National Test Agency( NTA) Is conducting JEE Mains 2020 Exam in two shifts, through an online mode. JEE Main 2020 is a computer-based test. In this article, we are going to share a JEE main 2020 question paper with solution for the afternoon session of 7th January 2020. These solutions are prepared by experts. We suggest you go through the article once, it will help you to understand your performance and with the help of this article, you will be able to understand that which concept is important, Type of exam questions, etc.
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Q1. The number of sp^{2} hybridized carbon atoms in aspartame.
Solution
The aspartame structure is given below:
The number of sp^{2} hybridised carbon atoms in aspartame is 9.
Q2. Find the bond order and magnetic nature of CN^{-}.
Solution
The bond order of CN^{-} is 3 and all the electrons are paired in this ion, thus it is diamagnetic in nature.
Q3. Which one of the two elements among the following pairs (F, Cl), (S, Se) and (Li, Na) will release more energy on gaining of an electron?
Solution
From the given pairs, the more energy on gaining of an electron will be released by the following elements:
Cl, S, and Li
Q4. 3 grams of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken in another container and 1/2 mL of 5M NaOH is added to this. Find the pH of this solution.
(log 3 = 0.4771, pKa = 4.74)
Solution
m mole of acetic acid in 20 mL = 2
m mole of HCl in 20 mL = 1
m mole of NaOH = 2.5
2 3/2 0 0
0.5 0 3/2
pH = 4.74 + log 3
pH = 4.74 + 0.48 = 5.22
Thus, the correct answer is 5.22
Q5. The standard enthalpy of formation(in kJ/mol) of ethane, it the enthalpies of combustion of ethane, dihydrogen, and C(graphite) are -1560kJ/mol, -286kJ/mol and -393.5kJ/mol respectively is:
Solution
The reactions we have:
...................(i)
..........................(ii)
........................(iii)
Multiply eq(ii) by 2 and eq(iii) by 3.
Now, H^{o}_{f} = 2 x eq(ii) + 3eq(iii) - 1
Thus, the equaton is given as:
Q6.
Select the correct options :
(1) A = C_{MPS}, B = C_{Average}, C = C_{RMS,} (2) A = C_{Average}, B = C_{MPS}, C = C_{RMS}
(3) A = C_{RMS}, B = C_{Average}, C = C_{MPS}, (4) A = C_{Average}, B = C_{MPS}, C = C_{RMS}
Solution
C_{RMS} > C_{Average} > C_{MPS.}
Opiton 1 is correct.
Q7.
Solution
Option 3 is correct.
Q8. A and B are in the given reaction ?
Solution
Option 1 is correct.
Q9. Amongs the following which is a redox reaction?
Solution
Option 1 is correct.
Q10.
Solution
Option 1 is correct.
Q11. There are two beakers (I) having a pure volatile solvent and (II) having a volatile solvent and non-volatile solute. If both beakers are placed together in a closed container then:
(1) Volume of solvent beaker will decrease and solution beaker will increase
(2) Volume of solvent beaker will increase and solution beaker will also increase
(3) Volume of solvent beaker will decrease and solution beaker will also decrease
(4) Volume of solvent beaker will increase and solution beaker will decrease
Solution
There will be lowering in vapour pressure in the second beaker.
Option 1 is correct.
Q12. Metal with low melting point containing impurities of high melting point can be purified by
(1) Zone refining (2) Vapor phase refining
(3) Distillation (4) Liquation
Solution
Concept-based.
Option 4 s correct.
Q13. Which of the following statements are correct?
(I) On the decomposition of H_{2}O_{2}, O_{2} gas is released .
(II) 2-ethylanthraquinol is used in preparation of H_{2}O_{2}
(III) On heating KClO_{3}, Pb(NO_{3})_{2}, NaNO_{3}, O_{2} gas is released.
(IV) In the preparation of sodium peroxoborate, H2O2 is treated with sodium metaborate.
(1) I, II, IV (2) II, III, IV (3) I, II, III, IV (4) I, II, III
Solution
Concept based.
Option 3 is correct.
Q14. Which of the following reactions are possible?
Solution
1. Vinyl halides (C) do not give Fridel Craft's reaction becuase of unstable carbocation formation.
2. Aryl halides (A) do not give Fridel Craft's reaction with Benzene because of formation of unstable phenyl carbocation.
Option 2 is correct.
Q15. The correct statement about gluconic acid is
(1) It is prepared by oxidation of glucose with HNO_{3}
(2) It is obtained by partial oxidation of glucose
(3) It is a dicarboxylic acid
(4) It forms hemiacetal or acetal
Solution
Gluconic acid is obtained by partial oxidation of glucose by Tollen's reagent or Fehling solution or Br_{2},H_{2}O. Gluconic acid can not form hemiacetal or acetal.
option 2 is correct.
Q16. The standard enthalpy of formation(in kJ/mol) of ethane, it the enthalpies of combustion of ethane, dihydrogen, and C(graphite) are -1560kJ/mol, -286kJ/mol and -393.5kJ/mol respectively is:
Solution
The reactions we have:
...................(i)
..........................(ii)
........................(iii)
Multiply eq(ii) by 2 and eq(iii) by 3.
Now, H^{o}_{f} = 2 x eq(ii) + 3eq(iii) - 1
Thus, the equaton is given as:
Q17. Flocculation value for As_{2}S_{3} sol by HCl is 30 mmoleL^{–1} . Calculate mass of H_{2}SO_{4} required in gram for 250 mL sol.
Solution
For 1L sol 30 mmol of HCl is required.
Thus, for 1L sol 15 mmol H_{2}SO_{4} is required
For 250 mL of sol:
(15/4) x 10^{-3} mmol H_{2}SO_{4} = 0.3675 g
Thus, the correct answer is 0.37
Q18.
Determine total number of atoms per unit formula of (A), (B) and (C).
Solution
(A) = CrO_{2}Cl_{2}
(B) = Na_{2}CrO_{4}
(C) = CrO_{5}
Thus, the correct answer is 18.
Q19. 0.6 g of urea on strong heating with NaOH evolves NH_{3}. Liberated NH_{3} will combine completely with which of the following HCl solution?
(1) 100 mL of 0.2 N HCl
(2) 400 mL of 0.2 N HCl
(3) 100 mL of 0.1 N HCl
(4) 200 mL of 0.2 N HCl
Solution
2 × mole of Urea = mole of NH3 ........(1)
mole of NH3 = mole of HCl ........(2)
mole of HCl = 2 × mole of Urea = 2 x (0.6/60) = 0.02 mol
100 x 0.2 x 1= 0.02 mol, from 1.
So, option 1 is correct.
Q20. Which of the following statements are incorrect?
(A) Co^{+3} with strong field ligand forms a high magnetic moment complex.
(B) For Co+3 if pairing energy(P) > then the complex formed will have a configuration
(C) For [Co(en)_{3}] ^{3+} is less than for [CoF_{6}] ^{3–}
(D) If = 18000 cm^{–1} for Co^{+3} then with same ligands for it = 16000 cm^{–1}
(1) A, D
(2) B, C
(3) A, B
(4) A, B, C, D
Solution
Concept-Based.
Option 1 is correct.
Q21. Stability order of following alkoxide ions is.
Solution
When negative charge is delocalised with electron withdrawing group like (NO_{2}) then stability increases.
(A) Negative charge is localised
(B) Negative charge is delocalised with NO_{2} group
(C) Negative charge is delocalised with carbon of alkene
Option 1 is correct.
Q22. For the complex [Ma_{2}b_{2}] if M is sp^{3} or dsp^{2} hybridised respectively then total number of optical isomers are respectively :
(1) 1, 1 (2) 2, 1 (3) 0, 0 (4) 1, 2
Solution
Both will not show optical isomerism.
Option 3 is correct.
Q.1 if the sum of the first 40 terms of the series 3+4+8+9+13+14+18+19+...... Is 102m then what is the value of m
Solution
Q2. What is the coefficient of in the expansion of is?
Solution
Q3. Find the area between the curve
Solution
Point of intersection (-1,4),(3,36)
Q4. Let A = [aij], B=[bij] are two 3 × 3 matrices such that & |B| = 81. Find |A| if = 3.
Solution:
Taking Common from and from
Taking Common from and from
Q5. Pair of tangents are drawn from origin to the circle then the square of the length of the chord of contact is?
Solution
Length of the tangent from the origin (0,0) is
The radius of the circle
Length of Chord of contact
Q6. The contrapositive of if and then
Solution:
Contrapositive of
Q7. Let y(x) is the solution of the differential equation , then find the value of x where curve cuts the x-axis
Solution:
Hence
Q8. are two solutions of then equal to
Solution:
Q9. If , then a number of ordered pairs (r, k) are –(where ).
Solution
r should be less than or equal to 35
Hence r can be 5 and 35
For r=5 we get k=-2,-5
For r=35 we get k=-3,3
We get 4 ordered pair
Q10.
Solution
Q11. Let f(x) is a five-degree polynomial that has critical points and then which one is incorrect.
(1) f(x) has minima at x = 1 & maxima at x = –1
(2) f(1) –4f(–1) = 4
(3) f(x) is maxima at x = 1 and minima at x = –1
(4) f(x) is odd
Solution
On solving
f(x) is maxima at x = 1 and minima at x = –1
Q12. If are unit vectors such that and and and then
Solution
Q13. Let if LMVT is applicable on the value of c is :
Solution
f(x) is a polynomial function
Therefore, it is continuous and differentiable in [0, 1]
Here f(0) = 11, f(1) = 1 – 4 + 8 + 11 = 16
Q14. There are 5 machines. The probability of a machine being faulted is . Probability of atmost two machines is faulted, is then the value of k is
Solution
Required probability = when no. the machine has fault + when only one machine has fault + when only two machines have the fault.
Q15. Let and are the roots of such that then which one is incorrect?
Solution
Q16. From any point P on line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find
the locus of midpoint of PQ.
Solution
Using midpoint
so locus is
Q17. If mean and variance of 2, 3, 16, 20, 13, 7, x, y are 10 and 25 respectively then find xy ?
Solution
Q18. If the system of equation x + y + z = 6, x + 2y + 3z = 10, 3x + 2y + z = has more than two solutions, then value of
Solution
x + y + z = 6 …….. (1)
x + 2y + 3z = 10 …….. (2)
3x + 2y + z = …….. (3)
from (1) and (2)
if z = 0 x + y = 6 and x + 2y = 10
y = 4, x = 2
(2, 4, 0)
if y = 0 x + z = 6 and x + 3z = 10
z = 2 and x = 4
(4, 0, 2)
so, 3x + 2y + z = must pass through (2, 4, 0) and (4, 0, 2)
= 14
and 12 + 2 =
= 1
= 13
Q19. If f(x) is defined in
Find k such that f(x) is continuous
Solution
Q20. If is foot of perpendicular drawn from P(1, 0, 3) on a line L and if line L is passing through (, 7, 1), then value of is
Solution
Since PQ is perpendicular to L, therefore
Q21.
Then find number of elements in the smallest subset of X which contain elements of both A and B
Solution
Q22.
If
Solution
Q23. a1, a2, a3 …..a9 are in GP where a1 < 0, a1 + a2 = 4, a3 + a4 = 16, if
then find the
Solution
a1 + a2 = 4 a1 + a1r = 4 ………(i)
a3 + a4 = 16 a1r^{2} + a1r^{3} = 16 …….(ii)
Q24.
Solution
as z is purely real
Q25. 3x + 4y = 12 is the tangent to the ellipse then the distance between focii of ellipse is-
Solution
condition of tangency
distance between foci =
Q1. What is the dimensional formula of , where B is the magnetic field and is permeability of free space?
Solution
The energy density of Electromagnetic wave=
So,
Q2. Moment of inertia of a colid disc as mass per unit area varies as A+Bx^{2}, given the radius of the body is 'a':-
solution
We take an elemental ring of mass dm at a distance x from the centre.
So Area of an elemental ring=
As moment of inertia for an elemental ring =
Total moment of inertia=
Q3. A lift can hold a maximum of 10 people, each of mass 68 kg, a mass of lift is 920 kg, friction of 6000 N is acting on it. Lift is moving with a velocity of 3 m/s. If a lift is moving upward, calculate the power supplied by the motor to the lift, it is moving with its maximum intensity.
Solution
The elevator moves upward hence frictional force (F) opposes their motion downward.
so, net force act on elevator =mg+F
F_{net}=( (10*68+920) x 10+6000)=22000 N
using power =F_{net} x speed=22000*3=66000 Watt
Using 1 HP=746 Watts
So power=88.47 HP
Q4. If weight of an object at pole is 196 N then weight at equator is [g = 10 m/s2 ; radius of earth = 6400 Km]
Solution
Given weight = 196N, therefore mass = 19.6Kg
At pole centripetal acceleration is zero ,there for w=mg
Due to rotation of the earth the effective weight is
Q5. In a house 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and Two heaters of 1 KW each is connected to 220 V Mains supply then find minimum fuse current
Option
1)5 2)20 3)25 4)1
Solution
Power =VI
So current
Minimum fuse current should be 20A
Q6. A block of mass 10kg is suspended from a string of length 4m. When pulled by a force F along horizontal from the midpoint. The upper half of the string makes 45° with vertical. The value of F is
options- 1)100N 2)90N 3)75N 4)70N
Solution
Equating the vertical force
Equating the horizontal force
Q7. Cascaded Carnot engine is an arrangement in which the heat sink of one engine is a source for others. If the high temperature for one engine is T1, low temperature for other engines is T2 (Assume work done by both engine is same) Calculate the lower temperature of the first engine.
options:
a) b) c) 0 d)
Solution
Since the heat sink of one engine is a source for other
So
And Work done by 1^{st} engine = work done by 2^{nd} engine
Let the lower temperature of the first engine=T
And in Carnot engine
So
Q8. The activity of a substance changes from 700 s–1 to 500 s–1 in 30 minutes. Find its half-life in minutes
options- a)66 b)50 c)62 d)56
Solution
Activity at any time is given as
For half-life
so
Now for
from( (i) and (ii)
Q9. In YDSE, the separation between slits is 0.15 mm, the distance between slits and screen is 1.5 m and the wavelength of light is 589 nm, then the fringe width is
Options- a)5.9 mm b)1.9 mm c)3.9 mm d)2.3 mm
Solution
Given d=0.15 mm
D=1.5 m
Fringe width
The fringe width is 5.9 mm
Q10. An ideal fluid is flowing in a pipe in streamline flow. A pipe has a maximum and minimum diameter of 6.4 cm and 4.8 cm respectively. Find out the ratio of minimum to maximum velocity.
Solution
By equation of continuity
If the area is large velocity is minimum
Therefor
Q11. There is an electric circuit as shown in the figure. Find the potential difference between points a and b.
Solution
A diode is forward biased
So the circuit will be as follows
The potential difference between a and b is
Q12. A particle of mass m and positive charge q is projected with a speed of v0 in the y-direction in the presence of electric and magnetic fields are in the x-direction. Find the instant of time at which the speed of particle becomes double the initial speed.
Options-
a) b) c) d)
Solution
Initially
As both electric and magnetic field is in the x-direction
And initial velocity is perpendicular to both electric and magnetic fields.
So particle will move in a helical motion
and magnitude of velocity does not change in y–z plane
and Speed will be increased due to electric filed only i.e in the x-direction
and finally, the speed of the particle becomes double the initial speed.
i.e
And in x-direction
So
Q13. In an adiabatic process, the volume is doubled then find the ratio of final average relaxation time and initial relaxation time
Solution
relaxation time=
and using gives
Q14. Two sources of sound moving with the same speed v and emitting a frequency of 1400 Hz are moving such that one source s1 is moving towards the observer and s2 is moving away from an observer. If the observer hears a beat frequency of 2 Hz. Then find the speed of the source. Given V_{sound} >> V_{Source} and V_{sound} = 350 m/s.
Solution
given f_{0}=1400Hz
Now,
Q15. An electron & a photon have the same energy E. Find the ratio of de Broglie wavelength of an electron to the wavelength of the photon. Given the mass of an electron is m & speed of light is c.
Solution
Q16. A ring is rotated about a diametric axis in a uniform magnetic field perpendicular to the plane of the ring. If initially, the plane of the ring is perpendicular to the magnetic field. Find the instant of time at which EMF will be maximum & minimum respectively:
Solution
We know that
And
Q17. The electric field in space is given by . A positively charged particle at is given velocity . The direction of force acting on a particle is:-
Solution
Force due to electric field is in direction
because at t=0,
Force due to the magnetic field is in direction
therefore, it is parallel to
therefore the net force is antiparallel to
Q18. The focal length of the convex lens in air is . Now the lens is submerged in a liquid of refractive index 1.42. Find the ratio of focal length in medium to focal length in the air:-
Solution
Q19. The hysteresis curve for the material is shown in the figure. Then for the material retentivity, coercivity and saturation magnetization respectively will be -
(1) 50 A/m, 1T, 1.5 T (3) 1 T, 50 A/m, 1.5 T
(2) 1.5 T, 50 A/m, 1T (4) 50 A/m, 1.5 T, 1 T
Solution
So, from the figure, we can see that the-