explanation

HAB = 0
HDB = HBC
[means TA = TB = 20]

79. 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.

1/2

1/4

3/4

2

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.

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.

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

The line perpendicular to the plane of rotation

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

The radius

The tangent to the orbit

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.

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.

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

6 V/m

9 V/m

3 V/m

12 V/m

Ε = CB = 3 × 108 × 20 × 10–9 = 6 V/m

Explanation

Ε = CB = 3 × 108 × 20 × 10–9 = 6 V/m

82. 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 :

Io

Io/2

Io/4

Io/8

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)

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)

83. 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 :

2

16

24

32

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

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

84. 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

50

25

37.5

53.5

PB=PB×B=75× 1/1.5 =50torr

Explanation

PB=PB×B=75× 1/1.5 =50torr

85.

4/5

2/5

-4/5

-2/5

Explanation

86. For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then :

A, b and c are in G.P.

B, c and a are in A.P.

B, c and a are in G.P.

A, b and c are in A.P.

Explanation

87. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower -bed, is

30

12.5

15

25

Explanation

88. 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

99J

90J

100J

1J

W = Q2 ( T1/T2 -1)
η = 1 - (T2/T1)

Explanation

W = Q2 ( T1/T2 -1)
η = 1 - (T2/T1)

89. 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

M1r1 :m2r2

M1 :m2

1:1

R1: r2

a∝r

Explanation

a∝r

90. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :

58th

46th

52th

60th

Total number of words which can be formed using
all the letters of the word 'SMALL'
= 5!/2! = 60
Now, 60th word is - SMLLA
59th word is - SMLAL
58th word is - SMALL

Explanation

Total number of words which can be formed using
all the letters of the word 'SMALL'
= 5!/2! = 60
Now, 60th word is - SMLLA
59th word is - SMLAL
58th word is - SMALL

VR's JEE Mock Test -1