JEE Main 2020 is conducting by National Test Agency. Through this exam, students will get admission into B.tech /B.E courses. JEE Main 2020 exam will be conducted in 2 shifts; Morning shift ( 9:30 am to 12:30 pm) and Afternoon shift (2:30 pm to 5:30 pm). In this article, our experts share the exam analysis as per the student's feedback of the 7th January Morning session paper and JEE main 2020 question paper with solutions. Aspirants who appeared for the JEE Mains exam can check the 7th January, Morning session question paper and solutions here. With the help of the solutions, students who appeared for the JEE Main 2020 - Jan 7th first shift exam will be able to check their performance in the exam. As we know the JEE main 2020 exam is divided into 3 sections physics, chemistry, and mathematics. Below you can check JEE main 2020 exam question and solutions - subject wise.
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Q1. The relative strength of interionic/intermolecular forces in decreasing order is:
(a) Ion-dipole > dipole-dipole > Ion-ion
(b) dipole-dipole > ion-dipole > ion-ion
(c) ion-ion > ion-dipole > dipole-dipole
(d) ion-dipole > ion-ion > dipole-dipole
Solution
The correct order of intermolecular forces is:
ion-ion > ion-dipole > dipole-dipole
Therefore, Option(c) is correct.
Q2. The oxidation number of potassium in K_{2}O , K_{2}O_{2} and KO2, respectively is :
Solution
For K_{2}O: 2x - 2 = 0
x = +1
For K_{2}O_{2}: 2x - 2 = 0 (In peroxide, the oxidation state of oxygen is -1)
x = +1
For KO_{2}: x - 2(1/2) = 0 (In supeoxide, the oxidation state of oxygen is -1/2)
x = +1
Q3. At 35^{o}C, the vapour pressure of CS_{2} is 512mm Hg and that of acetone is 344mm Hg. A solution of CS_{2} in acetone has a total vapour pressure of 600mm Hg. The false statement among the following is:
(a) CS_{2} and acetone are less attracted to each other than to themselves.
(b) Heat must be absorbed in order to produce the solution at 35^{o}C.
(c) Raoult's law is not obeyed by this system.
(d) A mixture of 100ml CS_{2} and 100ml acetone has a volume of <200ml.
Solution
We have:
P_{observed} = 600Hg
We know:
P_{calc} = P^{o}_{A}x_{A} + P^{o}_{B}x_{B}
= 512x + 344 - 344x
= 168x + 344
= 512 ( if x =1 for miximum value)
Thus, P_{calc} < P_{observed}
Thus, the process is deviated from Raoult's law, showing positive deviation from Raoult's law.
, i.e, heat absorbed
Therefore, Option(d) is correct.
Q4. The atomic radius of Ag is closest to:
(a) Ni
(b) Cu
(c) Au
(d) Hg
Solution
The atomic radius of Ag is closest to Au.
Therefore, Option(c) is correct.
Q5. The dipole moments of CCl_{4}, CHCl_{3} and CH_{4} are in the order:
Solution
Q6. In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is:
(a) Less efficient as it exchanges only anions
(b) More efficient as it can exchange only cations.
(c) Less efficient as the resins cannot be generated
(d) More efficient as it can exchange both cations and anions
Solution
The synthetic resins method is more efficient as it can exchange both cations and anions.
Therefore, Option(d) is correct.
Q7. Among the following statements that which was not proposed by Dalton was:
(a) Matter consists of invisible atoms
(b) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T and P.
(c) Chemical reactions involve the reorganization of atoms.These are neither created nor destroyed in a chemical reaction
Solution
Option(b) is correct as it was proposed by Gay Lussac's law of gaseous volume.
Therefore, Option(b) is correct.
Q8. The increasing order of pKb for the following compounds will be:
(a) NH2-CH=NH
(b)
(c) CH3NHCH3
Solution
Thus, the increasing order of basicity is:
(b) > (a) > (c)
pK_{b }order is given as:
(c) > (a) > (b)
The conjugate acid of guanidine is most resonance stabilized followed by imidine.
Q9. What is the product of following reaction:
Solution
The reactions occur as follows:
Q10. The number of orbitals associated with quantum numbers n = 5, m_{s} = +1/2 is:
Solution
We have:
n - 5, m_{s} = +1/2
Thus, values of l are from 0 to (n-1)
l = 0 to 4
Thus, values of l are 5s, 5p, 5d, 5f and 5g
Now, the total number of orbitals = n^{2} = 5^{2} = 25
Q11. The purest form of commercial iron is:
Solution
The purest form of iron is wrought iron.
Q12. The theory that can completely /properly explain the nature of bonding in [Ni(CO)_{4}]
(a) Werner's theory
(b) Crystal field theory
(c) Molecular Orbital theory
(d) Valence bond theory
Solution
The metal-carbonyl is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding ^{*} orbital of carbon monoxide.
Therefore, Option(c) is correct.
Q13. The IUPAC name of the complex [Pt(NH_{3})_{2}Cl(NH_{2}CH_{3})]Cl is:
Solution
The IUPAC name of the given compound is:
Diamminechloromethylammineplatinum(II)chloride.
Q14. 1-methyl ethylene oxide when treated with an excess of HBr produces:
Solution
Q15. Consider the following reaction:
The product 'X' is used:
(a) in proteins estimation as an alternative to ninhydrin
(b) as food-grade colourant
(c) in laboratory test for phenols
(d) in acid-base titration as an indicator
Solution
Q16. Match the following:
(i) Riboflavin (a) Beriberi
(ii) Thiamine (b) Scurvy
(iii) Pyridoxine (c) Cheilosis
(iv) Ascorbic acid (d) Convulsions
Solution
(i) Riboflavin - (c) Cheilosis
(ii) Thiamine - (a) Beriberi
(iii) Pyridoxine - (d) Convulsions
(iv) Ascorbic acid - (b) Scurvy
Q17. Given that the standard potentials(E^{o}) of Cu^{2+}|Cu and Cu^{+}|Cu are 0.34V and 0.522V respectively, the E^{o} of Cu^{2+}|Cu^{+} is:
Solution
We have:
ΔG = -nFE^{o}
-2xFx0.34V = -1xFxE^{o} -1xFx0.522V
-E^{o} = -2x0.34V + 0.522V
Thus, E^{o} = 0.158V
Q18. A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO_{3} to give fraction A. The leftover orgainc phase was extracted with dilute NaOH solution to give fraction. The final organic layer was labelled as fraction C. Fractions A, B and C contain respectively:
Solution
The reactions occur as follows:
Q19. The electron gain enthalpy(in kJ/mol) of fluorine, chlorine, bromine and iodine respectively are:
Solution
The electron gain enthalpy values are given below:
Fluorine = -333kJ/mol
Chlorine = -348kJ/mol
Bromine = -324kJ/mol
Iodine = -295kJ/mlol
Thus the correct order is:
Cl > F > Br > I
Q20. Consider the following reactions:
(a)
(b)
(c)
(d)
Which of these reaction(s) will not produce Saytzeff product?
Solution
The reactions occur as follows:
Therefore, Option(c) is correct.
Q21. Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H_{2}SO_{4} in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is:
Solution
For given solutions, we have:
Moles of NaOH 4/40 = 0.1 moles
Moles of H_{2}SO_{4} = 9.8/98 = 0.1 moles
Molarity of NaOH = 0.1/100L
And molarity of H_{2}SO_{4} = 0.1/100
Now, 40L of NaOH solution and 10L of H_{2}SO_{4} solution are added, thus we get:
Total volume = 50L
Milliequivalents of NaOH = 40x(0.1/100)x1 = 0.04
Milliequivalents of H_{2}SO_{4} = 10x(0.2/100)x2 = 0.02
Thus, Meq of NaOH left = 0.04 - 0.02 = 0.02
[OH^{-}] = 4x10^{-4}
pOH = -log[4x10^{-4}]
pOH = -log4 - log10^{-4}
pOH = -0.60 + 4 = 3.4
Further, we know:
pH = 14 - 3.4
pH = 10.5
Q22. During the nuclear explosion, one of the products of ^{90}Sr was absorbed in the bones of a newly born bany in place of Ca, how much time is years is required to reduce it by 90% if it is not lost metabolically.
(Half-life of ^{90}Sr is 6.93years).
Solution
We know:
K is given as:
Thus, t is given as:
Thus, t = 23.03 years
Q23. Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is:
Solution
We have:
Y X
white(ppt)
The structure of NaClO^{-}_{3} is given below:
The average bond order is given as:
B.O = 5/3 = 1.66
Q24. The number of chiral carbons in chloramphenicol is:
Solution
The structure of chloramphenicol is given below:
Thus, the number of chiral carbons in chloramphenicol is 2.
Q25. For the reaction:
ΔU = 2.1 Kcal, ΔS = 20calK^{-1} at 300K. Hence ΔG in Kcal is:
Solution
We know:
Thus, we have:
On putting the given values we get:
Q1. if f(a+b+1-x) = f(x) , for all , where a and b are fixed positive real numbers, then
is equal to:
Solution
...........1
..........2
Adding 1 and 2
Q2. Let and beta be two real roots of the equation where K 1 and are real numbers. if . Then value of is :
Solution
...... given
since are the roots
Q3. A total number of 6 digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear is :
Solution
5X6! / 2!
Q4. if is a tangent to both the parabola, , then b is equal to :
Solution
also ,
since ,
Therefore ,
Q5. Let be a root of the equation and the matrix , then the matrix is equal to:
Solution
Given
Q6. if is the solution of the differential equation, , such that then is equal to :
Solution
at y(0)=0
c=1
y(1)=1+ln2
Q7. if
then at is :
Solution
put
we get 4
Q8. let the function be continuous on [ -7, 0] and differentiable on (-7, 0). If f(-7)=-3 and for all then for all such functions f , f(-1)+ f(0) lies in the interval :
Solution
Directly use
we get
Q9. A vector lies in the plane of the vectors and . If bisects the angle between and . Then
Solution
Q10. If distance between the foci of an ellipse is 6 and the distance between directrices is 12, then the length of its latus rectum is:
Solution
given
.......(1)
Also,
.......(2)
from (1) and (2)
since ,
Substitute the values of 'e' and 'a' in above equation .
length of latus rectum =
Q11. The greatest positive integer k, for which is factor of the sum , is :
Solution
Hence, K=63
Q12. Let P be the plane passing through the pts. (2,1,0), (4,1,1) and (5,0,1) and R be any point ( 2,1,6). Then the image of R in the plane P is:
Solution
Points of plane P ( 2,1,0), (4,1,1) and ( 5,0,1) and point R ( 2,1,6) .
Then the image of R in the plane P is:
Q13 . If the system of linear equation
Where a, b , c are non-zero and distinct; has a non-zero solution then:
Solution
For non zero solutions D = 0
on solving
Q14. The area of the region enclosed by the circle which is not common to the region bounded by the parabola and the straight line , is :
Solution
Q15. The logical statement is equivalent to :
Solution
which is equal to
Q16. An unbiased coin is tossed 5 times . Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3,4,5 .Otherwise X takes the value -1. Then the expected value of X is :
Solution
Q17. Let , then 'k' is :
Solution
Q18. if then the point (x,y) lies on a:
Solution
Q19. If is equal to :
Solution
Therefore ,
Solving the above equations,
Therefore,
Q20. Five numbers are in A.P whose sum is 25 and product is 2520. If one of these five numbers is , then the greatest number amongst them is :
Solution
Five terms are in AP
let terms are a-2d, a-d,a,a+d,a+2d
Sum of the terms
5a=25
a=5
Product of the terms( a-2d) (a-d)(a+d)(a+2d)a
d= 11/2 and -11/2 satisfies the condition
so a+ 2d = 16
Q21. Let A(1,0) , B( 6,2) and be the vertices of a triangle ABC. If P is a point inside the such that triangles and have equal area. Then the length of the line segment PQ, where Q is the point :
Solution
A(1,0) B(6,2) C(3/2,6)
Point P is the centroid of triangle ABC
P(17/6,8/3)
Distance between PQ is 5
Q22. If the variance of the first 'n' natural numbers is 10 and the variance of the first 'm' even natural numbers is 16, then m+n is equal to :
Solution
Therefore n = 11
m= 7
Q23. is equal to :
Solution
for, put x=2 in the above equation, to get
Q24. If the sum of the coefficients of all even powers of 'x' in the product is is 61 then 'n' is equal to:
Solution
Q25. Let S be the set of points where the function is not differentiable . Then is equal to :
Solution
Q1. Which of the following gives a reversible operation?
Solution
For reversible operation, NOT gate is used. If an input is then output=. The following circuit represents the NOT gate
Q2. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 1000 kg. If the frictional force on the elevator is 4000N , then the speed of the elevator at full load is close to: (1 HP=746 W, g=10 ms^{-2}).
Solution
The elevator moves upward hence frictional force (F) opposes their motion downward.
so, net force act on elevator =mg+F
F_{net}=(1000 x 10+4000)=14000 N
using power =F_{net} x speed
So speed==3.19
Q3. An LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', then the correct equivalence would be:-
a)
b)
c)
d)
Solution
For the spring-mass damping system, the governing equation is given by
For an LCR damped oscillator, the equation is
Comparing 1 and 2
So the answer is option 1
Q4. As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest, the bob starts falling vertically. When is has covered a distance h, the angular speed of the wheel will be:-
Solution
Q5. Three-point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown. The centre of mass of the system is at a point:
Solution:-
let m1=1 kg, m2=1.5 kg and m3=2.5 kg
x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4
and
Let point A be origin and mass m_{1}=1.0 kg be at origin.
So,
and
so centre of mass of the system is at (0.9,2).
Q6. A parallel plate capacitor has plates of area A separated by distance d with a dielectric which has a dielectric constant that varies as
where x is the distance measured from one of the plates.
If then the total capacitance of the system is best given by expression by
A)
B)
C)
D)
Solution
If K filled between the plates -
Given
For
here,
since
So
Q7. The time period of revolution of electron in its ground state orbit in a hydrogen atom is . The frequency of the electron in its first excited state (in s^{-1}) is:-
Solution
Velocity of the n_{th} orbit
radius of the n_{th} orbit
and
Q8. The current I_{1} (in A) flowing through resistor in the following circuit is:-
Solution
As
and
From equation (1) and (2)
and
Q9. A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches a height of R(R=radius of r=earth), it ejects a rocket of mass so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G=gravitational constant; M= mass of the earth):
a)
b)
c)
d)
Solution
Before the rocket rejection
Apply energy conservation
After the rocket rejection
Apply momentum conservation
Along y-axis
Along x-axis
And since
So Kinetic energy of the rocket
Q10. A long solenoid of radius R carries a time(t) dependent current . A ring of radius 2R is placed coaxially near its middle. During the time interval , the induced current (I_{R}) and the induced EMF (V_{R}) in the ring changes as:-
Solution
using
SO
Now considering solenoid as ideal solenoid extended up to infinite and ring as its centre
Since induced emf is changing so current will also be changing
because
So, since direction of emf is changing so direction of current is also changing.
And it will be zero at t=0.5 sec
Q11. The radius of gyration of a uniform rod of length l, about an axis passing through a point away from the centre of the rod and perpendicular to it, is:-
Solution
Moment of inertia of rod about an axis perpendicular through COM
And
Radius of Gyration
Q12. Two moles of an ideal gas with are mixed 3 moles of another ideal gas with . The value of for the mixture is:-
Solution
For ideal gas:-
For first case:-
So,
For second case:-
and
Now,
Q13. A liter of dry air at STP expands adiabatically to a volume of 3l. If , then work done by air is [take air to be an ideal gas]:-
Solution
At STP, Pressure = P_{1} = 100000 Pascal
Volume = V_{1} = 1 liter (As given in question)
So,
Q14. If the magnetic field in a plane electromagnetic wave is given by ; then what will be the expression for the electric field?
Solution
given,
wave is propagating in -x direction, i.e., in - i direction.
the direction of EMW wave is in the direction of .
Since B is in j direction and EMW is in -i direction. Therefore E is in (k) direction.
Q15. A polarizer-analyzer set is adjusted such that the intensity of light coming out of the analyzer is just 10% of the original intensity. Assuming that the polarizer-analyzer set does not absorb any light, the angle by which the analyzer needs to be rotated further to reduce the output intensity to be zero is:
a) 45^{o}
b) 71.6^{o}
c) 90^{o}
d) 18.4^{o}
^{Solution}
output intensity is given by
Initial output intensity=10% of I_0
I.e
Final output intensity=O
means new angle is
the angle by which the analyser needs to be rotated further is
Q16. Speed of transverse wave of a straight wire (mass=6.0 g, length=60 cm and area of cross-section=1.0 mm^{2}) is 90 ms^{-1}. If the young's modulus of wire is , the extension of wire over its natural length is:
Solution
The linear mass density
Q17. Two infinite planes each with uniform surface charge density + are kept in each such a way that the angle between them is 30^{0}. The electric field in the region shown between them is given by:
Solution
Assuming sheet to be non conducting
So,
So,
Where y and x are corresponding unit vectors
Q18. If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to :
a. 12mm
b. 33mm
c. 22mm
d. 2mm
Solution
The magnification is given by
Q19. Visible height of wavelength 600010^{-8 }cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction min. is at 60 from the central max. If the first minimum is produced at , the is close to,
a. 25^{}
b. 45^{}
c. 20^{}
d. 30^{}
Solution
So,
For first minima,
Q20. Consider a coil of wire carrying current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by i. The magnetic flux through the area given by _{o}. Which of the following is correct?
a. i = -
b. i >
c. i <
d. i =
Solution
Flux going right comes back to the left (forms closed loop)
Flux inside the coil come bach through outside
Q21 A particle (m=1 kg) slides down a frictionless block(AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P (height 1 m), the kinetic energy of the particle (in J) is (take g=10 ms^{-2}):-
Solution
Loss of potential energy= Gain in kinetic energy
Q22. A carnot engine separates between two reservoirs of temperature 900 K and 300 K. The engine performing 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir in a cycle is:-
Solution:-
given, W=1200 J
So,
So,
Q23. A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is .The quantity of flux through the loop ABCDEFA (in Wb) is:-
Solution
Total area vectot=ara of ABCD+area of DEFA=
Total Magnetic flux=
Q24. A beam of electromagnetic radiation of intensity is comprised of wavelength . It falls normally on a metal (work function =2 eV) of surface area 1 cm^{2}. If one in 10^{3} photons ejects an electron, total number of electrons ejected in 1s is 10^{x} (hc=1240 eVnm, 1 eV=1.6 x 10^{-19}J), then x is :-
Solution
Energy in 1 second =
Q25. A non isotropic solid metal cube has coefficients of linear expansion as along the x-axis and along the y-axis anf z-axis. If the coefficient of volume expansion of the solid is , then the value of C is:-
Solution
Since, it is cube, so all it's side is equal. Let the length of cube be L.
Given, and
Let the coefficient of volume expansion be
After increase in temperature:-
Increase in volume of cube=
and we can also write it as:- \
from equation 1 and 2 we get:-
From question:-
So, C=60
The number of questions has been reduced from 30 to 25 for each subject. However, the duration of the exam remains the same as before at 3 hours.
5 questions have been included with answers as numerical values. There will be no options for those questions.
These 5 questions will not have negative marks
Maximum marks a candidate can score has been reduced to 300. Last year it was 360 marks. The maximum marks are equally divided into 100 each in Physics, Chemistry, and Mathematics.
Stay tuned with us for further updates, and solutions would be available soon.