JEE Mains 2020 January session is almost over now, the JEE main 2020 exam was conducted from January 7th to January 9th in 6 shifts. The last paper of JEE Main was conducted today January 9 in two shifts. First shift timings were 9:30 am to 12:30 pm and second shift timings were 2:30 pm to 5:30 pm. In this article, you will get JEE main 2020 question paper with solution (Jan 9th Second shift). As we all know that NTA is conducting JEE main 2020 exam, this time paper has been more balanced and providing an equal chance to all aspirants. The JEE mains 2020 Jan 9 second shift exam solutions are created by experts on the basis of memory-based JEE mains question available from aspirants. Stay tuned with us for more updates and solutions.
Exam date and Shift | Article URL |
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7th Jan 2020 - Shift 1 | Click here |
7th Jan 2020 - Shift 2 | Click here |
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8th Jan 2020 - Shift 2 | Click here |
9th Jan 2020 - Shift 1 | Click here |
9th Jan 2020 - Shift 2 | Click here |
Q1) In which compound C–Cl bond length is shortest?
(1) Cl–CH=CH_{2}
(2) Cl–CH=CH–CH_{3}
(3) Cl–CH=CH–OCH_{3}
(4) Cl–CH=CH–NO_{2}
Solution
Resonance form of Cl–CH=CH–NO_{2} is more stable than resonance form of any other given compounds. Hence, double bond character in carbon-chlorine bond is maximum and bond length is shortest.
Option 4 is correct.
Q2) Amongst the following which has minimum conductivity.
(1) Distilled water
(2) Sea water
(3) Saline water used for intra venous injection
(4) Well-water
Solution
Distilled water is de-ionised water.
option (1) is correct.
Q3) Number of sp^{2} hybrid orbitals in a molecule of benzene.
1. 6
2. 12
3. 18
4. 24
Solution: 18 sp^{2} hybrid orbitals are present in a molecule of benzene.
Therefore, the option number (3) is correct.
Q4) Biochemical oxygen demand (BOD) is defined as _______ in ppm of O2.
(1) Required to sustain life
(2) The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water.
(3) The amount of oxygen required by anaerobic bacteria to break down the inorganic matter present in a certain volume of a sample of water.
(4) Required photochemical reaction to degrade waste
Solution
The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water, is called Biochemical Oxygen Demand (BOD).
Option 2 is correct.
Q5) Which of the following reaction will not form racemic mixture as product?
Solution
Opton 2 is correct.
Q6)
Percentage carbon in compound A is:
Solution
Compounds is:
Q7) 0.1 ml of an ideal gas has volume 1 dm^{3} in a locked box with frictionless piston. The gas is in thermal equilibrium with excess of 0.5 m aqueous ethylene glycol at its freezing point. If the piston is released all of a sudden at 1 atm then determine the final volume of gas in dm^{3} (R = 0.08 atm L mol^{–1} K^{–1} K_{f} = 2.0 K molal^{–1}).
Solution
After releasing piston P_{1}V_{1} = P_{2}V_{2}
2.176 x 1 = 1 x V_{2}
V_{2} = 2.176 dm^{3}
Q8) One litre sea water (d = 1.03g/cm^{3}) contains 10.3 mg O_{2} gas. Determine concentration of O_{2} in ppm.
Solution
Thus, the correct answer is 10.
Q9) Lacto bacillus has generation time 60 min. at 300 K and 40 min. at 400 K. Determine activation energy in mol kJ/mol. (R = 8.3 J K^{–1}mol^{–1}) [ln(2/3) = -0.4].
Solution
ln(3/2) × 8.3 × 1200 = Ea
E_{a} = 0.4 × 8.3 × 1200
E_{a} = 3984 J/mol.
E_{a} = 3.984 kJ/mol
Thus, the correct answer is 3.98
Q10) Total number of Cr–O bonds in Chromate ion and dichromate ion is:
Solution
Total number of Cr and O bonds is 12.
Q11)
Solution
Q12)
Compound A will be:
Solution
Q13) The order of basic character is :
(1) I > II > III > IV
(2) IV > III > I > II
(3) II > I > III > IV
(4) IV > I > II > III
Solution
Basic strength depends upon availability of lone pairs. Greater the resonance of lone pairs lesser the basic strength.
Option 2 is correct.
Q14)
Solution
Laboratory Test | Molisch's test | Barfoed test |
Biuret test |
Yes | Lactose, Glucose, Fructose | Glucose |
Albumin |
No | Sucrose |
Alanine |
Therefore, Option(1) is correct.
Q15) 1. 5 g of Zn reacts with
(I) Excess of NaOH (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is
(1) 2 : 1
(2) 1 : 2
(3) 1 : 1
(4) 3 : 1
Solution
According to stoichiometry in both the reactions, equal number of moles of H_{2} are evolved.
Option 3 is correct.
Q16) Given K_{sp} for Cr(OH)_{3} is 6 × 10^{–31} then determine [OH^{–}].
(Neglect the contribution of OH^{– }ions from H_{2}O)
Solution
s 3s
Option 1 is correct.
Q17) Select the correct statements among the followings
(A) LiCl does not dissolve in pyridine
(B) Li does not react ethyne to form ethynide.
(C) Li and Mg react slowly with water.
(D) Among alkali metals Li has highest hydration tendency.
Solution
Concept Based
Option 1 is correct.
Q18) Given an element having following ionisation enthalpies IE1 = 496 kJ/mol and IE2 = 4562 kJ/mol one mole hydroxide of this element is treated separated with HCl and H_{2}SO_{4} respectively. Moles of HCl and H_{2}SO_{4} reacted respectively is
(1) 1, 0.5
(2) 0.5, 1
(3) 2, 0.5
(4) 0.5, 2
Solution
According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form
hydroxide of the type M(OH).
Q19) Reactant A represented by square is in equilibrium with product B represented by circles. Then value of
equilibrium constant is
(1) 1
(2) 2
(3) 3
(4) 4
Solution
Option 2 is correct.
Q20) Given following complexes
(I) Na_{4}[Fe(CN)_{6}] (II) [Cr(H_{2}O)_{6}] Cl_{2}
(III) (NEt_{4})_{2} [CoCl_{4}] (IV) Na_{3}[Fe(C_{2}O_{4})_{3}]
Correct order of spin only magnetic moment for the above complexes is.
(1) (II) > (III) > (IV) > I
(2) (II) > (IV) > (III) > (I)
(3) (I) > (IV) > (III) > (II)
(4) (II) > (I) > (IV) > (III)
Solution
Option 1 is correct.
Q21) Select the correct option :
(1) Entropy is function of temperature and also entropy change is function of temperature.
(2) Entropy is a function of temperature & entropy change is not a function of temperature.
(3) Entropy is not a function of temperature & entropy change is a function of temperature.
(4) Both entropy & entropy change are not a function of temperature.
Solution
Option 1 is correct.
Q22) A compound (A ; B_{3}N_{3}H_{3}Cl_{3}) reacts with LiBH_{4} to form inorganic benzene (B). (A) reacts with (C) to form B_{3}N_{3}H_{3}(CH_{3})_{3}. (B) and (C) are respectively.
(1) Boron nitride, MeMgBr
(2) Boron nitride, MeBr
(3) Borazine, MeBr
(4) Borazine, MeMgBr
Solution
Option 4 is correct.
Q23) In a box a mixture containing H2, O2 and CO along with charcoal is present then variation of pressure with the time will be as follows :
Solution
Concept based
Option 3 is correct
Q24) Given complex [Co(NH_{3})_{4}Cl_{2}]. In it if Cl - Co-Cl bond angle is 90º then it is :
(1) Cis-isomer
(2) Trans- isomers
(3) Meridional and trans
(4) Cis and trans both
Solution
Option 1 is correct.
Q25) Monomer(s) of which of the given polymer is chiral?
(1) Buna-S
(2) Neoprene
(3) Nylon-6,6
(4) PHBV
Solution
In PHBV, both monomers have chiral centre.
Option 4 is correct.
Q1. AB is focal chord at parabola , where then equation ale tangent at B is
Solution:
Q2. IF and
Solution:
Use
Q3. If the statement is false then p and q respectively is ?
Solution:
Q4. If , y(1) = 1 and y(x) = e, then x is
Solution:
Q5. If the minimum value of term free from x for is L_{1 } and L_{2} in. Find .
Solution:
Q6. Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is
Solution:
Q7. Let a angle between equal to If is perpendicular to then find the value of .
Solution:
Q8. Let circles (x – 0)^{2} + (y – 4)^{2} = k and (x – 3)^{2} + (y – 0)^{2} = 12 touches each other than find the maximum value of 'k'
Solution:
Two circles touch each other if
Distance between C2(3, 0) and C1(0, 4) is either
Also
Maximum value of K is 36
Q9.
Solution:
Q10. Let the distance between plane passing through lines and and plane 23x – 10y – 2z + 48 = 0 is then k is equal to
Solution:
and
Lines must be intersecting
Hence points are (-3,-2,1)
The distance of plane contains given lines from given plane is same as distance between point (–3, –2,1) from given plane.
Required distance equal to
Q11. then find the area bounded by f(x) and g(x) from
Solution:
Required area = Area of trapezium ABCD -
Q12. z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be
Solution:
z = x + iy
|x| + |y| = 4
Minimum value of |z| =
Maximum value of |z| = 4
So |z| can't be
Q13. If f(x) = and a – 2b + c = 1, than
(1) f(50) = 1
(2) f(–50) = – 1
(3) f(50) = 501
(4) f(50) = – 501
Solution:
Q14. Let a_{n} is a positive term of a GP and
Solution:
From above, r = 2
add both
Q15. Let probability distribution is
then value of p(x > 2) is
Solution:
Q16. Let and then
Q17. Let both root of equation ax^{2} – 2bx + 5 = 0 are and root of equation x^{2} – 2bx – 10 = 0 are and . Find the value of
Solution:
ax^{2} – 2bx + 5 = 0 having equal roots or and
Put in the second equation
Q18. then correct choice is
(1) F(x) has local minimum at x = 1
(2) F(x) has local maximum at x = 1
(3) F(x) has point of inflection at x = 1
(4) F(x) has no critical point
Solution:
Q19. Let and find at
Solution:
Q20. If 7x + 6y – 2z = 0; 3x + 4y + 2z = 0 and x – 2y – 6z = 0 then which option is correct
(1) no. solution
(2) only trivial solution
(3) Infinite non trivial solution for x = 2z
(4) Infinite non trivial solution for y = 2z
Solution:
so infinite non-trivial solution exist
Q21. then ordered pair is
Solution:
Q22. If = A then the value of x at which is discontinuous (where [.] denotes greatest integer function)
Solution:
Q23. Let x + 6y = 8 is tangent to standard ellipse where the minor axis is , then eccentricity of an ellipse is
Solution:
Q24. If f(x) and g(x) are continuous functions, fog is identity function, g'(b) = 5 and g(b) = a then f'(a) is
Solution:
Q-1
A mass m attached to spring of natural length and spring constant k. One end of string is attached to centre of disc in horizontal plane which is being rotated by constant angular speed . Find extension per unit length in spring (given )
Solution-
As natural lentgh=l_{0}
Let elongation=x
Mass m is moving with angular velocity in a radius r
where
Due to elongation x spring force is given by
And
as
So
using
So
Q-2
A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time period of oscillation
Solution-
Q-3
A string of mass per unit length is fixed at both ends under the tension 540 N. If the string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the string?
(1) 2.1 m (2) 1.1 m (3) 4.8 m (4) 4.2 m
Solution-
Fundamental frequency = 490 – 420 = 70 Hz
Q-4
Solution:-
Q-5
Solution:-
Q-6
Solution:-
Q-7
Solution:-
Q-8
Solution:-
Q-9
Solution:-
Q-10
Solution:-
So
Q-11
Find the current supplied by the battery
(1) 0.1 A (2) 0.3 A (3) 0.4 A (4) 0.5 A
Solution-
Both diodes are in reverse biased
So new circuits can be drawn as
So
Q-12
An AC source is connected to the LC series circuit with V = 10 sin (314t). Find the current in the circuit as a function of time? (L = 40 mH, C = 100 F)
(1) 10 sin (314t) (2) 5.2 sin (314t) (3) 0.52 sin (314t) (4) 0.52 cos (314t)
Solution-
As
But R=0
Q-13
Solution:-
Q-14
Solution:-
Q-15
Solution:-