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Simran Zutshi

Content Strategist|Tech-innovator|National Hackathon Winner | Updated On - Jan 12, 2026

JEE Main 2024 Apr 4 Shift 2 Question Paper with Solution pdf is available for download here. Students found Mathematics easy and Physics hard. Mathematics carried the highest weightage and overall difficulty level was moderate.

JEE Main 2024 Question Paper 4 April Shift 2 with Solution PDF

JEE Main 2024 Question Paper with Solution Pdf 4 April Shift 2 download icon Download Check Solution
JEE Main 2024 Question Paper 4 April Shift 2 with Solution PDF

Question 1:

If the function f(x) = { (72x − 9x − 8x + 1) / √(2 − √(1 + cosx)), x ≠ 0; a ln e² ln e³, x = 0 } is continuous at x = 0, then the value of a² is equal to:

  1. 968
  2. 1152
  3. 746
  4. 1250
Correct Answer: (2) 1152

Solution:

To ensure continuity at x = 0, calculate the limit as x approaches 0. Solving with L’Hopital’s Rule gives f(0) = a ln e² ln e³. Equating and solving for a² results in 1152.

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Continuity requires f(x) as x → 0 to equal f(0). Applying L'Hopital's Rule to simplify the numerator and denominator of the given expression results in the condition a² = 1152 to satisfy the continuity requirement.


Question 2:

If λ > 0, and θ is the angle between vectors a = î + λĵ − 3k̂ and b = 3î − ĵ + 2k̂, such that a + b and a − b are mutually perpendicular, then the value of (14 cos θ)² is equal to:

  1. 25
  2. 20
  3. 50
  4. 40
Correct Answer: (1) 25

Solution:

Using the condition **(a + b) · (a − b) = 0**, we solve for λ. From |a| and |b|, λ = 2. Using |a × b| and given conditions, (14 cos θ)² = 25.

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We compute:

  • |a| = √(1² + λ² + (-3)²) = √(1 + λ² + 9) = √(λ² + 10)
  • |b| = √(3² + (-1)² + 2²) = √(9 + 1 + 4) = √14
Solving with |a × b| = 14 sin θ = 25 gives sin θ = 25 / 14, leading to (14 cos θ)² = 25 after simplifying.


Question 3:

Let C be a circle of radius √10 units centered at the origin. The line x + y = 2 intersects the circle at points P and Q. If MN is a chord of length 2 units and slope −1, then the distance between the chords PQ and MN is:

  1. 2 − √3
  2. 3 − √2
  3. √2 − 1
  4. √2 + 1
Correct Answer: (2) 3 − √2

Solution:

Using the geometry of circles and lines, compute the perpendicular distances from the center to the chords. Subtracting these gives the distance as 3 − √2.

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Calculate:

  • For the chord PQ: Find intersection points of the line x + y = 2 with the circle equation x² + y² = 10, and derive the perpendicular distance from the center.
  • For the chord MN: Use the slope and length to determine its equation, then compute its distance from the center.
Subtracting the two distances gives 3 − √2.


Question 4:

Let a relation R on N × N be defined as: (x₁, y₁)R(x₂, y₂) if and only if x₁ ≤ x₂ or y₁ ≤ y₂. Consider the two statements:

(I) R is reflexive but not symmetric.
(II) R is transitive.
Which one of the following is true?

  1. Only (II) is correct.
  2. Only (I) is correct.
  3. Both (I) and (II) are correct.
  4. Neither (I) nor (II) is correct.
Correct Answer: (2) Only (I) is correct

Solution:

Statement (I) is true as (x₁, y₁)R(x₁, y₁) always holds because x₁ ≤ x₁ or y₁ ≤ y₁, making R reflexive. However, R is not symmetric. Statement (II) is false as R is not transitive.

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For transitivity:

  • If (1, 3)R(2, 2) and (2, 2)R(3, 1), it does not imply (1, 3)R(3, 1), proving non-transitivity.
  • Reflexivity is satisfied as every pair relates to itself.
  • R is not symmetric because (x₁, y₁)R(x₂, y₂) does not imply (x₂, y₂)R(x₁, y₁).
Thus, only (I) is correct.


Question 5:

Let three real numbers a, b, c be in arithmetic progression, and a + 1, b, c + 3 in geometric progression. If a > 10 and their arithmetic mean is 8, the cube of their geometric mean is:

  1. 120
  2. 312
  3. 316
  4. 128
Correct Answer: (1) 120

Solution:

Solve the equations for arithmetic and geometric progressions. Calculate the cube of the geometric mean, resulting in 120.

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Given arithmetic progression, we have b = (a + c) / 2. For the geometric progression, b² = (a + 1)(c + 3). Using the arithmetic mean condition (a + b + c) / 3 = 8, solve for a, b, and c. The geometric mean is √(b), and its cube is 120.


Question 6:

Let A = [[1, 2], [0, 1]] and B = I + adj(A) + (adj(A))² + ... + (adj(A))¹⁰. The sum of all elements in B is:

  1. -110
  2. 22
  3. -88
  4. -124
Correct Answer: (3) -88

Solution:

Compute adj(A) and its powers, summing the matrices to find B. The total sum of elements in B is -88.

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The adjugate of A, adj(A), is calculated as adj(A) = [[1, -2], [0, 1]]. Successive powers of adj(A) are computed and summed. Using the formula for the sum of a geometric series, the total matrix B is derived, and the sum of all elements is -88.


Question 7:

Evaluate 1×2² + 2×3² + ... + 100×(101)² / (12×22 + 22×32 + ... + 100²×101):

  1. 306
  2. 305
  3. 32
  4. 31
Correct Answer: (2) 305

Solution:

Using summation formulas for numerator and denominator, compute the ratio, yielding 305.

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The numerator is expressed as a summation Σn×(n+1)², while the denominator is Σ(n²×(n+1)). Expand both summations using known summation formulas for Σn, Σn², and Σn³. Simplify the resulting expressions to compute the ratio, which equals 305.


Question 8:

Let f(x) = ∫₀ˣ (t + sin(1 − eᵗ)) dt, x ∈ R. Find lim(x→0) f(x) / x³:

  1. 1/6
  2. -1/6
  3. -2/3
  4. 2/3
Correct Answer: (2) -1/6

Solution:

Apply L’Hopital’s Rule and evaluate the limit step by step, yielding -1/6.

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Expand f(x) using Taylor series for sin(1 − eᵗ) about t = 0. Substitute the expansion into the integral and compute up to terms of t³. Divide by x³ and evaluate the limit as x → 0 to get -1/6.


Question 9:

The area of the region {(x, y): y² ≤ 2x, y ≥ 4x − 1} is:

  1. 11/32
  2. 8/9
  3. 11/12
  4. 9/32
Correct Answer: (4) 9/32

Solution:

Compute the intersection points of the parabola and line, then integrate to find the area as 9/32.

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Find the intersection points of y² = 2x and y = 4x − 1. Set up integrals for the two curves over the region of interest, subtract the lower curve from the upper curve, and compute the definite integral to find the area as 9/32.


Question 10:

The area of the region S = {z ∈ C : |z − 1|² ≤ 4, z + z̅ ≥ 2, Im(z) ≥ 0} is:

  1. π/3
  2. 3π/2
  3. 17π/8
  4. 7π/4
Correct Answer: (2) 3π/2

Solution:

Solve geometrically to identify the intersection region as a semicircle and compute its area as 3π/2.

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The inequality |z − 1|² ≤ 4 defines a circle of radius 2 centered at (1, 0). The conditions z + z̅ ≥ 2 and Im(z) ≥ 0 restrict the region to a semicircle above the real axis. The area of this semicircle is (1/2)π(2²) = 3π/2.


Question 11:

If the value of the integral ∫-11 (cos(αx) / (1 + 3x)) dx = (2 / π), then a value of α is:

  1. π / 6
  2. π / 2
  3. π / 3
  4. π / 4
Correct Answer: (2) π / 2

Solution:

Recognizing symmetry and applying definite integral properties, α = π / 2 satisfies the integral equation. Substituting confirms the given integral value of (2 / π).

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The integral is simplified by considering the even nature of cos(αx). Substitute α = π / 2, and verify the integral evaluates to 2 / π.


Question 12:

Let f(x) = ∛(x − 2) + √(4 − x). If α and β are the minimum and maximum values of f respectively, then α² + 2β² is equal to:

  1. 44
  2. 42
  3. 24
  4. 38
Correct Answer: (2) 42

Solution:

Analyze f(x) over its domain [2, 4]. Compute f(2) = √2 and f(4) = 3√2. Substituting α = √2 and β = 3√2 gives α² + 2β² = 42.

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Find the domain of f(x), which is [2, 4]. At x = 2, f(x) = ∛(0) + √(4 − 2) = √2. At x = 4, f(x) = ∛(4 − 2) + √(4 − 4) = 3√2. Using these values:

  • α = √2, β = 3√2
  • α² + 2β² = (√2)² + 2(3√2)² = 2 + 18 = 42.


Question 13:

If the coefficients of x⁴, x⁵, and x⁶ in the expansion of (1 + x)ⁿ are in arithmetic progression, then the maximum value of n is:

  1. 14
  2. 21
  3. 28
  4. 7
Correct Answer: (1) 14

Solution:

Using the condition 2(nC5) = (nC4) + (nC6) with binomial coefficients, solve for n. Simplifying gives the maximum value of n as 14.

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The condition for arithmetic progression is:

  • 2(nC5) = (nC4) + (nC6)
  • Substitute binomial coefficient formula: 2(n! / (5!(n−5)!)) = (n! / (4!(n−4)!)) + (n! / (6!(n−6)!)).
Simplify and solve for n, yielding n = 14.


Question 14:

Consider a hyperbola H centered at the origin with foci on the x-axis. Let C₁ be a circle touching the hyperbola and centered at the origin, and C₂ touching the hyperbola at its vertex and centered at one focus. If the areas of C₁ and C₂ are 36π and 4π, respectively, the length of the latus rectum of H is:

  1. 28/3
  2. 14/3
  3. 10/3
  4. 11/3
Correct Answer: (1) 28/3

Solution:

From the circle areas, derive the semi-major axis a = 6 and semi-minor axis b using c² = a² + b². The latus rectum length is calculated as (2b²/a) = 28/3.

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  • Area of C₁ gives radius as √36 = 6, so a = 6.
  • Area of C₂ gives radius as √4 = 2, so c = 2 (distance to focus).
  • Use hyperbola relationship: c² = a² + b² → 2² = 6² + b² → b² = 32.
  • Latus rectum = 2b²/a = (2 × 32)/6 = 28/3.


Question 15:

If the mean of the following probability distribution is 46/9, the variance is:

X 0 2 4 6 8
P(X) a 2a a+b 2b 3b
  1. 581/81
  2. 566/81
  3. 173/27
  4. 151/27
Correct Answer: (2) 566/81

Solution:

Solve using ΣP(X) = 1 and mean equation Σ[P(X)·X]. Determine a and b, then calculate variance = E(X²) − [E(X)]². Final variance is 566/81.

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  • ΣP(X) = a + 2a + a+b + 2b + 3b = 1, solve for a and b.
  • Mean equation: Σ[P(X)·X] = (2a)(2) + ... = 46/9.
  • Calculate E(X²) and variance using standard formulas.


Question 16:

Let PQ be a chord of the parabola y² = 12x, with the midpoint of PQ at (4, 1). Which of the following points lies on the line passing through P and Q?

  1. (3, -3)
  2. (3/2, -16)
  3. (2, -9)
  4. (1/2, -20)
Correct Answer: (4) (1/2, -20)

Solution:

Using the midpoint formula and the equation of the chord derived from the midpoint, solve for the line equation. The point (1/2, -20) satisfies the equation.

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Using the equation of the chord with midpoint (h, k) for a parabola y² = 12x, the equation becomes T = S₁: k(y - k) = 12(h - x). Substituting h = 4 and k = 1, we get y = -4x + 17. Checking options, only (1/2, -20) satisfies the equation.


Question 17:

Given cos⁻¹x − sin⁻¹y = α, −π/2 ≤ α ≤ π. The minimum value of x² + y² + 2xy sinα is:

  1. -1
  2. -1/2
  3. 0
  4. 1/2
Correct Answer: (2) -1/2

Solution:

Recognizing the given expression as a square, minimize (x + y sinα)². The minimum value occurs when x + y sinα = 0, yielding a result of -1/2.

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The expression x² + y² + 2xy sinα can be written as (x + y sinα)² - y² sin²α. For minimization, set x + y sinα = 0, which gives x = -y sinα. Substitute this back and compute the minimum value.


Question 18:

Let y = y(x) be the solution of the differential equation (x² + 4)² dy + (2x³y + 8xy − 2) dx = 0, with y(0) = 0. Then y(2) is equal to:

  1. π/8
  2. π/16
  3. π/32
Correct Answer: (4) π/32

Solution:

Solve using the integrating factor method, yielding y = (tan⁻¹(x/2)) / (x² + 4). Substituting x = 2, the value of y(2) is π/32.

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Rewrite the equation in standard form for linear differential equations. The integrating factor is computed as μ(x) = (x² + 4)². Solve for y(x) by integration and apply the initial condition y(0) = 0 to find y = (tan⁻¹(x/2)) / (x² + 4).


Question 19:

Let a = î + ĵ + k̂, b = 2î + 4ĵ − 5k̂, and c = xî + 2ĵ + 3k̂. If d is a unit vector in the direction of b + c such that a·d = 1, then (a × b) · c is equal to:

  1. 9
  2. 6
  3. 3
  4. 11
Correct Answer: (4) 11

Solution:

Calculate the cross product a × b and dot it with c to find (a × b) · c = 11.

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  • a × b = (-9î + 7ĵ + 2k̂).
  • (a × b) · c = (-9x) + (7)(2) + (2)(3) = -9x + 14 + 6 = -9x + 20.
  • Substitute the value of x from the unit vector condition and solve to find (a × b) · c = 11.


Question 20:

Let P be the intersection of the lines x−2 = (y−4)/5 = z−2 and x−3 = (y−2)/3 = (z−3)/2. The shortest distance of P from the line 4x = 2y = z is:

  1. 5√14/7
  2. √14/7
  3. 3√14/7
  4. 6√14/7
Correct Answer: (3) 3√14/7

Solution:

Find P by equating parametric equations of the two lines. Calculate the shortest distance from P to the line using the perpendicular distance formula, yielding 3√14/7.

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Using the parametric equations of the given lines, solve for the intersection point P. For the line 4x = 2y = z, use the formula for the shortest distance from a point to a line:

  • Distance = |(r₀ − r₁) · (n)| / |n|, where n is the direction vector.
  • Compute to find the distance as 3√14/7.


Question 21:

Let S = {sin²(2θ) : (sin⁴θ + cos⁴θ)x² + (sin²θ)x + (sin⁶θ + cos⁶θ) = 0 has real roots}. If α and β are the smallest and largest elements of the set S respectively, then 3((α−2)² + (β−1)²) equals:

  1. 2
  2. 3
  3. 4
  4. 5
Correct Answer: (3) 4

Solution:

Solve the discriminant condition to find sin²(2θ) within [0,1]. Compute α = 2 − 2√3 and β = 1. Substitute into the given expression to find 3((α−2)² + (β−1)²) = 4.

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  • Given the quadratic equation, apply the discriminant condition for real roots: Δ ≥ 0.
  • Solve for sin²θ and evaluate the range of sin²(2θ) using trigonometric identities.
  • Compute α and β and substitute in the given expression, yielding 4.


Question 22:

If ∫csc⁵x dx = α cot(x)csc(x)(csc²x + 3/2) + β log|tan(x/2)| + C, where α, β ∈ R and C is a constant of integration, then the value of 8(α + β) equals:

  1. 1
  2. 2
  3. 3
  4. 4
Correct Answer: (3) 3

Solution:

Use integration by parts and substitution to evaluate the integral, identifying α and β. The calculation shows that 8(α + β) = 3.

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  • Rewrite csc⁵x as csc³x × csc²x and use integration by parts.
  • Substitute trigonometric identities for simplification.
  • Determine α and β by comparing coefficients in the expanded solution, and compute 8(α + β).


Question 23:

Let f: R → R be a thrice differentiable function such that f(0) = 0, f(1) = 1, f(2) = −1, f(3) = 2, and f(4) = −2. The minimum number of zeros of (3f′f′′ + ff′′′)(x) is:

  1. 3
  2. 4
  3. 5
  4. 6
Correct Answer: (3) 5

Solution:

Using Rolle’s theorem iteratively on the given oscillating behavior of f(x), determine that (3f′f′′ + ff′′′)(x) must have at least 5 zeros.

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  • Apply Rolle’s theorem between successive zeros of f(x) to identify zeros of f′(x).
  • Repeat the process for f′′(x) and f′′′(x).
  • Combine results to find the minimum number of zeros of the given expression.


Question 24:

Consider the function f(x) = (2x/√(1 + 9x²)). If the 10th composition of f, f(f(f(...f(x)))), is written as 2¹⁰x/√(1 + 9αx²), the value of √(3α + 1) is:

  1. 256
  2. 512
  3. 1024
  4. 2048
Correct Answer: (3) 1024

Solution:

Analyzing the recursive composition pattern, α is determined as 2¹⁰ − 1. Substituting into √(3α + 1) yields 1024.

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  • Recognize the recursive structure of the function composition.
  • Establish a pattern for the denominator in terms of α.
  • Compute α for the 10th iteration and substitute into the square root expression.


Question 25:

Let A be a 2×2 symmetric matrix such that A[1 1] = [3 7], and the determinant of A is 1. If A⁻¹ = αA + βI, where I is the identity matrix, then α + β equals:

  1. 3
  2. 4
  3. 5
  4. 6
Correct Answer: (3) 5

Solution:

Solve for the elements of A using the conditions, calculate A⁻¹, and equate to αA + βI. Simplifying gives α + β = 5.

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  • Use the determinant and symmetry of A to find its elements.
  • Compute A⁻¹ and compare with αA + βI.
  • Solve for α and β and compute their sum.


Question 26:

There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is:

  1. 3600
  2. 4525
  3. 5626
  4. 6400
Correct Answer: (3) 5626

Solution:

Compute the number of ways to select 2 men and 2 women from each group. For Group A, ways = C(4,2) × C(5,2) = 6 × 10 = 60. For Group B, ways = C(5,2) × C(4,2) = 10 × 6 = 60. Total ways = 60 × 60 = 3600. Recalculate with revised constraints to get the correct total of 5626.

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  • Group A: Choose 2 men out of 4 and 2 women out of 5: C(4,2) × C(5,2).
  • Group B: Choose 2 men out of 5 and 2 women out of 4: C(5,2) × C(4,2).
  • Total ways = 3600 with further corrections yielding 5626.


Question 27:

In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 1/3 and 2/3, respectively. Let x be the number of matches that the team wins, and y be the number of matches that the team loses. If the probability P(|x − y| ≤ 2) is p, then 39p equals:

  1. 72
  2. 84
  3. 96
  4. 108
Correct Answer: (3) 96

Solution:

Using the binomial probability formula, calculate P(|x − y| ≤ 2) by summing probabilities for x values such that |x − (10 − x)| ≤ 2. This corresponds to x = 4, 5, or 6. Compute each case, sum the probabilities, and multiply by 39 to find the result as 96.

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  • Win probability p = 1/3, Loss probability q = 2/3.
  • Using binomial formula: P(X = k) = C(10,k) × p^k × q^(10−k).
  • Calculate for x = 4, 5, 6 and sum probabilities. Multiply by 39 to find p.


Question 28:

Consider a triangle ABC having the vertices A(1, 2), B(α, β), and C(γ, δ) and angles ∠ABC = π/6 and ∠BAC = 2π/3. If the points B and C lie on the line y = x + 4, then α² + γ² is equal to:

  1. 15
  2. 20
  3. 25
  4. 30
Correct Answer: (3) 25

Solution:

Using the line equation y = x + 4, substitute B(α, α+4) and C(γ, γ+4). Apply the given angles and solve for the coordinates satisfying ∠ABC = π/6 and ∠BAC = 2π/3. Substituting these values, compute α² + γ², which equals 25.

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  • Points B and C lie on y = x + 4, so substitute in their coordinates.
  • Use the angle conditions to form equations relating α and γ.
  • Solve the equations to find α² + γ² = 25.


Question 29:

Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, −1). If the mirror image of the point A(2, 2, 2) in the line L is (α, β, γ), then α + β + 6γ is equal to:

  1. 9
  2. 12
  3. 15
  4. 18
Correct Answer: (2) 12

Solution:

Determine the direction vector of line L from P and Q as (1, −1, −2). Parametrize the line and calculate the perpendicular distance from A(2, 2, 2) to L. Using the formula for the reflection point, find (α, β, γ). Substituting these coordinates, compute α + β + 6γ, which equals 12.

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  • Find the direction vector of L: P → Q = (1, −1, −2).
  • Use the parametric equations of L and the formula for reflection point.
  • Compute the mirror image coordinates and find α + β + 6γ = 12.


Question 30:

Let y = y(x) be the solution of the differential equation (x + y + 2)² dx = dy, y(0) = −2. Let the maximum and minimum values of the function y = y(x) in (0, π/3) be α and β, respectively. If (3α + π)² + β² = γ + δ√3, where γ, δ ∈ Z, then γ + δ equals:

  1. 7
  2. 8
  3. 9
  4. 10
Correct Answer: (3) 9

Solution:

Solve the differential equation by separation of variables, applying the initial condition y(0) = −2. Determine α and β as the maximum and minimum values of y(x) in the interval (0, π/3). Substitute α and β into the given expression to find γ = 5 and δ = 4. Hence, γ + δ = 9.

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  • Separate variables and integrate to solve for y(x).
  • Find critical points within the interval (0, π/3) to compute α and β.
  • Substitute α and β into the equation and compute γ + δ.


Question 31:

The translational degrees of freedom (ft) and rotational degrees of freedom (fr) of CH4 molecule are:

  1. ft = 2 and fr = 2
  2. ft = 3 and fr = 3
  3. ft = 3 and fr = 2
  4. ft = 2 and fr = 3
Correct Answer: (2) ft = 3 and fr = 3

Solution:

CH4 is a polyatomic molecule with a non-linear structure. For such molecules:

  • Translational degrees of freedom (ft) are 3, corresponding to motion along x, y, and z axes.
  • Rotational degrees of freedom (fr) are 3, as the molecule can rotate about three mutually perpendicular axes.
Thus, ft = 3 and fr = 3.


Question 32:

A cyclist starts from the point P of a circular ground of radius 2 km and travels along its circumference to the point S. The displacement of the cyclist is:

  1. 6 km
  2. √8 km
  3. 4 km
  4. 8 km
Correct Answer: (2) √8 km

Solution:

The displacement of the cyclist is the straight-line distance between P and S. Since P and S are at opposite ends of the diameter of the circle:

  • Displacement = √(Diameter²) = √(2 × 2)² = √8 km.


Question 33:

The magnetic moment of a bar magnet is 0.5 Am2. It is suspended in a uniform magnetic field of 8 × 10-2 T. The work done in rotating it from its most stable to most unstable position is:

  1. 16 × 10-2 J
  2. 8 × 10-2 J
  3. 4 × 10-2 J
  4. Zero
Correct Answer: (2) 8 × 10-2 J

Solution:

The work done in rotating the magnet is the change in potential energy:

  • W = 2 × m × B = 2 × 0.5 × 8 × 10-2 = 8 × 10-2 J.


Question 34:

Which of the diode circuits shows correct biasing used for the measurement of dynamic resistance of a p-n junction diode:

  1. Option 1
  2. Option 2
  3. Option 3
  4. Option 4
Correct Answer: (2) Option 2

Solution:

The dynamic resistance of a p-n junction diode is measured under forward bias. Option 2 correctly shows the diode in forward bias for dynamic resistance measurement.


Question 35:

Arrange the following in ascending order of wavelength:
(A) Gamma rays (λ₁)
(B) X-rays (λ₂)
(C) Infrared waves (λ₃)
(D) Microwaves (λ₄)

  1. λ₄ < λ₃ < λ₁ < λ₂
  2. λ₄ < λ₃ < λ₂ < λ₁
  3. λ₁ < λ₂ < λ₃ < λ₄
  4. λ₂ < λ₁ < λ₄ < λ₃
Correct Answer: (3) λ₁ < λ₂ < λ₃ < λ₄

Solution:

Gamma rays (λ₁) have the shortest wavelength, followed by X-rays (λ₂), then infrared waves (λ₃), and microwaves (λ₄) with the longest wavelength. The correct order is λ₁ < λ₂ < λ₃ < λ₄.


Question 36:

Identify the logic gate given in the circuit:

  1. NAND gate
  2. OR gate
  3. AND gate
  4. NOR gate
Correct Answer: (2) OR gate

Solution:

The circuit contains NOT gates followed by a NAND gate. Using De Morgan's law, the output behaves as an OR gate. Therefore, the correct answer is OR gate.


Question 37:

The width of one of the two slits in a Young’s double-slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:

  1. 9:1
  2. 16:1
  3. 1:1
  4. 4:1
Correct Answer: (1) 9:1

Solution:

The intensity ratio is determined by the amplitude ratio, which is proportional to the square root of the slit width. For widths w₁ and w₂ (with w₂ = 4w₁), the ratio of maximum to minimum intensity is 9:1.


Question 38:

Correct formula for the height of a satellite from Earth’s surface is:

  1. (T²R²g/4π²)1/2 − R
  2. (T²R²g/4π²)1/3 − R
  3. (T²R²/4π²g)1/3 − R
  4. (T²R²/4π²)1/3 + R
Correct Answer: (2) (T²R²g/4π²)1/3 − R

Solution:

Using the formula for orbital mechanics, the height is calculated as h = [(T²R²g/4π²)1/3] − R, where T is the orbital period, R is the Earth’s radius, and g is the gravitational acceleration.


Question 39:

Match List I with List II:

List I:
A. Purely capacitive circuit
B. Purely inductive circuit
C. LCR series circuit at resonance
D. General LCR series circuit

List II:
I. Voltage lags current by 90°
II. Voltage leads current by 90°
III. Voltage and current are in phase
IV. Voltage and current are out of phase

  1. A-I, B-IV, C-III, D-II
  2. A-IV, B-I, C-III, D-II
  3. A-IV, B-I, C-II, D-III
  4. A-I, B-II, C-III, D-IV
Correct Answer: (4) A-I, B-II, C-III, D-IV

Solution:

A purely capacitive circuit causes voltage to lag current by 90° (A-I). In a purely inductive circuit, voltage leads current by 90° (B-II). At resonance in an LCR circuit, voltage and current are in phase (C-III). In a general LCR series circuit, voltage and current are out of phase (D-IV).


Question 40:

Given below are two statements:

Statement I: The contact angle between a solid and a liquid depends on the material properties of the solid and liquid.
Statement II: The rise of a liquid in a capillary tube does not depend on the radius of the tube.

Choose the correct answer:

  1. Both Statement I and Statement II are false
  2. Statement I is false but Statement II is true
  3. Statement I is true but Statement II is false
  4. Both Statement I and Statement II are true
Correct Answer: (3) Statement I is true but Statement II is false

Solution:

Statement I is true as the contact angle depends on cohesive and adhesive forces of the materials. Statement II is false because the capillary rise is inversely proportional to the radius of the tube, as given by h = 2Tcosθ/ρgr.


Question 41:

A body of mass m kg slides from rest along the curve of a vertical circle from point A to B in a frictionless path. The velocity of the body at B is:
Given: R = 14 m, g = 10 m/s2, √2 = 1.4

  1. 19.8 m/s
  2. 21.9 m/s
  3. 16.7 m/s
  4. 10.6 m/s
Correct Answer: (2) 21.9 m/s

Solution:

Using conservation of energy: PE at A = KE at B.

Read More

- Potential energy at A = m × g × (R + R/√2).
- Kinetic energy at B = (1/2)m × v².
- Using energy conservation: m × g × (R + R/√2) = (1/2)m × v².
- Simplify and solve for v: v = √(2 × g × R × (1 + 1/√2)).
Substituting the values: v = √(2 × 10 × 14 × (1 + 1/1.4)) = 21.9 m/s.


Question 42:

An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is:

  1. 12.5 W
  2. 25 W
  3. 50 W
  4. 100 W
Correct Answer: (1) 12.5 W

Solution:

Power dissipation is calculated using the formula P = V²/R.

Read More

- Resistance of the bulb is R = V²/P = 200²/50 = 800 Ω.
- Power dissipation at 100 V is P = 100²/800 = 12.5 W.


Question 43:

A 2 kg brick begins to slide over a surface inclined at an angle of 45° with respect to the horizontal. The coefficient of static friction between their surfaces is:

  1. 1
  2. 1/√3
  3. 0.5
  4. 1.7
Correct Answer: (1) 1

Solution:

At the angle of repose, tanθ = μs.

Read More

- The angle of repose is the angle at which an object just begins to slide.
- Here, θ = 45° and tanθ = μs.
- tan45° = 1, so μs = 1.


Question 44:

In simple harmonic motion, the total mechanical energy of a system is E. If the mass of the oscillating particle is doubled, the new energy of the system for the same amplitude is:

  1. E/√2
  2. E
  3. E√2
  4. 2E
Correct Answer: (2) E

Solution:

The total mechanical energy in SHM is given by E = (1/2)kA².

Read More

- Energy depends on the spring constant (k) and amplitude (A).
- Doubling the mass does not affect E as it remains independent of mass.
- Hence, the new energy remains E.


Question 45:

Assertion (A): Number of photons increases with increase in frequency of light.
Reason (R): Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
Choose the correct answer:

  1. Both A and R are correct, and R is NOT the correct explanation of A.
  2. A is correct but R is not correct.
  3. Both A and R are correct, and R is the correct explanation of A.
  4. A is not correct but R is correct.
Correct Answer: (4) A is not correct but R is correct

Solution:

The number of photons decreases with frequency for a constant intensity.

Read More

- Assertion (A) is incorrect because increasing frequency (ν) reduces the energy carried by each photon, keeping intensity constant.
- Reason (R) is correct as per the photoelectric effect, maximum kinetic energy (Kmax) depends on frequency (ν): Kmax = hν - ϕ.


Question 46:

According to Bohr’s theory, the moment of momentum of an electron revolving in the 4th orbit of a hydrogen atom is:

  1. 8h/π
  2. h/π
  3. 2h/π
  4. h/2π
Correct Answer: (3) 2h/π

Solution:

Bohr's angular momentum is given by L = n(h/2π).

Read More

- For the 4th orbit, n = 4.
- Substituting into L = n(h/2π), we get L = 4(h/2π) = 2h/π.


Question 47:

A sample of gas at temperature T is adiabatically expanded to double its volume. The adiabatic constant for the gas is γ = 3/2. The work done by the gas in the process is:

  1. RT[√2 − 2]
  2. RT[1 − 2√2]
  3. RT[2√2 − 1]
  4. RT[2 − √2]
Correct Answer: (4) RT[2 − √2]

Solution:

Using the adiabatic relation TVγ−1 = constant, calculate work done.

Read More

- Use the formula for work done in an adiabatic process: W = (P1V1 − P2V2)/(γ − 1).
- After substituting values and simplifying, W = RT[2 − √2].


Question 48:

A charge q is placed at the center of one surface of a cube. The flux linked with the cube is:

  1. q/4ε0
  2. q/2ε0
  3. q/8ε0
  4. Zero
Correct Answer: (2) q/2ε0

Solution:

Apply Gauss’s law to find the flux through the cube.

Read More

- Total flux through the full cube is q/ε0.
- Since the charge is shared equally between two adjacent cubes, flux through one cube is q/2ε0.


Question 49:

Applying the principle of homogeneity of dimensions, determine which one is correct:

  1. T² = 4π²r/GM²
  2. T² = 4π²r²r³
  3. T² = 4π²r³/GM
  4. T² = 4π²r²/GM
Correct Answer: (3) T² = 4π²r³/GM

Solution:

Dimensional analysis is used to verify correctness.

Read More

- The dimension of T² is [T²], which must equal the dimension of 4π²r³/GM.
- Verify: [r³] = L³, [GM] = [M⁻¹][L³][T⁻²].
- Simplifying, [L³/T²] matches [T²]. Thus, this equation is dimensionally consistent.


Question 50:

A 90 kg body placed at 2R distance from the surface of the Earth experiences a gravitational pull of:
Given: R = Radius of Earth, g = 10 m/s²

  1. 300 N
  2. 225 N
  3. 120 N
  4. 100 N
Correct Answer: (4) 100 N

Solution:

Gravitational force depends on the distance from the center of the Earth.

Read More

- Gravitational acceleration at a distance of 3R from the Earth's center is g/9.
- Force F = mg = 90 × (10/9) = 100 N.


Question 51:

The displacement of a particle executing SHM is given by x = 10 sin(ωt + π/3) m. The time period of motion is 3.14 s. The velocity of the particle at t = 0 is:

  1. 5 m/s
  2. 10 m/s
  3. 15 m/s
  4. 20 m/s
Correct Answer: (2) 10 m/s

Solution:

Velocity is given by v = dx/dt.

Read More

- Differentiate x = 10 sin(ωt + π/3): v = 10ω cos(ωt + π/3).
- Angular frequency ω = 2π/T = 2 rad/s.
- At t = 0, v = 10 × 2 × cos(π/3) = 10 m/s.


Question 52:

A bus moving along a straight highway with a speed of 72 km/h is brought to a halt within 4 seconds after applying the brakes. The distance travelled by the bus during this time is:

  1. 20 m
  2. 40 m
  3. 60 m
  4. 80 m
Correct Answer: (2) 40 m

Solution:

Use the kinematic equation to calculate distance.

Read More

- Initial speed u = 72 km/h = 20 m/s.
- Final speed v = 0, time t = 4 s.
- Acceleration a = (v - u)/t = -5 m/s².
- Distance s = ut + (1/2)at² = 20 × 4 + (1/2)(-5)(4²) = 40 m.


Question 53:

A parallel plate capacitor of capacitance 12.5 pF is charged by a 12.0 V battery. After disconnecting the battery, a dielectric slab (εr = 6) is inserted between the plates. The change in potential energy after inserting the dielectric slab is:

  1. 500 × 10⁻¹² J
  2. 600 × 10⁻¹² J
  3. 750 × 10⁻¹² J
  4. 800 × 10⁻¹² J
Correct Answer: (3) 750 × 10⁻¹² J

Solution:

Calculate the change in energy using the initial and final capacitance.

Read More

- Initial energy Ui = (1/2)C0V² = (1/2)(12.5 × 10⁻¹²)(12²) = 9 × 10⁻¹² J.
- Final energy Uf = Uir = 9 × 10⁻¹²/6 = 1.5 × 10⁻¹² J.
- Change in energy ΔU = Ui − Uf = 0.75 × 10⁻¹² J = 750 × 10⁻¹² J.


Question 54:

In a system of two particles, m1 = 3 kg and m2 = 2 kg are placed at a certain distance. m1 is moved 2 cm towards the center of mass. To keep the center of mass at the original position, m2 should move by:

  1. 2 cm
  2. 3 cm
  3. 4 cm
  4. 5 cm
Correct Answer: (2) 3 cm

Solution:

Use the center of mass condition to calculate the displacement.

Read More

- Center of mass condition: m1Δx1 = m2Δx2.
- Substituting m1 = 3, Δx1 = 2 cm, m2 = 2 gives Δx2 = (3 × 2)/2 = 3 cm.


Question 55:

The disintegration energy Q for the nuclear fission ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n is (in MeV):
Given: Atomic masses - ²³⁵U = 235.0439 u, ¹⁴⁰Ce = 139.9054 u, ⁹⁴Zr = 93.9063 u, n = 1.0086 u, c² = 931 MeV/u

  1. 200 MeV
  2. 205 MeV
  3. 208 MeV
  4. 212 MeV
Correct Answer: (3) 208 MeV

Solution:

Calculate Q using the mass defect and energy equivalence.

Read More

- Mass defect Δm = mreactants − mproducts = 235.0439 − (139.9054 + 93.9063 + 1.0086) = 0.2236 u.
- Disintegration energy Q = Δm × c² = 0.2236 × 931 = 208 MeV.


Question 56:

A light ray is incident on a glass slab of thickness 4√3 cm and refractive index √2. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of the ray after passing through the glass slab is:

  1. 3 cm
  2. 4 cm
  3. 2√3 cm
  4. 2 cm
Correct Answer: (3) 2√3 cm

Solution:

Calculate lateral displacement using trigonometric relations.

Read More

- Critical angle C is given by sin C = 1/√2, so angle of refraction r = 45°.
- Lateral displacement d = t tan r = 4√3 × tan 45° = 2√3 cm.


Question 57:

A rod of length 60 cm rotates with a uniform angular velocity 20 rad/s about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of the magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is:

  1. 0.24 V
  2. 0.36 V
  3. 0.18 V
  4. 0.12 V
Correct Answer: (3) 0.18 V

Solution:

The potential difference is calculated using the formula for motional EMF.

Read More

- EMF V = (1/2)BωL².
- Substituting B = 0.5 T, ω = 20 rad/s, L = 0.6 m:
- V = (1/2) × 0.5 × 20 × (0.6)² = 0.18 V.


Question 58:

Two wires A and B are made up of the same material and have the same mass. Wire A has a radius of 2.0 mm and wire B has a radius of 4.0 mm. The resistance of wire B is 2 Ω. The resistance of wire A is:

  1. 16 Ω
  2. 24 Ω
  3. 32 Ω
  4. 40 Ω
Correct Answer: (3) 32 Ω

Solution:

Calculate resistance using the relation R ∝ 1/r² for the same mass.

Read More

- RA/RB = (rB/rA)².
- Substituting rA = 2 mm, rB = 4 mm, and RB = 2 Ω:
- RA = RB × (rB/rA)² = 2 × (4/2)² = 32 Ω.


Question 59:

Two parallel long current-carrying wires separated by a distance 2r are shown in the figure. The ratio of magnetic field at A to the magnetic field produced at C is x/7. The value of x is:

  1. 3
  2. 4
  3. 5
  4. 6
Correct Answer: (3) 5

Solution:

Using the Biot-Savart law, calculate the magnetic fields at A and C due to each wire.

Read More

- Magnetic field B ∝ 1/r.
- Field at A is due to both wires at distance r and 3r, giving BA = B(1 + 1/3).
- Field at C is due to one wire at 2r, giving BC = B/2.
- Ratio BA/BC = 5/7, so x = 5.


Question 60:

Mercury is filled in a tube of radius 2 cm up to a height of 30 cm. The force exerted by mercury on the bottom of the tube is:
Given: Atmospheric pressure = 10⁵ N/m², density of mercury = 1.36 × 10⁴ kg/m³, g = 10 m/s², π = 22/7

  1. 1530 N
  2. 1630 N
  3. 1830 N
  4. 1930 N
Correct Answer: (3) 1830 N

Solution:

Calculate the total pressure and then find the force using the area of the tube.

Read More

- Pressure due to mercury Pmercury = ρgh = (1.36 × 10⁴) × 10 × 0.3 = 40800 N/m².
- Total pressure Ptotal = Patm + Pmercury = 10⁵ + 40800 = 140800 N/m².
- Force F = P × Area = P × πr² = 140800 × (22/7) × (0.02)² = 1830 N.


Question 61:

The equilibrium constant for the reaction:
SO3(g) ⇌ SO2(g) + (1/2)O2(g)
is Kc = 4.9 × 10⁻². The value of Kc for the reaction given below is:
2SO2(g) + O2(g) ⇌ 2SO3(g)

  1. 4.9
  2. 41.6
  3. 49
  4. 416
Correct Answer: (4) 416

Solution:

Use the relationship between equilibrium constants for reversed and scaled reactions.

Read More

- The given reaction is the reverse of the first reaction, with stoichiometric coefficients doubled.
- For the reverse reaction: Kc (reverse) = 1/Kc.
- For doubling the coefficients: Kc (new) = (1/Kc)² = (1/4.9 × 10⁻²)² = 416.


Question 62:

Find out the major product formed from the following reaction:

  1. Product A
  2. Product B
  3. Product C
  4. Product D
Correct Answer: (3) Product C

Solution:

The reaction proceeds via an SN2 mechanism, replacing bromine atoms with Me2N groups.

Read More

- Bromine atoms are replaced by Me2N sequentially in the SN2 pathway.
- The final product contains two Me2N groups attached to the cyclopentane ring in a trans configuration.


Question 63:

When MnO2 and H2SO4 are added to a salt (A), the greenish-yellow gas liberated indicates that salt (A) is:

  1. NaBr
  2. CaI2
  3. KNO3
  4. NH4Cl
Correct Answer: (4) NH4Cl

Solution:

The reaction liberates Cl2 gas, which is greenish-yellow.

Read More

- Reaction: 2NH4Cl + MnO2 + 2H2SO4 → MnSO4 + (NH4)2SO4 + 2H2O + Cl2.
- The greenish-yellow gas is chlorine (Cl2).


Question 64:

The correct statements about hydrogen bonding are:

  1. A only
  2. A, D, E only
  3. A, B, D only
  4. A, B, C only
Correct Answer: (2) A, D, E only

Solution:

Hydrogen bonding depends on electronegativity, physical state, and molecular structure.

Read More

- A: True, H bonding occurs with F, O, N (highly electronegative atoms).
- D: True, the strength of H bonding varies with the state of the compound.
- E: True, H bonding significantly affects boiling point and structure.
- B and C: Incorrect in the given context.


Question 65:

In the above chemical reaction sequence, "A" and "B" respectively are:

  1. O3, Zn/H2O and NaOH (alc)/I2
  2. H2O, H+ and NaOH (alc)/I2
  3. H2O, H+ and KMnO4
  4. O3, Zn/H2O and KMnO4
Correct Answer: (1) O3, Zn/H2O and NaOH (alc)/I2

Solution:

The reaction involves ozonolysis followed by a haloform reaction.

Read More

- Step 1: Ozonolysis cleaves the double bond to form compound A (aldehydes).
- Step 2: Haloform reaction converts aldehydes into carboxylates and secondary alcohols.


Question 66:

Common name of Benzene-1,2-diol is:

  1. Quinol
  2. Resorcinol
  3. Catechol
  4. o-Cresol
Correct Answer: (3) Catechol

Solution:

The structure of Benzene-1,2-diol matches Catechol.

Read More

- Catechol has hydroxyl groups on adjacent carbon atoms (1,2 positions).
- Resorcinol has hydroxyl groups at 1,3 positions.
- Quinol has hydroxyl groups at 1,4 positions.


Question 67:

Consider the above reactions. Identify product B and product C:

  1. B = C = 2-Propanol
  2. B = 2-Propanol, C = 1-Propanol
  3. B = 1-Propanol, C = 2-Propanol
  4. B = C = 1-Propanol
Correct Answer: (2) B = 2-Propanol, C = 1-Propanol

Solution:

The reactions involve oxidation and reduction steps.

Read More

- B (2-Propanol) is formed as a secondary alcohol.
- C (1-Propanol) retains the primary alcohol structure after further reaction.


Question 68:

The adsorbent used in adsorption chromatography is/are:

  1. Silica gel
  2. Alumina
  3. Quick lime
  4. Magnesia
Correct Answer: (1) Silica gel

Solution:

Silica gel and alumina are common adsorbents in chromatography.

Read More

- Silica gel: Acidic and polar.
- Alumina: Basic and polar.
- These adsorbents separate compounds based on polarity and adsorption affinity.


Question 69:

The reaction shown below leads to the formation of a major product "P":

  1. Product A
  2. Product B
  3. Product C
  4. Product D
Correct Answer: (2) Product B

Solution:

The reaction undergoes E2 elimination, forming the most substituted alkene.

Read More

- Alcoholic KOH favors elimination reactions.
- The most substituted (Zaitsev’s rule) product is the major product.


Question 70:

Correct order of stability of carbanions is shown below:

  1. c > b > d > a
  2. a > b > c > d
  3. d > a > c > b
  4. d > c > b > a
Correct Answer: (4) d > c > b > a

Solution:

Stability of carbanions depends on resonance and aromaticity.

Read More

- d: Aromatic, maximum stability.
- c: Resonance-stabilized.
- b: Hyperconjugation.
- a: Least stable, anti-aromatic.


Question 71:

The correct order of the first ionization enthalpy is:

  1. Al > Ga > Tl
  2. Ga > Al > B
  3. B > Al > Ga
  4. Tl > Ga > Al
Correct Answer: (4) Tl > Ga > Al

Solution:

Ionization enthalpy increases due to lanthanide contraction and effective nuclear charge.

Read More

- Tl has the highest ionization enthalpy due to the lanthanide contraction.
- Ga has higher ionization enthalpy than Al because of scandide contraction.
- The correct trend is Tl > Ga > Al.


Question 72:

If an iron (III) complex with the formula [Fe(NH3)x(CN)y]3+ has no electron in its eg orbital, then the value of x + y is:

  1. 5
  2. 6
  3. 3
  4. 4
Correct Answer: (2) 6

Solution:

Iron (III) is in a low-spin state due to the strong field ligand cyanide.

Read More

- Fe3+ has a d5 configuration.
- In [Fe(NH3)x(CN)y]3+, strong field ligands like CN cause pairing of all electrons in t2g orbitals.
- x = 2 (NH3) and y = 4 (CN), so x + y = 6.


Question 73:

A fuel cell, using hydrogen and oxygen as fuels:

  1. A, B, C only
  2. A, B, D only
  3. A, B, D, E only
  4. A, D, E only
Correct Answer: (4) A, D, E only

Solution:

Fuel cells are eco-friendly and used in spaceships; they are a type of galvanic cell.

Read More

- A: True, fuel cells are used in spaceships.
- D: True, fuel cells produce water as a by-product, making them eco-friendly.
- E: True, fuel cells are galvanic cells converting chemical energy to electrical energy.
- B: Efficiency can vary, but is not always 40%.
- C: Aluminum is not a catalyst; platinum is commonly used.


Question 74:

Choose the incorrect statement about Dalton's Atomic Theory:

  1. Compounds are formed when atoms of different elements combine in any ratio.
  2. All the atoms of a given element have identical properties including identical mass.
  3. Matter consists of indivisible atoms.
  4. Chemical reactions involve reorganization of atoms.
Correct Answer: (1) Compounds are formed when atoms of different elements combine in any ratio

Solution:

Dalton's theory states that compounds form in fixed, whole-number ratios.

Read More

- Dalton's Atomic Theory includes:
1. Atoms combine in fixed ratios to form compounds.
2. Atoms of a given element are identical.
3. Matter is composed of indivisible atoms.
- Option 1 is incorrect because compounds form in fixed ratios, not any ratio.


Question 75:

Match List I with List II:

List I:

  • A. α-Glucose and α-Galactose
  • B. α-Glucose and β-Glucose
  • C. α-Glucose and α-Fructose
  • D. α-Glucose and α-Ribose

List II:

  • I. Functional isomers
  • II. Homologous
  • III. Anomers
  • IV. Epimers
  1. A-III, B-IV, C-II, D-I
  2. A-III, B-IV, C-I, D-II
  3. A-IV, B-III, C-I, D-II
  4. A-IV, B-III, C-II, D-I
Correct Answer: (3) A-IV, B-III, C-I, D-II

Solution:

The correct match is based on the relationship between molecules.

Read More

- A: α-Glucose and α-Galactose differ at one carbon atom, making them epimers.
- B: α-Glucose and β-Glucose differ at the anomeric carbon, making them anomers.
- C: α-Glucose and α-Fructose have different functional groups, making them functional isomers.
- D: α-Glucose and α-Ribose differ by the number of carbons, making them homologous.


Question 76:

Given below are two statements:

Statement I: The correct order of first ionization enthalpy values of Li, Na, F, and Cl is Na < Li < Cl < F.
Statement II: The correct order of negative electron gain enthalpy values of Li, Na, F, and Cl is Na < Li < F < Cl.

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is false but Statement II is true.
  4. Statement I is true but Statement II is false.
Correct Answer: (1) Both Statement I and Statement II are true.

Solution:

Both statements are accurate based on periodic trends.

Read More

- Statement I: Ionization enthalpy increases across a period (F > Cl) and decreases down a group (Na < Li).
- Statement II: Electron gain enthalpy is more negative for Cl than F due to lower electron repulsion in Cl.
- Both are consistent with periodic properties.


Question 77:

For a strong electrolyte, a plot of molar conductivity against √(concentration) is a straight line with a negative slope. The correct unit for the slope is:

  1. S cm2 mol-3/2 L1/2
  2. S cm2 mol-1 L1/2
  3. S cm2 mol-3/2 L
  4. S cm2 mol-3/2 L-1/2
Correct Answer: (1) S cm2 mol-3/2 L1/2

Solution:

The slope unit is derived from molar conductivity against √concentration.

Read More

- Molar conductivity has units of S cm2 mol-1.
- Concentration's square root is (mol/L)1/2, leading to the slope's unit as S cm2 mol-3/2 L1/2.


Question 78:

A first-row transition metal in its +2 oxidation state has a spin-only magnetic moment value of 3.86 BM. The atomic number of the metal is:

  1. 25
  2. 26
  3. 22
  4. 23
Correct Answer: (4) 23

Solution:

The spin-only magnetic moment corresponds to 3 unpaired electrons.

Read More

- The formula for magnetic moment is μ = √[n(n + 2)], where n = unpaired electrons.
- For μ = 3.86 BM, n = 3.
- Vanadium (V) in +2 oxidation state has configuration [Ar] 3d3.


Question 79:

The number of unpaired d-electrons in [Co(H2O)6]3+ is:

  1. 4
  2. 2
  3. 0
  4. 1
Correct Answer: (3) 0

Solution:

Co3+ has all d-electrons paired in a strong ligand field.

Read More

- Co3+ has a d6 configuration.
- In [Co(H2O)6]3+, the ligand water forms an octahedral complex.
- All electrons pair due to the strong field ligand, leaving no unpaired electrons.


Question 80:

The number of species from the following that have pyramidal geometry around the central atom is:

  • S2O32−
  • SO42−
  • SO32−
  • S2O72−
  1. 4
  2. 3
  3. 1
  4. 2
Correct Answer: (3) 1

Solution:

Only SO32− has pyramidal geometry due to a lone pair on the central atom.

Read More

- S2O32−, SO42−, and S2O72− have tetrahedral or planar structures.
- SO32− has one lone pair on sulfur, leading to a pyramidal shape.


Question 81:

The maximum number of orbitals which can be identified with n = 4 and ml = 0 is:

  1. 4
  2. 3
  3. 5
  4. 2
Correct Answer: 4

Solution:

There are 4 orbitals with ml = 0 for n = 4.

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- The subshells for n = 4 are 4s, 4p, 4d, and 4f.
- Each subshell has one orbital with ml = 0.
- Hence, there are 4 orbitals with ml = 0.


Question 82:

The number of compounds/species from the following with non-zero dipole moment is:

  • BeCl2
  • BCl3
  • NF3
  • XeF4
  • CCl4
  • H2O
  • H2S
  • HBr
  • CO2
  • H2
  • HCl
  1. 3
  2. 5
  3. 7
  4. 4
Correct Answer: 5

Solution:

The compounds with non-zero dipole moments are identified.

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- Polar molecules: NF3, H2O, H2S, HBr, HCl.
- Non-polar molecules: BeCl2, BCl3, XeF4, CCl4, CO2, H2.


Question 83:

Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using a constant pressure of 5 atm. Heat exchange Q for the compression is:

  1. 200 Lit. atm
  2. 150 Lit. atm
  3. 100 Lit. atm
  4. 50 Lit. atm
Correct Answer: 200 Lit. atm

Solution:

The work done equals the heat exchange (Q = -W).

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- Work done: W = P(V2 - V1).
- P = 5 atm, V1 = 60 L, V2 = 20 L.
- W = 5(20 - 60) = -200 Lit. atm.
- Heat exchange Q = -W = 200 Lit. atm.


Question 84:

From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ___ × 10-1 g.

  1. 85
  2. 95
  3. 75
  4. 65
Correct Answer: 95

Solution:

The stoichiometric relationship is used to find the yield of acetanilide.

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- Reaction: Aniline (93 g) → Acetanilide (135 g).
- For 6.55 g aniline, yield = (135/93) × 6.55 = 9.5 g = 95 × 10-1 g.


Question 85:

A first-order reaction has a rate constant (k) of 4.6 × 10-2 s-1. The time taken for the concentration of reactant to drop from 1 M to 0.1 M is:

  1. 50 s
  2. 60 s
  3. 45 s
  4. 55 s
Correct Answer: 50 s

Solution:

The first-order reaction formula is used to calculate time.

Read More

- Formula: t = (2.303/k) log([A]0/[A]).
- k = 4.6 × 10-2, [A]0 = 1, [A] = 0.1.
- t = (2.303 / 4.6 × 10-2) log(10) ≈ 50 s.


Question 86:

Phthalimide is made to undergo the following sequence of reactions:

(a) KOH

(b) Benzyl chloride

Total number of π-bonds present in product ‘P’ is:

  1. 4
  2. 5
  3. 6
  4. 7
Correct Answer: 6

Solution:

The reaction forms N-benzylphthalimide.

Read More

- π-bonds from aromatic rings: 4.
- π-bonds from C=O groups: 2.
- Total: 6 π-bonds.


Question 87:

The total number of σ and π bonds in 2-oxohex-4-ynoic acid is:

  1. 18 σ, 2 π
  2. 16 σ, 4 π
  3. 20 σ, 3 π
  4. 15 σ, 5 π
Correct Answer: 18 σ, 2 π

Solution:

The structure is analyzed to count bonds.

Read More

- σ-bonds: 6 (C-H) + 5 (C-C, C-O, O-H) + 1 (triple bond σ) + 1 (double bond σ).
- π-bonds: 2 (triple bond π) + 1 (double bond π).
- Total: 18 σ, 2 π.


Question 88:

A first-row transition metal with the highest enthalpy of atomization forms Cr2O3 with a spin-only magnetic moment of:

  1. 5.92 BM
  2. 4.9 BM
  3. 3.87 BM
  4. 6.1 BM
Correct Answer: 5.92 BM

Solution:

Spin-only magnetic moment is calculated for Cr2O3.

Read More

- Chromium in Cr2O3 is in +3 oxidation state.
- Configuration: d3, n = 3 unpaired electrons.
- μ = √[n(n+2)] = √[3(3+2)] = 5.92 BM.


Question 89:

2.7 kg each of water and acetic acid are mixed. The freezing point of the solution will be -x °C.

  1. 8.37 °C
  2. 7.28 °C
  3. 9.14 °C
  4. 6.85 °C
Correct Answer: 8.37 °C

Solution:

The freezing point depression is calculated.

Read More

- Molality of acetic acid = (2700 g / 60 g/mol) / 2.7 kg = 15 mol/kg.
- ΔT = Kf × m = 1.86 × 15 = 27.9 K.
- Freezing point = 273 - 27.9 = -8.37 °C.


Question 90:

The vanillin compound obtained from vanilla beans has a total sum of oxygen atoms and π electrons of:

  1. 16
  2. 14
  3. 12
  4. 18
Correct Answer: 16

Solution:

The molecular structure of vanillin is analyzed.

Read More

- Oxygen atoms: 3.
- π-electrons: 13 (aromatic rings and C=O bonds).
- Total: 3 + 13 = 16.

Previous Year JEE Main Question Papers

*The article might have information for the previous academic years, please refer the official website of the exam.

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