JEE Main 2023 Question Paper Jan 29 Shift 1 is available for download here. NTA conducted JEE Main 2023 Jan 29 Shift 1 from 9 AM to 12 PM for B.E./B.Tech paper. Based on the initial JEE Main 2023 Paper Analysis Jan 29 Shift 1, the paper was reported as easy to moderate. Students found Physics easy, Mathematics tough and Chemistry moderate. Candidates can download the memory-based JEE Main 2023 Question Paper PDF with Solution and Answer Key for Jan 29 Shift 1 using the link below.
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JEE Main 2023 Jan 29 Shift 1 Questions with Solutions
Question 1:
The domain of
f(x) = logx+1(x − 2) / e2 logxx − (2x + 3), x ∈ R is:
(1) R − {−1, 3}
(2) (2, ∞) − {3}
(3) (−1, ∞) − {3}
(4) R − {3}
View Solution
Question 2:
Let f : R → R be a function such that
f(x) = (x² + 2x + 1) / (x + 1).
(1) f(x) is many-one in (−∞, −1)
(2) f(x) is many-one in (1, ∞)
(3) f(x) is one-one in [1, ∞) but not in (−∞, ∞)
(4) f(x) is one-one in (−∞, ∞)
View Solution
Question 3:
For two non-zero complex numbers z₁ and z₂, if
Re(z₁z₂) = 0 and Re(z₁ + z₂) = 0,
then which of the following are possible?
Choose the correct answer from the options given below:
(1) B and D
(2) B and C
(3) A and B
(4) A and C
Answer: (2) B and C
View Solution
Question 4:
Let λ ≠ 0 be a real number. Let α, β be the roots of the equation
14x2 − 3λx + 3λ = 0,
and α, γ be the roots of the equation
35x2 − 53x + 4λ = 0.
Then 3α/β and 4α/γ are the roots of the equation:
(1) 7x2 + 245x − 250 = 0
(2) 7x2 − 245x + 250 = 0
(3) 49x2 − 245x + 250 = 0
(4) 49x2 + 245x + 250 = 0
View Solution
Question 5:
Consider the following system of equations:
- αx + 2y + z = 1
- 2αx + 3y + z = 1
- 3x + αy + 2z = β
For some α, β ∈ R. Then which of the following is NOT correct:
(1) It has no solution if α = −1 and β ≠ 2.
(2) It has no solution for α = −1 and for all β ∈ R.
(3) It has no solution for α = 3 and for all β ≠ 2.
(4) It has a solution for all α ≠ −1 and β = 2.
Answer: (2)
View Solution
Question 6:
Let α and β be real numbers. Consider a 3 × 3 matrix A such that:
A2 = 3A + αI,
A4 = 21A + βI.
Then:
(1) α = 1
(2) α = 4
(3) β = 8
(4) β = −8
Answer: (4) β = −8
View Solution
Question 7:
Let x = 2 be a root of the equation x2 + px + q = 0 and
f(x) = {
1 − cos(x2 − 4px + q − 8q2 + 16) / (x − 2p)2, x ≠ 2p,
0, x = 2p.
}
Then
limx→2p [f(x)],
where [·] denotes the greatest integer function, is:
(1) 2
(2) 1
(3) 0
(4) −1
Answer: (3) 0
View Solution
Question 8:
Let
f(x) = x + a / (π / 2 − 4 sin x) + b / (π / 2 − 4 cos x), x ∈ R
be a function which satisfies
f(x) = x + ∫0π/2 sin(x + y)f(y) dy.
Then (a + b) is equal to:
(1) −π(π + 2)
(2) −2π(π + 2)
(3) −2π(π − 2)
(4) −π(π − 2)
Answer: (2) −2π(π + 2)
View Solution
Question 9:
Let
A = {(x, y) ∈ R2 : y ≥ 0, 2x ≤ y ≤ π/4 − (x − 1)2}
B = {(x, y) ∈ R2 : 0 ≤ y ≤ min{2x, π/4 − (x − 1)2}}.
Then the ratio of the area of A to the area of B is:
(1) (π−1)/(π+1)
(2) π/(π−1)
(3) π/(π+1)
(4) (π+1)/(π−1)
Answer: (1) (π−1)/(π+1)
View Solution
Question 10:
Let Δ be the area of the region
{(x, y) ∈ R2 : x2 + y2 ≤ 21, y2 ≤ 4x, x ≥ 1}.
Then
(1/2)Δ − 21 sin−1(2/√7) is equal to:
(1) 2√3 − 1/3
(2) √3 − 2/3
(3) 2√3 − 2/3
(4) √3 − 4/3
Answer: (4) √3 − 4/3
View Solution
Question 11:
A light ray emits from the origin making an angle of 30 degrees with the positive x-axis. After getting reflected by the line x + y = 1, if this ray intersects the x-axis at Q, then the abscissa of Q is:
(1) (√2)/3 − 1
(2) 2/3 + √3
(3) 2/3 − √3
(4) √3/2 (√3 + 1)
Answer: (2) 2/3 + √3
View Solution
Question 12:
Let B and C be the two points on the line y + x = 0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y − 2x = 2 such that triangle ABC is an equilateral triangle. Then, the area of triangle ABC is:
(1) √3
(2) 2√3
(3) √8/3
(4) √10/3
Answer: (3) √8/3
View Solution
Question 13:
Let the tangents at the points A(4, −11) and B(8, −5) on the circle x² + y² − 3x + 10y − 15 = 0 intersect at the point C. Then the radius of the circle, whose center is C and the line joining A and B is its tangent, is equal to:
(1) 3√3 / 4
(2) 2√13
(3) 13
(4) 2√13 / 3
Answer: (4) 2√13 / 3
View Solution
Question 14:
Let [x] denote the greatest integer. Consider the function f(x) = max{x², 1 + [x]}, where [x] denotes the greatest integer ≤ x. Then the value of the integral ∫₂⁰ f(x) dx is:
(1) 5 + 4√2/3
(2) 8 + 4√2/3
(3) 1 + 5√2/3
(4) 4 + 5√2/3
Answer: (1) 5 + 4√2/3
View Solution
Question 15:
If the vectors a = λi + μj + 4k, b = −2i + 4j − 2k, and c = 2i + 3j + k are coplanar, and the projection of a on vector b is √54 units, then the sum of all possible values of λ + μ is equal to:
(1) 0
(2) 6
(3) 24
(4) 18
Answer: (3) 24
View Solution
Question 16:
Fifteen football players of a club are given 15 T-shirts with their names written on the back. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is:
(1) 5/24
(2) 2/15
(3) 1/6
(4) 5/36
Answer: (1) 5/24
View Solution
Question 17:
Let f(θ) = 3 sin^4(3π/2 − θ) + sin^4(3π + θ) − 2(1 − sin^2(2θ)), and S = {θ ∈ [0, π] : f′(θ) = −√3/2}. If 4β = Σθ∈S θ, then f(β) is equal to:
(1) 11/8
(2) 5/4
(3) 9/8
(4) 3/2
Answer: (2) 5/4
View Solution
Question 18:
If p, q, and r are three propositions, then which of the following combinations of truth values of p, q, and r makes the logical expression {(p ∨ q) ∧ ((¬p) ∨ r)} → ((¬q) ∨ r) false?
(1) p = T, q = F, r = T
(2) p = T, q = T, r = F
(3) p = F, q = T, r = F
(4) p = T, q = F, r = F
Answer: (3) p = F, q = T, r = F
View Solution
Question 19:
Three rotten apples are accidentally mixed with seven good apples, and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If μ and σ² represent the mean and variance of X, respectively, then 10(μ² + σ²) is equal to:
(1) 20
(2) 250
(3) 25
(4) 30
Answer: (1) 20
View Solution
Question 20:
Let y = f(x) be the solution of the differential equation y(x + 1) dx − x² dy = 0, y(1) = e. Then limₓ→0⁺ f(x) is equal to:
(1) 0
(2) 1/e
(3) e²
(4) 2e
Answer: (1) 0
View Solution
Question 21:
Let the coordinates of one vertex of triangle ABC be A(0, 2, α) and the other two vertices lie on the line x + α/5 = (y - 1)/2 = (z + 4)/3. For α ∈ Z, if the area of triangle ABC is 21 square units and the line segment BC has length 2√21 units, then α² is equal to:
(1) 9
(2) 16
(3) 25
(4) 36
Answer: (3) 25
View Solution
Question 22:
Let the equation of the plane P containing the line x + 10 = (8 - y)/2 = z be ax + by + 3z = 2(a + b), and the distance of the plane P from the point (1, 27, 7) be c. Then a² + b² + c² is equal to:
(1) 355
(2) 200
(3) 150
(4) 400
Answer: (1) 355
View Solution
Question 23:
Suppose f is a function satisfying f(x + y) = f(x) + f(y) for all x, y ∈ N and f(1) = 1/5. If the sum from n = 1 to m of [f(n) / (n(n + 1)(n + 2))] equals 1/12, then m is equal to:
(1) 5
(2) 10
(3) 15
(4) 20
Answer: (2) 10
View Solution
Question 24:
Let a1, a2, a3, ... be a geometric progression (GP) of increasing positive numbers. If the product of the fourth and sixth terms is 9 and the sum of the fifth and seventh terms is 24, then a1a9 + a2a4a9 + a5 + a7 is equal to:
Answer: 60
View Solution
Question 25:
Let a, b, and c be three non-zero, non-coplanar vectors. Let the position vectors of four points A, B, C, and D be a - b + c, λa - 3b + 4c, -a + 2b - 3c, and 2a + 4b + 6c respectively. If vectors AB, AC, and AD are coplanar, then λ is:
Answer: 2
View Solution
Question 26:
If all the six-digit numbers x1x2x3x4x5x6 with 0 < x1 < x2 < x3 < x4 < x5 < x6 are arranged in increasing order, then the sum of the digits in the 72nd number is:
Answer: 32
View Solution
Question 27:
Let f : R → R be a differentiable function that satisfies the relation f(x + y) = f(x) + f(y) − 1 for all x, y ∈ R. If f′(0) = 2, then |f(−2)| is equal to:
(1) 1
(2) 2
(3) 3
(4) 4
Answer: (3) 3
View Solution
Question 28:
If the coefficient of x⁹ in (a x³ + 1/(β x¹¹)) and the coefficient of x⁻⁹ in (a x − 1/(β x³))¹¹ are equal, then (αβ)² is equal to:
(1) 1
(2) 4
(3) 9
(4) 16
Answer: (1) 1
View Solution
Question 29:
Suppose the coefficients of three consecutive terms in the binomial expansion of (1 + 2x)ⁿ are in the ratio 2 : 5 : 8. Then the coefficient of the term which is in the middle of these three terms is:
(1) 560
(2) 1120
(3) 1680
(4) 2240
Answer: (2) 1120
View Solution
Question 30:
Five-digit numbers are formed using the digits {1, 2, 3, 5, 7} with repetitions allowed, and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is:
(1) 1430
(2) 1436
(3) 1500
(4) 1600
Answer: (2) 1436
View Solution
Question 31:
Match List-I with List-II:
| List-I (Organelle/Structure) | List-II (Function) |
|---|---|
| (A) Ribosome | (I) Protein synthesis |
| (B) Mitochondria | (II) Energy production |
| (C) Lysosome | (III) Intracellular digestion |
| (D) Golgi apparatus | (IV) Protein modification and packaging |
Choose the correct answer from the options given below:
View Solution
Question 32:
In a cuboid of dimensions 2L × 2L × L, a charge q is placed at the center of the surface S having area 4L2. The flux through the opposite surface to S is:
View Solution
Question 33:
Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is:
View Solution
Question 34:
A single current-carrying loop of wire carrying current I flows in the anticlockwise direction (seen from the +z direction) and lies in the xy plane. The plot of the ĵ component of magnetic field (By) at a distance a (less than the radius of the coil) and on the yz plane vs z coordinate looks like:
View Solution
Question 35:
The magnitude of magnetic induction at the mid-point O due to the current arrangement shown in the figure is:
View Solution
Question 36:
Find the mutual inductance in the arrangement, when a small circular loop of radius R is placed inside a large square loop of side L (L ≫ R). The loops are coplanar and their centers coincide:
View Solution
Question 37:
Which of the following are true?
A. Speed of light in vacuum depends on the direction of propagation.
B. Speed of light in a medium is independent of the wavelength of light.
C. Speed of light is independent of the motion of the source.
D. Speed of light in a medium is independent of intensity.
Choose the correct answer from the options given below:
View Solution
Question 38:
In a Young’s double slit experiment, two slits are illuminated with light of wavelength 800 nm. The first minimum is detected at P. The value of slit separation a is:
View Solution
Question 39:
A stone is projected at an angle of 30° to the horizontal. The ratio of kinetic energy at projection to its kinetic energy at the highest point is:
View Solution
Question 40:
A block of mass m slides down a plane inclined at an angle of 30° with an acceleration of g/4. The coefficient of kinetic friction is:
View Solution
Question 41:
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of the car will be, if the friction coefficient between tyres and road is 0.34. (Take g = 10 m/s²):
View Solution
Question 42:
Two particles of equal mass m move in a circle of radius r under the action of their mutual gravitational attraction. The speed of each particle will be:
View Solution
Question 43:
The surface tension of a soap bubble is 2 × 10⁻² N/m. Work done to increase the radius of the bubble from 3.5 cm to 7 cm will be:
View Solution
Question 44:
Assertion A: If dQ and dW represent the heat supplied to the system and the work done on the system respectively, then according to the first law of thermodynamics dQ = dU − dW.
Reason R: First law of thermodynamics is based on the law of conservation of energy.
In the light of the above statements, choose the correct answer from the options given below:
View Solution
Question 45:
A bicycle tyre is filled with air at a pressure of 270 kPa at 27°C. The approximate pressure of the air in the tyre when the temperature increases to 36°C is:
View Solution
Question 46:
A person observes two moving trains, A reaching the station and B leaving the station with equal speed of 30 m/s. If both trains emit sounds with frequency 300 Hz, the approximate difference of frequencies heard by the person will be:
View Solution
Question 47:
If the height of transmitting and receiving antennas are 80 m each, the maximum line of sight distance will be:
View Solution
Question 48:
The threshold wavelength for photoelectric emission from a material is 5500 Å. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from:
- A. 75 W infra-red lamp
- B. 10 W infra-red lamp
- C. 75 W ultra-violet lamp
- D. 10 W ultra-violet lamp
Choose the correct answer from the options given below:
View Solution
Question 49:
If a radioactive element with a half-life of 30 min undergoes beta decay, the fraction of the radioactive element that remains undecayed after 90 min is:
View Solution
Question 50:
Which of the following statements is not correct in the case of light emitting diodes (LEDs)?
- A. It is a heavily doped p-n junction.
- B. It emits light only when it is forward biased.
- C. It emits light only when it is reverse biased.
- D. The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used.
Choose the correct answer from the options given below:
View Solution
Question 51:
A radioactive element 24292X emits two α-particles, one electron, and two positrons. The product nucleus is represented by 234PY. The value of P is ——.
View Solution
Question 52:
Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is —– degrees.
View Solution
1. Resultant Amplitude Formula:
Aresultant = √(A₁² + A₂² + 2A₁A₂ cos φ).
2. Given Data:
- A₁ = A₂ = 8 cm, Aresultant = 8 cm.
3. Substitute Values:
8 = √(8² + 8² + 2 × 8 × 8 cos φ).
64 = √(128 + 128 cos φ).
Square both sides:
64 = 128(1 + cos φ).
Solve for cos φ:
cos φ = −1/2.
4. Phase Difference:
φ = 120°.
Question 53:
A body cools from 60°C to 40°C in 6 minutes. If the temperature of the surroundings is 10°C, then after the next 6 minutes, its temperature will be —— °C.
View Solution
1. Newton’s Law of Cooling:
Average rate of cooling:
(T − Ts)/∆t = k(T − Ts),
where Ts is the surrounding temperature.
2. First Interval:
(60 − 40)/6 = k[(60 + 40)/2 − 10],
k = 20 / (6 × 50).
3. Second Interval:
(40 − T)/6 = k[(40 + T)/2 − 10].
Solve:
T = 28°C.
Question 54:
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of the center of mass of the sphere will be —– ms⁻¹.
View Solution
1. Kinetic Energy of a Rolling Sphere:
Total kinetic energy:
KE = (1/2)mv² + (1/2)Iω²,
where I = (2/5)mr² for a sphere.
2. Relation Between v and ω:
For pure rolling: ω = v/r.
3. Substitute Values:
KE = (1/2)mv² + (1/2)(2/5)mr²(v²/r²).
Simplify:
KE = (1/2)mv² + (1/5)mv² = (7/10)mv².
4. Solve for v:
2240 = (7/10) × 2 × v²,
v² = 1600, v = 40 ms⁻¹.
Final Answer: 40 ms⁻¹
Question 55:
A 0.4 kg mass takes 8 seconds to reach the ground when dropped from a certain height P above the surface of Earth. The loss of potential energy in the last second of fall is —— J. (Take g = 10 m/s²)
View Solution
1. Height Traveled in the Last Second:
The distance traveled in the nth second of free fall is:
hₙ = u + (1/2)g(2n − 1).
Here, u = 0, g = 10 m/s², n = 8:
h₈ = (1/2) × 10 × (2 × 8 − 1) = (1/2) × 10 × 15 = 75 m.
2. Loss of Potential Energy:
The loss of potential energy is given by:
ΔPE = mgh₈.
Substituting m = 0.4 kg, g = 10 m/s², and h₈ = 75 m:
ΔPE = 0.4 × 10 × 75 = 300 J.
Final Answer: 300 J
Question 56:
A tennis ball is dropped onto the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. The ball comes in contact with the floor for 0.2 s. The average acceleration during contact is —— m/s². (Given g = 10 m/s²)
View Solution
1. Velocity Just Before Impact:
Using v² = u² + 2gh, where u = 0, h = 9.8 m:
v = √(2 × 10 × 9.8) = √196 = 14 m/s.
2. Velocity Just After Rebound:
Using v² = u² + 2gh, where u = 0, h = 5.0 m:
v = √(2 × 10 × 5.0) = √100 = 10 m/s.
3. Change in Velocity During Contact:
Total change in velocity:
Δv = vbefore impact + vafter rebound = 14 + 10 = 24 m/s.
4. Average Acceleration:
Average acceleration:
a = Δv/Δt = 24/0.2 = 120 m/s².
Final Answer: 120 m/s²
Question 57:
A point charge q₁ = 4q₀ is placed at the origin. Another point charge q₂ = −q₀ is placed at x = 12 cm. The proton is placed on the x-axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is — cm.
View Solution
1. Force Balance Condition: - The electrostatic force on the proton is zero when:
F₁ = F₂.
Using Coulomb’s law:
k · |4q₀| / r² = k · |q₀| / (12 − r)².
2. Simplify: - Cancel k and q₀:
4 / r² = 1 / (12 − r)².
Take the square root:
2 / r = 1 / (12 − r).
Cross-multiply:
2(12 − r) = r ⇒ 24 − 2r = r.
Solve for r:
3r = 24 ⇒ r = 8 cm.
3. Position of Proton: - The proton is 12 + r = 24 cm from the origin.
Final Answer: 24 cm
Question 58:
In a metre bridge experiment, the balance point is obtained if the gaps are closed by 2 Ω and 3 Ω. A shunt of X Ω is added to the 3 Ω resistor to shift the balancing point by 22.5 cm. The value of X is — Ω.
View Solution
1. Initial Balance Point: - The ratio of resistances gives the balance length:
l₁ / l₂ = R₁ / R₂,
where l₁ + l₂ = 100 cm.
2. Initial Condition: - For R₁ = 2 Ω and R₂ = 3 Ω:
l₁ / l₂ = 2 / 3, so l₁ = (2/5) × 100 = 40 cm.
3. After Adding Shunt: - The effective resistance of R₂ with a shunt X:
R'₂ = (R₂X) / (R₂ + X) = (3X) / (3 + X).
- The new balance point shifts by 22.5 cm:
l'₁ = 40 + 22.5 = 62.5 cm.
4. New Condition: - The new ratio is:
l'₁ / l'₂ = R₁ / R'₂.
Substituting l'₁ = 62.5, l'₂ = 37.5, R₁ = 2 Ω, and R'₂ = (3X) / (3 + X):
62.5 / 37.5 = 2 / ((3X) / (3 + X)).
Simplify:
5 / 3 = 2(3 + X) / 3X.
Cross-multiply and solve for X:
15X = 18 + 6X ⇒ 9X = 18 ⇒ X = 2 Ω.
Final Answer: 2 Ω
Question 59:
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released, the radius of the loop starts shrinking at a constant rate of 2 cm/s. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be — mV.
View Solution
1. Magnetic Flux: - The magnetic flux through the loop is:
Φ = B · A = B · πr²,
where B = 0.8 T and r = 10 cm = 0.1 m.
2. Rate of Change of Flux: - The emf induced is:
E = − dΦ/dt.
Differentiate Φ with respect to time:
E = − d/dt(Bπr²) = −B · 2πr (dr/dt).
3. Substitute Values: - B = 0.8 T, r = 0.1 m, dr/dt = −2 cm/s = −0.02 m/s:
E = 0.8 · 2π · 0.1 · 0.02 = 0.010 V.
4. Convert to mV:
E = 10 mV.
Final Answer: 10 mV
Question 60:
Three identical polaroids P₁, P₂, and P₃ are placed one after another. The pass axis of P₂ and P₃ are inclined at angles of 60° and 90° with respect to the axis of P₁. The source S has an intensity of 256 W/m². The intensity of light at point O is —— W/m².
View Solution
1. Intensity After First Polaroid (P₁): - When unpolarized light passes through a polaroid, its intensity is reduced by half:
I₁ = I₀ / 2 = 256 / 2 = 128 W/m².
2. Intensity After Second Polaroid (P₂): - The intensity after P₂ is given by Malus’s Law:
I₂ = I₁ cos²(60°).
- Substituting cos 60° = 1/2:
I₂ = 128 · (1/2)² = 128 · 1/4 = 32 W/m².
3. Intensity After Third Polaroid (P₃): - The intensity after P₃ is again reduced according to Malus’s Law:
I₃ = I₂ cos²(30°),
where the relative angle between P₂ and P₃ is 30° (since 90° − 60° = 30°).
- Substituting cos 30° = √3/2:
I₃ = 32 · (√3/2)² = 32 · 3/4 = 24 W/m².
Final Answer: 24 W/m²
Question 61:
For 1 mol of gas, the plot of pV vs p is shown below. p is the pressure and V is the volume of the gas. What is the value of compressibility factor at point A?
(1) 1 − a/RT V
(2) 1 + b/V
(3) 1 − b/V
(4) 1 + a/RT V
View Solution
Question 62:
The shortest wavelength of hydrogen atom in the Lyman series is λ. The longest wavelength in the Balmer series of He+ is:
(1) 5/9 λ
(2) 9/5 λ
(3) 36/5 λ
(4) 5/9 λ
View Solution
Question 63:
Which of the following salt solutions would coagulate the colloid solution formed when FeCl3 is added to NaOH solution, at the fastest rate?
(1) 10 mL of 0.2 mol dm−3 AlCl3
(2) 10 mL of 0.1 mol dm−3 Na2SO4
(3) 10 mL of 0.1 mol dm−3 Ca3(PO4)2
(4) 10 mL of 0.15 mol dm−3 CaCl2
View Solution
Question 64:
The bond dissociation energy is highest for:
(1) Cl2
(2) I2
(3) Br2
(4) F2
View Solution
Question 65:
The reaction representing the Mond process for metal refining is:
(1) Ni + 4CO Δ → Ni(CO)4
(2) 2K[Au(CN)2] + Zn Δ → K2[Zn(CN)4] + 2Au
(3) Zr + 2I2 Δ → ZrI4
(4) ZnO + C Δ → Zn + CO
View Solution
Question 66:
Which of the given compounds can enhance the efficiency of a hydrogen storage tank?
(1) Li/P4
(2) SiH4
(3) NaNi5
(4) Di-isobutylaluminium hydride
View Solution
Question 67:
The correct order of hydration enthalpies is:
(A) K+
(B) Rb+
(C) Mg2+
(D) Cs+
(E) Ca2+
Choose the correct answer from the options given below:
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Question 68:
The magnetic behavior of Li2O, Na2O2, and KO2, respectively, is:
(1) diamagnetic, paramagnetic, and diamagnetic
(2) paramagnetic, paramagnetic, and diamagnetic
(3) paramagnetic, diamagnetic, and paramagnetic
(4) diamagnetic, diamagnetic, and paramagnetic
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Question 69:
"A" obtained by Ostwald's method involving air oxidation of NH3, upon further air oxidation produces "B". "B" on hydration forms an oxoacid of nitrogen along with evolution of "A". The oxoacid also produces "A" and gives a positive brown ring test.
(1) NO2, N2O5
(2) NO2, N2O4
(3) NO, NO2
(4) N2O3, NO2
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Question 70:
The standard electrode potential (M3+/M2+) for V, Cr, Mn, and Co are -0.26 V, -0.41 V, +1.57 V, and +1.97 V, respectively. The metal ions which can liberate H2 from a dilute acid are:
(1) V2+ and Mn2+
(2) Cr2+ and Co2+
(3) V2+ and Cr2+
(4) Mn2+ and Co2+
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Question 71:
Correct statement about smog is:
(1) NO2 is present in classical smog
(2) Both NO2 and SO2 are present in classical smog
(3) Photochemical smog has a high concentration of oxidizing agents
(4) Classical smog also has a high concentration of oxidizing agents
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Question 72:
Chiral complex from the following is:
(1) cis–[PtCl2(en)2]2+
(2) trans–[PtCl2(en)2]2+
(3) cis–[PtCl2(NH3)2]
(4) trans–[Co(NH3)4Cl2]+
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Question 73:
Identify the correct order for the given property for the following compounds:
(The exact compounds and property details are not fully shown in the question excerpt, but the final conclusion is provided.)
Choose the correct answer from the options given below:
(1) (B), (C), and (D) only
(2) (A), (C), and (E) only
(3) (A), (C), and (D) only
(4) (A), (B), and (E) only
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Question 74:
The increasing order of pKa for the following phenols is:
(1) 2,4-Dinitrophenol
(2) 4-Nitrophenol
(3) 2,4,5-Trimethylphenol
(4) Phenol
(5) 3-Chlorophenol
Answer: (2) 2,4-Dinitrophenol < 4-Nitrophenol < 3-Chlorophenol < Phenol < 2,4,5-Trimethylphenol
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Question 75:
Match the reactions in List-I with the reagents in List-II:
| List-I (Reaction) | List-II (Reagents) |
|---|---|
| (A) Hoffmann Degradation | (I) Conc. KOH, heat |
| (B) Clemenson Reduction | (II) CHCl3, NaOH / H3O+ |
| (C) Cannizzaro Reaction | (III) Br2, NaOH |
| (D) Reimer-Tiemann Reaction | (IV) Zn-Hg / HCl |
(1) (A) – III, (B) – IV, (C) – II, (D) – I
(2) (A) – II, (B) – IV, (C) – I, (D) – III
(3) (A) – III, (B) – IV, (C) – I, (D) – II
(4) (A) – II, (B) – I, (C) – III, (D) – IV
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Question 76:
The major product 'P' for the following sequence of reactions is:
(1) Some other derivative
(2) Yet another derivative
(3) Ph-CH2-CH2-CH2-CH2-NH2
(4) A substituted ring compound
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Question 77:
During the borax bead test with CuSO4, a blue-green colour of the bead was observed in the oxidizing flame due to the formation of:
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Question 78:
Match List I with List II:
| List I (Antimicrobials) | List II (Names) |
|---|---|
| (A) Narrow Spectrum Antibiotic | (I) Furacin |
| (B) Antiseptic | (II) Sulphur Dioxide |
| (C) Disinfectants | (III) Penicillin-G |
| (D) Broad Spectrum Antibiotic | (IV) Chloramphenicol |
Answer: (1) (A) – III, (B) – I, (C) – II, (D) – IV
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Question 79:
Number of cyclic tripeptides formed with 2 amino acids A and B is:
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Question 80:
Compound that will give positive Lassaigne's test for both nitrogen and halogen is:
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Question 81:
Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is x × 10−1. The value of x is —— (Nearest integer). Assume complete dissociation.
Question 82:
The number of molecules or ions from the following, which do not have an odd number of electrons, are ——:
- (A) NO2
- (B) ICl4−
- (C) BrF3
- (D) ClO2
- (E) NO2+
- (F) NO
Question 83:
Consider the following reaction approaching equilibrium at 27°C and 1 atm pressure:
A + B ⇌ C + D
Kf = 103, Kr = 102
The standard Gibbs energy change (ΔrG°) at 27°C is (–) —— kJ mol−1 (Nearest integer).
(Given: R = 8.3 J K−1 mol−1 and ln(10) = 2.3)
Answer: 6
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Question 84:
Solid lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15°C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at –0.8°C. The solubility product of PbCl2 formed is — × 10−6 at 298 K (Nearest integer).
Given: Kb = 0.5 K kg mol−1, Kf = 1.8 kg mol−1
Answer: 13
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Question 85:
Water decomposes at 2300 K:
2 H2O(g) ⇌ 2 H2(g) + O2(g)
The percent of water decomposing at 2300 K and 1 bar is —— (Nearest integer). Equilibrium constant for the reaction is 2 × 10^-3 at 2300 K.
Answer: 2
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Question 86:
The following figure shows the dependence of molar conductance of two electrolytes on concentration. Lambda°m is the limiting molar conductivity. The number of incorrect statements from the following is ——.
(A) Lambda°m for electrolyte A is obtained by extrapolation.
(B) For electrolyte B, sqrt(c) vs. Lambda(m) graph is a straight line with intercept equal to Lambda°m.
(C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B.
(D) Lambda°m for any electrolyte A or B can be calculated using lambda° for individual ions.
Answer: 2
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Question 87:
For a certain chemical reaction X → Y, the rate of formation of the product is plotted against time as shown in the figure. The number of correct statements from the following is ——.
(A) Overall order of this reaction is one.
(B) Order of this reaction cannot be determined.
(C) In region-I and III, the reaction is of first and zero order respectively.
(D) In region-II, the reaction is of first order.
(E) In region-II, the order of the reaction is in the range of 0.1 to 0.9.
Answer: 2
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Question 88:
The sum of bridging carbonyls in W(CO)6 and Mn2(CO)10 is ——.
Answer: 0
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Question 89:
Following chromatogram was developed by adsorption of compound A on a 6 cm TLC glass plate. The retardation factor of the compound A is —— × 10−1.
Answer: 6
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Question 90:
17 mg of a hydrocarbon (Molecular Formula: C10H16) takes up 8.40 mL of H2 gas measured at 0°C and 760 mmHg. Ozonolysis of the hydrocarbon yields certain products. The number of double bonds present in the hydrocarbon is ——.
Answer: 3
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Also Check: JEE Main 2024 Question Paper with Solution PDF Download
JEE Main 2023 Jan 29 Shift 1 Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Answer Key PDF |
|---|---|
| Aakash BYJUs | Check Here |
| Vedantu | Check Here |
JEE Main 2023 Paper Analysis Jan 29 Shift 1
JEE Main 2023 Paper Analysis for the exam scheduled on January 29 Shift 1 has been updated here. Candidates can check subject-wise paper analysis for the exam scheduled on January 29 Shift 1 here along with the topics with the highest weightage.














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