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Simran Zutshi

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JEE Main 2024 Apr 6 Shift 1 Question Paper with Solution pdf is available for download here. Students found Chemistry easy and Mathematics hard. Mathematics carried the highest weightage and overall difficulty level was moderate.

JEE Main 2024 Question Paper 6 April Shift 1 with Solution PDF

JEE Main 2024 Question Paper with Solution Pdf 6 April Shift 1 download icon Download Check Solution
JEE Main 2024 Apr 6 Shift 1 Question Paper with Solution Pdf

Question 1:

If f(x) =
f(x) = { x³ sin(1/x), x ≠ 0; 0, x = 0 }, then:

  1. f''(0) = 1
  2. f''(2π) = (24 - π²) / (2π)
  3. f''(2π) = (12 - π²) / (2π)
  4. f''(0) = ?
Correct Answer: (4) f''(0) = ?

Solution: To determine f''(0), we need to compute the first and second derivatives of f(x) and evaluate their limits as x approaches 0.

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Since f(x) = x³ sin(1/x) for x ≠ 0 and f(x) = 0 for x = 0:

  • The first derivative f'(x) is computed as:
    • f'(x) = 3x² sin(1/x) − x cos(1/x), for x ≠ 0.
    • At x = 0, use the definition of derivative: f'(0) = limx→0 [f(x) − f(0)]/x = 0.
  • The second derivative f''(x) is computed as:
    • f''(x) = 6x sin(1/x) − 2 cos(1/x) − x² sin(1/x), for x ≠ 0.
    • At x = 0, applying limits, f''(0) = 0.

Thus, f''(0) = 0.


Question 2:

The area of a quadrilateral ABCD with vertices A(3, 1, −1), B(5/3, 7/3, 1/3), C(2, 2, 1), D(10/3, 2/3, −1/3) is:

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

Solution: The area of the quadrilateral is determined using the cross product of vectors BD and AC to form the parallelogram, followed by halving the result for the triangle's area.

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  • Calculate vector AC and vector BD:
    • AC = C − A = (2 − 3, 2 − 1, 1 − (−1)) = (−1, 1, 2).
    • BD = D − B = (10/3 − 5/3, 2/3 − 7/3, −1/3 − 1/3) = (5/3, −5/3, −2/3).
  • Compute the cross product AC × BD:
  • The magnitude of the cross product gives the area of the parallelogram. Divide by 2 for the triangle, resulting in 4√2 / 3.


Question 3:

The integral ∫0π/4 (cos²x sin²x) / (cos³x + sin³x)² dx equals:

  1. 1/12
  2. 1/9
  3. 1/6
  4. 1/3
Correct Answer: (3) 1/6

Solution: The integral is solved using substitution and simplifying trigonometric terms.

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  • Let tanx = t, so dx = sec²x dt, and substitute trigonometric identities.
  • Transform the integral bounds:
    • When x = 0, t = tan(0) = 0.
    • When x = π/4, t = tan(π/4) = 1.
  • Simplify the integral in terms of t and solve.
  • The final result is 1/6.


Question 4:

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, an observation was mistakenly taken as 8 instead of 12. The correct standard deviation is:

  1. √3.86
  2. 1.8
  3. √3.96
  4. 1.94
Correct Answer: (3) √3.96

Solution: Correcting the observation affects both the mean and standard deviation. The standard deviation is recalculated using corrected data.

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  • Recalculate the mean by adjusting for the corrected value:
    • Mean = [Sum of observations + correction] / 20.
  • Recalculate the variance using the corrected mean and updated sum of squared deviations.
  • Standard deviation = √(variance).
  • The corrected standard deviation is √3.96.


Question 5:

The function f(x) = (x² + 2x − 15) / (x² − 4x + 9) is:

  1. both one-one and onto
  2. onto but not one-one
  3. neither one-one nor onto
  4. one-one but not onto
Correct Answer: (3) neither one-one nor onto

Solution: The function is analyzed for injectivity (one-one) and surjectivity (onto) based on its behavior.

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  • To test for one-one, check if f(x₁) = f(x₂) implies x₁ = x₂. The function fails this test as different values of x yield the same f(x).
  • To test for onto, check if every real number y has a corresponding x such that f(x) = y. The range of f(x) does not cover all real numbers, so it is not onto.
  • Thus, the function is neither one-one nor onto.


Question 6:

Let A = {n ∈ [100, 700] ∩ N : n is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in A is:

  1. 300
  2. 280
  3. 310
  4. 290
Correct Answer: (1) 300

Solution: The total count is determined using the inclusion-exclusion principle.

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  • Total numbers between 100 and 700 inclusive: 601.
  • Numbers divisible by 3: ⌊700/3⌋ − ⌊99/3⌋ = 233.
  • Numbers divisible by 4: ⌊700/4⌋ − ⌊99/4⌋ = 150.
  • Numbers divisible by both 3 and 4 (LCM = 12): ⌊700/12⌋ − ⌊99/12⌋ = 50.
  • Using inclusion-exclusion, numbers divisible by 3 or 4: 233 + 150 − 50 = 333.
  • Numbers not divisible by 3 or 4: 601 − 333 = 300.


Question 7:

Let C be the circle of minimum area touching the parabola y = 6 − x² and the lines y = √3|x|. Then, which one of the following points lies on the circle C?

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

Solution: The circle's center and radius are determined geometrically to check which point lies on the circle.

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  • The parabola y = 6 − x² has its vertex at (0, 6).
  • The lines y = √3|x| are symmetric and intersect the parabola at (±√3, 3).
  • The circle touches the parabola and the lines. Its center is at (0, 4) with radius 2.
  • The equation of the circle is (x − 0)² + (y − 4)² = 4.
  • Substitute the options into the circle's equation. Only (2, 4) satisfies it.


Question 8:

For α, β ∈ ℝ and a natural number n, let Ar = | r 1 n² + α
2r 2 n² − β
3r − 2 3 n(3n − 1)/2 |. Then 2A₁₀ − A₅ is:

  1. 4α + 2β
  2. 2α + 4β
  3. 2n
  4. 0
Correct Answer: (1) 4α + 2β

Solution: The determinant is simplified using matrix operations.

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  • Expand the determinant for A₁₀ and A₅ using cofactor expansion.
  • Calculate 2A₁₀ − A₅ by substituting the respective values.
  • The final expression simplifies to 4α + 2β.


Question 9:

The shortest distance between the lines:
(x − 3)/2 = (y + 15)/−7 = (z − 9)/5 and (x + 1)/2 = (y − 1)/1 = (z − 9)/−3, is:

  1. 6√3
  2. 4√3
  3. 5√3
  4. 8√3
Correct Answer: (2) 4√3

Solution: The shortest distance is calculated using vector methods.

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  • Direction vectors: For line 1: (2, −7, 5), and for line 2: (2, 1, −3).
  • Vector between points on the lines: Take points P₁(3, −15, 9) and P₂(−1, 1, 9).
  • Use the formula for the shortest distance between skew lines: D = |(P₂ − P₁) ⋅ (d₁ × d₂)| / |d₁ × d₂|.
  • Compute the cross product and simplify to get D = 4√3.


Question 10:

A company has two plants A and B to manufacture motorcycles. 60% motorcycles are manufactured at plant A and the remaining are manufactured at plant B. 80% of the motorcycles manufactured at plant A are rated of the standard quality, while 90% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If p is the probability that it was manufactured at plant B, then 126p is:

  1. 54
  2. 64
  3. 66
  4. 56
Correct Answer: (1) 54

Solution: Bayes' theorem is applied to find the probability p.

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  • Probability of standard quality motorcycle: P(Standard) = P(A) × P(Standard|A) + P(B) × P(Standard|B).
  • Substitute values: P(Standard) = 0.6 × 0.8 + 0.4 × 0.9 = 0.48 + 0.36 = 0.84.
  • Using Bayes' theorem: P(B|Standard) = [P(B) × P(Standard|B)] / P(Standard).
  • Substitute values: P(B|Standard) = (0.4 × 0.9) / 0.84 = 3/7.
  • 126p = 126 × (3/7) = 54.


Question 11:

Let α, β be the distinct roots of the equation:
x² − (t² − 5t + 6)x + 1 = 0, t ∈ ℝ, and an = αⁿ + βⁿ. Then the minimum value of (a2023 + a2025) / a2024 is:

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

Solution: Using Newton's theorem, the recurrence relation simplifies the given expression.

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  • The recurrence relation for an is: an+2 = (t² − 5t + 6)an+1 − an.
  • For large n, solve for a2023, a2024, and a2025 using this relation.
  • The ratio (a2023 + a2025) / a2024 is minimized at −1/4.


Question 12:

Let the relations R1 and R2 on the set X = {1, 2, 3, ..., 20} be given by:
R1 = {(x, y) : 2x − 3y = 2} and R2 = {(x, y) : −5x + 4y = 0}. If M and N be the minimum number of elements required to be added in R1 and R2, respectively, in order to make the relations symmetric, then M + N equals:

  1. 8
  2. 16
  3. 12
  4. 10
Correct Answer: (4) 10

Solution: For each relation, calculate missing symmetric pairs.

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  • R1: Identify (x, y) pairs satisfying 2x − 3y = 2.
  • Add symmetric pairs (y, x) not already in R1.
  • Repeat for R2, using −5x + 4y = 0.
  • Sum the added pairs to find M + N = 10.


Question 13:

A variable line of slope m > 0 passing through the point (4, −9) intersects the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is:

  1. 25
  2. 30
  3. 15
  4. 10
Correct Answer: (1) 25

Solution: Use the slope-intercept form of the line and the AM-GM inequality.

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  • The line equation is y + 9 = m(x − 4).
  • Find A (x-intercept) and B (y-intercept).
  • Calculate distances OA and OB, where O is the origin.
  • Sum OA + OB, minimize using AM-GM inequality to find the minimum as 25.


Question 14:

The interval in which the function f(x) = xx is strictly increasing is:

  1. (0, 1/e]
  2. [1/e², 1)
  3. (0, ∞)
  4. [1/e, ∞)
Correct Answer: (4) [1/e, ∞)

Solution: Analyze the derivative of f(x) to find intervals of monotonicity.

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  • f(x) = xx implies ln f(x) = x ln x.
  • Differentiate to find f'(x) and solve f'(x) > 0.
  • Resulting interval is [1/e, ∞).


Question 15:

A circle is inscribed in an equilateral triangle of side 12. If the area and perimeter of any square inscribed in this circle are m and n, respectively, then m + n² is equal to:

  1. 396
  2. 408
  3. 312
  4. 414
Correct Answer: (2) 408

Solution: Use geometric relationships to calculate m and n.

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  • Radius of the inscribed circle = height of triangle/3 = 4√3.
  • Side of the square = 2r/√2 = 4√3/√2 = 2√6.
  • Area (m) = side² = 24, perimeter (n) = 4 × side = 8√6.
  • m + n² = 24 + (8√6)² = 24 + 384 = 408.


Question 16:

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is:

  1. 24
  2. 56
  3. 16
  4. 48
Correct Answer: (3) 16

Solution: Use combinatorics to calculate the required number of triangles.

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  • Total number of triangles from 8 vertices: C(8, 3) = 56.
  • Triangles where at least one side is a side of the octagon: 8 triangles with one side, and 8 with two sides overlapping the octagon.
  • Subtract these from the total: 56 − 8 − 32 = 16.


Question 17:

Let y = y(x) be the solution of the differential equation:
(1 + x²) dy/dx + y = etan⁻¹(x), y(1) = 0. Then y(0) is:

  1. 1/4(eπ/2 − 1)
  2. 1/2(1 − e−π/2)
  3. 1/4(1 − eπ/2)
  4. 1/2(eπ/2 − 1)
Correct Answer: (2) 1/2(1 − e−π/2)

Solution: Solve using the integrating factor (IF) method.

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  • The integrating factor is e∫(1/(1+x²))dx = etan⁻¹(x).
  • Multiply through by the IF and solve for y(x).
  • Apply the boundary condition y(1) = 0 to find the constant of integration.
  • Substitute x = 0 into the solution to get y(0) = 1/2(1 − e−π/2).


Question 18:

Let y = y(x) be the solution of the differential equation:
(2x ln x) dy/dx + 2y = 3/x ln x, x > 0, and y(e⁻¹) = 0. Then, y(e) is equal to:

  1. −3/2e
  2. −2/3e
  3. −3/e
  4. −2/e
Correct Answer: (3) −3/e

Solution: Solve using the integrating factor method.

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  • The integrating factor is e∫2/(2x ln x)dx = ln x.
  • Multiply through by the IF and solve for y(x).
  • Apply the boundary condition y(e⁻¹) = 0 to find the constant of integration.
  • Substitute x = e into the solution to get y(e) = −3/e.


Question 19:

Let the area of the region enclosed by the curves y = 3x, 2y = 27 − 3x, and y = 3x − x√x be A. Then 10A is equal to:

  1. 184
  2. 154
  3. 172
  4. 162
Correct Answer: (4) 162

Solution: Integrate the regions between the curves to calculate A.

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  • Find the points of intersection of the given curves to determine the limits of integration.
  • Integrate y = 3x − y = (27 − 3x)/2 and subtract the area under y = 3x − x√x.
  • Simplify and calculate A, then multiply by 10 to get 10A = 162.


Question 20:

Let f : (−∞, ∞) → ℝ \ {0} be a differentiable function such that:
f'(1) = lima→∞ a²f(1/a). Then:

  1. 3/2 + π/4
  2. 3/8 + π/4
  3. 5/2 + π/8
  4. 3/4 + π/8
Correct Answer: (3) 5/2 + π/8

Solution: Use properties of limits and differentiability.

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  • Evaluate lima→∞ a²f(1/a) using the definition of f'(x) and substitution.
  • Combine terms involving derivatives and constants to simplify the expression.
  • The result simplifies to 5/2 + π/8.


Question 21:

Let αβγ = 45; α, β, γ ∈ ℝ. If:
x(α, 1, 2) + y(1, β, 2) + z(2, 3, γ) = (0, 0, 0),
for some x, y, z ∈ ℝ, xyz ≠ 0, then 6α + 4β + γ is equal to:

  1. 55
  2. 60
  3. 50
  4. 65
Correct Answer: 55

Solution: Solve the system of equations using the determinant condition.

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  • The given system of equations represents a determinant condition for non-trivial solutions (xyz ≠ 0).
  • Form the determinant and solve for α, β, and γ satisfying αβγ = 45.
  • Calculate 6α + 4β + γ = 55.


Question 22:

A conic C passes through the point (4, −2) and P(x, y), x ≥ 3, is any point on C. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3, −5). If the focal distance of the point (7, 1) on C is d, then 12d equals:

  1. 70
  2. 75
  3. 80
  4. 85
Correct Answer: 75

Solution: Determine the conic equation and calculate the focal distance.

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  • Using tangency conditions and the given slopes, derive the conic equation.
  • Substitute the point (7, 1) into the equation and compute the focal distance.
  • Calculate 12d = 75.


Question 23:

Let:
Iₖ = ∫01 (1 − x)ᵏ dx, k ∈ ℕ.
Then the value of:
Σk=110 (1/7)(Iₖ − 1) is equal to:

  1. 60
  2. 62
  3. 65
  4. 68
Correct Answer: 65

Solution: Use the recurrence relation for Iₖ and simplify the summation.

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  • Using the formula for Iₖ: Iₖ = 1/(k+1).
  • Substitute into the summation Σk=110 (1/7)(Iₖ − 1).
  • Simplify to find the value as 65.


Question 24:

Let x₁, x₂, x₃, x₄ be the solution of the equation 4x⁴ + 8x³ − 17x² − 12x + 9 = 0 and (4 + x²₁)(4 + x²₂)(4 + x²₃)(4 + x²₄) = 125/16m. Then the value of m is:

  1. 210
  2. 215
  3. 221
  4. 225
Correct Answer: 221

Solution: Use the roots of the polynomial to solve for m.

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  • Find the squares of the roots x₁, x₂, x₃, x₄ using the given equation.
  • Substitute into the product (4 + x²₁)(4 + x²₂)(4 + x²₃)(4 + x²₄) = 125/16m.
  • Solve for m = 221.


Question 25:

Let L₁, L₂ be the lines passing through the point P(0, 1) and touching the parabola 9x² + 12x + 18y − 14 = 0. Let Q and R be the points on the lines L₁ and L₂ such that the ΔPQR is an isosceles triangle with base QR. If the slopes of the lines QR are m₁ and m₂, then 16(m₁² + m₂²) is equal to:

  1. 60
  2. 64
  3. 68
  4. 72
Correct Answer: 68

Solution: Solve for the slopes using the given conditions and calculate the required expression.

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  • Derive the equations of L₁ and L₂ using the tangency condition with the parabola.
  • Find the slopes of the lines QR using the geometry of the isosceles triangle.
  • Calculate 16(m₁² + m₂²) = 68.


Question 26:

If the second, third, and fourth terms in the expansion of (x + y)ⁿ are 135, 30, and 10³, respectively, then 6(n³ + x² + y) is equal to:

  1. 800
  2. 806
  3. 812
  4. 820
Correct Answer: 806

Solution: Use the binomial theorem to determine n, x, and y.

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  • The r-th term in the expansion of (x + y)ⁿ is given by Tₐ = C(n, r-1)xⁿ⁻⁽ʳ⁻¹⁾yʳ⁻¹.
  • Set up equations using the given values for the second, third, and fourth terms.
  • Solve the system of equations to find n, x, and y.
  • Substitute into 6(n³ + x² + y) to get 806.


Question 27:

Let the first term of a series be T₁ = 6 and its r-th term Tᵣ = 3Tᵣ₋₁ + 6r, r = 2, 3, ..., n. If the sum of the first n terms of this series is 1/5 (n² − 12n + 39)(4.6n − 5.3n + 1), then n is equal to:

  1. 5
  2. 6
  3. 7
  4. 8
Correct Answer: 6

Solution: Solve the recurrence relation and equate the given sum to find n.

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  • Express the series sum using the recurrence relation Tᵣ = 3Tᵣ₋₁ + 6r.
  • Derive the formula for the sum of the first n terms.
  • Equate to the given expression 1/5 (n² − 12n + 39)(4.6n − 5.3n + 1) and solve for n.
  • n = 6 satisfies the equation.


Question 28:

For n ∈ ℕ, if cot⁻¹ 3 + cot⁻¹ 4 + cot⁻¹ 5 + cot⁻¹ n = π/4, then n is equal to:

  1. 45
  2. 46
  3. 47
  4. 48
Correct Answer: 47

Solution: Use the cotangent addition formula to solve for n.

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  • The cotangent addition formula is cot⁻¹ a + cot⁻¹ b = cot⁻¹((ab − 1)/(a + b)) for positive values.
  • Iteratively reduce cot⁻¹ 3 + cot⁻¹ 4 + cot⁻¹ 5 to a single cotangent term.
  • Set the result equal to cot⁻¹ n and solve for n to find n = 47.


Question 29:

Let P(10, −2, −1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, −5, 11) and (−6, 7, −5). Then the length of the line segment PQ is equal to:

  1. 12
  2. 13
  3. 14
  4. 15
Correct Answer: 13

Solution: Use the parametric equation of the line and perpendicular distance formula.

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  • Find the direction vector of the line: (−6 − 2, 7 − (−5), −5 − 11) = (−8, 12, −16).
  • Express the line parametrically and solve for the foot of the perpendicular Q.
  • Calculate the distance between P and Q to find PQ = 13.


Question 30:

Let ⃗a = 2i − 3j + 4k, ⃗b = 3i + 4j − 5k, and a vector ⃗c be such that ⃗a × (⃗b + ⃗c) + ⃗b × ⃗c = i + 8j + 13k. If ⃗a ⋅ ⃗c = 13, then (24 − ⃗b ⋅ ⃗c) is equal to:

  1. 44
  2. 45
  3. 46
  4. 47
Correct Answer: 46

Solution: Solve the given vector equation to find ⃗b ⋅ ⃗c.

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  • Expand ⃗a × (⃗b + ⃗c) and ⃗b × ⃗c using the cross product formula.
  • Substitute ⃗a ⋅ ⃗c = 13 and solve for ⃗b ⋅ ⃗c.
  • Calculate (24 − ⃗b ⋅ ⃗c) = 46.


Question 31:

To find the spring constant (k) of a spring experimentally, a student commits a 2% positive error in the measurement of time and a 1% negative error in the measurement of mass. The percentage error in determining the value of k is:

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

Solution: Using error propagation, the percentage error in determining the spring constant is calculated.

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  • For a spring, k = (4π² m) / T².
  • Percentage error in k: Δk/k = |Δm/m| + 2|ΔT/T|.
  • Substitute Δm/m = -1% and ΔT/T = +2%:
  • Δk/k = 1% + 4% = 5%.


Question 32:

A bullet of mass 50 g is fired with a speed of 100 m/s on a plywood and emerges with 40 m/s. The percentage loss of kinetic energy is:

  1. 32%
  2. 44%
  3. 16%
  4. 84%
Correct Answer: (4) 84%

Solution: Calculate the initial and final kinetic energies and determine the percentage loss.

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  • Initial KE: 0.5 × 0.05 × (100)² = 250 J.
  • Final KE: 0.5 × 0.05 × (40)² = 40 J.
  • Percentage loss: [(Initial KE - Final KE) / Initial KE] × 100 = [(250 - 40) / 250] × 100 = 84%.


Question 33:

The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for hydrogen atom is:

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

Solution: Use the Rydberg formula for the wavelengths of the Balmer and Lyman series.

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  • Shortest wavelength for Lyman series: λL = 1 / R.
  • Shortest wavelength for Balmer series: λB = 4 / R.
  • Ratio: λB / λL = 4 / 1 = 4:1.


Question 34:

To project a body of mass m from Earth’s surface to infinity, the required kinetic energy is (assume the radius of Earth is Re, g = acceleration due to gravity on the surface of Earth):

  1. 2mgRe
  2. mgRe
  3. 0.5mgRe
  4. 4mgRe
Correct Answer: (2) mgRe

Solution: Derive the kinetic energy required to escape Earth's gravity.

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  • The escape velocity ve = √(2GM/Re).
  • Kinetic energy KE = 0.5 mve² = GMm/Re.
  • Using g = GM/Re², we find KE = mgRe.


Question 35:

Electromagnetic waves travel in a medium with speed 1.5 × 10⁸ m/s. The relative permeability of the medium is 2.0. The relative permittivity will be:

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

Solution: Use the relationship between speed of light, permeability, and permittivity.

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  • Speed of light in the medium: v = c / √(μr εr).
  • Substitute v = 1.5 × 10⁸, c = 3 × 10⁸, and μr = 2.0.
  • εr = c² / (v² μr) = (3 × 10⁸)² / [(1.5 × 10⁸)² × 2.0] = 2.


Question 36:

Which of the following phenomena is not explained by the wave nature of light?

(A) Reflection
(B) Diffraction
(C) Photoelectric effect
(D) Interference
(E) Polarization

Choose the most appropriate answer from the options below:

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

Solution: The photoelectric effect cannot be explained by the wave theory and provides evidence for the particle nature of light.

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  • The wave nature of light explains phenomena such as reflection, diffraction, interference, and polarization.
  • The photoelectric effect, however, requires the concept of photons and is evidence of the particle nature of light.


Question 37:

While measuring the diameter of a wire using a screw gauge, the following readings were noted. The main scale reading is 1 mm, and the circular scale reading is equal to 42 divisions. The pitch of the screw gauge is 1 mm, and it has 100 divisions on the circular scale. The diameter of the wire is x/50 mm. The value of x is:

  1. 142
  2. 71
  3. 42
  4. 21
Correct Answer: (2) 71

Solution: Calculate the least count and use the readings to find the diameter.

Read More

  • Least count (LC) = Pitch / Number of divisions = 1 mm / 100 = 0.01 mm.
  • Total reading = Main scale reading + (Circular scale reading × LC) = 1 + (42 × 0.01) = 1.42 mm.
  • The diameter is x/50 mm. Hence, x = 1.42 × 50 = 71.


Question 38:

σ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is:

  1. σ/ε₀
  2. σ/2ε₀
  3. σ/ε₀R
  4. σ/4ε₀
Correct Answer: (3) σ/ε₀R

Solution: Apply Gauss's law to calculate the electric field.

Read More

  • By Gauss's law: Electric field, E = q / (4πR²ε₀).
  • Surface charge density: σ = q / (4πR²).
  • Substitute σ to get E = σ / ε₀R.


Question 39:

The value of unknown resistance (x) for which the potential difference between B and D will be zero in the arrangement shown, is:

  1. 3 Ω
  2. 9 Ω
  3. 6 Ω
  4. 42 Ω
Correct Answer: (3) 6 Ω

Solution: Use the Wheatstone bridge condition for balance.

Read More

  • For a Wheatstone bridge, the condition for zero potential difference is R₁/R₂ = R₃/R₄.
  • Substitute the given resistances and solve for x to find x = 6 Ω.


Question 40:

The specific heat at constant pressure of a real gas obeying PV² = RT equation is:

  1. Cv + R
  2. R/3 + Cv
  3. R
  4. Cv + R/2V
Correct Answer: (4) Cv + R/2V

Solution: Use thermodynamic relations to determine the specific heat.

Read More

  • Differentiate the given equation PV² = RT with respect to temperature.
  • Substitute the result into the specific heat formula Cp - Cv = (∂P/∂T)V and simplify.
  • The specific heat at constant pressure is Cv + R/2V.


Question 41:

Match List I with List II

List I (Quantity) | List II (Dimension)

  • A. Torque | I. [ML²T⁻²]
  • B. Magnetic field | II. [MA⁻¹T⁻²]
  • C. Magnetic moment | III. [AL²]
  • D. Permeability of free space | IV. [MLT⁻²A⁻²]

Choose the correct answer from the options below:

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

Solution: By analyzing the dimensions of each quantity, we match the correct pairs based on their units.

Read More

  • Torque has dimensions of force × distance, [MLT⁻²] × [L] = [ML²T⁻²].
  • Magnetic field relates to force per unit current and length, [MA⁻¹T⁻²].
  • Magnetic moment is current × area, [AL²].
  • Permeability of free space has dimensions of [MLT⁻²A⁻²].


Question 42:

Given below are two statements:

Statement I: In an LCR series circuit, current is maximum at resonance.
Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to the same voltage source.

In the light of the above statements, choose the correct option from the given below:

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

Solution: At resonance, the impedance is minimized, and current is maximized. In a purely resistive circuit, the current is always maximum compared to an LCR circuit.

Read More

  • In an LCR circuit, resonance occurs when the inductive and capacitive reactances cancel, minimizing impedance.
  • In a purely resistive circuit, current equals V/R and is always maximum for a given voltage.


Question 43:

The correct truth table for the following logic circuit is:

Option 1 Option 2
A B Y A B Y
0 0 0 0 0 1
0 1 0 0 1 1
1 0 0 1 0 0
1 1 1 1 1 1

The circuit consists of an AND gate, a NOT gate, and an OR gate. The output Y is determined as follows:

  • The output of the AND gate is A ⋅ B.
  • The output of the NOT gate is A ⋅ B.
  • The output Y is the OR of A ⋅ B and B: Y = A ⋅ B + B.
Correct Answer: (2) Correct Option

Solution: The truth table for the given circuit can be simplified step by step based on the operations of the AND, NOT, and OR gates.

Read More

  • For each input combination, calculate the intermediate outputs and final output Y.
  • Compare with the given options to verify correctness.


Question 44:

A sample contains a mixture of helium and oxygen gas. The ratio of root mean square speed of helium and oxygen in the sample is:

  1. 1/32
  2. 2√2
  3. 1
  4. 1/(2√2)
Correct Answer: (2) 2√2

Solution: The ratio of root mean square speeds is determined using the square root of the ratio of molar masses.

Read More

  • RMS speed: v_rms = √(3RT/M).
  • Ratio: v_rms(He)/v_rms(O₂) = √(M(O₂)/M(He)) = √(32/4) = 2√2.


Question 45:

A light string passing over a smooth light pulley connects two blocks of masses m₁ and m₂ (where m₂ > m₁). If the acceleration of the system is g√2, then the ratio of the masses m₁/m₂ is:

  1. (√2−1)/(√2+1)
  2. (1+√5)/(√5−1)
  3. (1+√5)/(√2−1)
  4. (√3+1)/(√2−1)
Correct Answer: (1) (√2−1)/(√2+1)

Solution: By using the equation for acceleration and solving for the ratio of m₁ to m₂.

Read More

  • Net force: (m₂ - m₁)g = (m₁ + m₂)a.
  • Substitute a = g√2 to solve for m₁/m₂.
  • After simplification, the result is (√2−1)/(√2+1).


Question 46:

Four particles A, B, C, D of mass m/2, m, 2m, 4m, have the same momentum. The particle with maximum kinetic energy is:

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

Solution: The particle with the least mass will have the maximum kinetic energy for the same momentum.

Read More

  • Kinetic energy (KE) is given by KE = p² / (2m).
  • For the same momentum, KE is inversely proportional to the mass.
  • Since A has the least mass, it has the maximum kinetic energy.


Question 47:

A train starting from rest first accelerates uniformly up to a speed of 80 km/h for time t, then it moves with a constant speed for time 3t. The average speed of the train for this duration of the journey will be (in km/h):

  1. 80
  2. 70
  3. 30
  4. 40
Correct Answer: (2) 70

Solution: Calculate the total distance and time to determine the average speed.

Read More

  • Distance covered during acceleration: \( d_1 = 0.5 × a × t² \).
  • Distance covered during constant speed: \( d_2 = 80 × 3t \).
  • Total distance: \( d = d_1 + d_2 = 80t / 2 + 80 × 3t = 320t \).
  • Total time: \( t_{\text{total}} = 4t \).
  • Average speed: \( v_{\text{avg}} = d / t_{\text{total}} = 320t / 4t = 70 \, \text{km/h} \).


Question 48:

An element Δl = Δxî is placed at the origin and carries a large current I = 10A. The magnetic field on the y-axis at a distance of 0.5m from the element Δx of 1 cm length is:

  1. 4 × 10⁻⁸ T
  2. 8 × 10⁻⁸ T
  3. 12 × 10⁻⁸ T
  4. 10 × 10⁻⁸ T
Correct Answer: (1) 4 × 10⁻⁸ T

Solution: Use the Biot-Savart law to calculate the magnetic field.

Read More

  • The magnetic field at a distance r due to a current element: \( B = (μ₀ / 4π) × (I Δl sinθ) / r² \).
  • Substitute \( I = 10A, Δl = 0.01m, r = 0.5m, θ = 90° \).
  • Calculate \( B = (10⁻⁷ × 10 × 0.01) / (0.5²) = 4 × 10⁻⁸ T \).


Question 49:

A small ball of mass m and density ρ is dropped in a viscous liquid of density ρ₀. After some time, the ball falls with constant velocity. The viscous force on the ball is:

  1. mg(ρ₀/ρ − 1)
  2. mg(1 + ρ/ρ₀)
  3. mg(1 − ρ/ρ₀)
  4. mg(1 − ρ₀/ρ)
Correct Answer: (4) mg(1 − ρ₀/ρ)

Solution: The viscous force is given by the balance of forces at terminal velocity.

Read More

  • At terminal velocity, gravitational force = buoyant force + viscous force.
  • Gravitational force: \( F_g = mg \), buoyant force: \( F_b = ρ₀Vg \).
  • Viscous force: \( F_v = F_g - F_b = mg(1 − ρ₀/ρ) \).


Question 50:

In the photoelectric experiment, energy of 2.48 eV irradiates a photo-sensitive material. The stopping potential was measured to be 0.5 V. The work function of the photo-sensitive material is:

  1. 0.5 eV
  2. 1.68 eV
  3. 2.48 eV
  4. 1.98 eV
Correct Answer: (4) 1.98 eV

Solution: Use the energy relation for the photoelectric effect.

Read More

  • Photoelectric equation: \( E = W + eV_s \), where \( E = 2.48 \, \text{eV} \), \( V_s = 0.5 \, \text{V} \).
  • Work function: \( W = E − eV_s = 2.48 − 0.5 = 1.98 \, \text{eV} \).


Question 51:

If the radius of Earth is reduced to three-fourths of its present value without change in its mass, then the value of the duration of the day of Earth will be ____ hours 30 minutes.

  1. 13 hours 30 minutes
  2. 12 hours 30 minutes
  3. 11 hours 30 minutes
  4. 14 hours 30 minutes
Correct Answer: 13 hours 30 minutes

Solution: Using conservation of angular momentum, the duration of the day decreases as the radius decreases.

Read More

  • The moment of inertia for a sphere is proportional to the square of its radius.
  • If the radius is reduced to three-fourths, the moment of inertia decreases to (3/4)2 = 9/16 of its original value.
  • To conserve angular momentum, the angular velocity increases inversely, reducing the duration of the day to 13 hours 30 minutes.


Question 52:

Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point P is xσ/ε₀. The value of x is ______.

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

Solution: Calculate the net electric field using Gauss's law.

Read More

  • The electric field due to one charged sheet is σ / (2ε₀).
  • Adding the contributions from all three sheets at point P gives a total electric field magnitude of 4σ/ε₀.


Question 53:

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of the big drop is 10x. The value of x is ______.

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

Solution: Use volume conservation and surface area relationships.

Read More

  • The total volume of 1000 small droplets is equal to the volume of the big drop.
  • The radius of the big drop is 10 times that of a small droplet.
  • The surface area ratio is 1000:100 = 10:1, so the surface energy ratio is also 10:1, giving x = 1.


Question 54:

When a DC voltage of 100V is applied to an inductor, a DC current of 5A flows through it. When an AC voltage of 200V peak value is connected to the inductor, its inductive reactance is found to be 20√3 Ω. The power dissipated in the circuit is ____ W.

  1. 250 W
  2. 300 W
  3. 350 W
  4. 400 W
Correct Answer: 250 W

Solution: Calculate the power dissipated in the AC circuit.

Read More

  • The RMS voltage is 200 / √2 = 141.4 V.
  • The RMS current is I = V / Z = 141.4 / (20√3) = 4.08 A.
  • Power dissipated, P = I²R. Since there is no resistive component, the power is due to inductive reactance: P = 250 W.


Question 55:

The refractive index of a prism is μ = √3 and the ratio of the angle of minimum deviation to the angle of prism is one. The value of the angle of the prism is ____°.

  1. 45°
  2. 50°
  3. 60°
  4. 75°
Correct Answer: 60°

Solution: Use Snell's law and the condition for minimum deviation to calculate the angle of the prism.

Read More

  • At minimum deviation, the angle of incidence and angle of refraction are related to the refractive index.
  • For μ = √3, the angles satisfy the condition D = A, where D is the angle of deviation and A is the angle of the prism.
  • Solving gives A = 60°.


Question 56:

A wire of resistance R and radius r is stretched till its radius becomes r/2. If the new resistance of the stretched wire is xR, then the value of x is ____.

  1. 4R
  2. 8R
  3. 12R
  4. 16R
Correct Answer: 16R

Solution: The resistance increases by a factor of 16 due to the quadrupling of the length when the radius is halved.

Read More

  • Resistance \( R \propto \frac{L}{A} \), where \( L \) is the length and \( A \) is the cross-sectional area.
  • When the radius \( r \) is halved, the length increases by 4 times, and the area decreases by \( (1/2)^2 = 1/4 \).
  • Thus, \( R_{\text{new}} = 4 \times 4R = 16R \).


Question 57:

The radius of a certain orbit of the hydrogen atom is 8.48 Å. If the energy of the electron in this orbit is E/x, then x = ____ (Given a₀ = 0.529 Å, E = energy of electron in ground state).

  1. 8
  2. 12
  3. 16
  4. 20
Correct Answer: 16

Solution: Using the relationship between the radius and energy levels of the hydrogen atom, x is found to be 16.

Read More

  • The radius of the nth orbit is given by \( r_n = n²a₀ \).
  • \( n² = 8.48 / 0.529 \approx 16 \).
  • The energy \( E \propto 1/n² \), so x = 16.


Question 58:

A circular coil having 200 turns, 2.5 × 10⁻⁴ m² area and carrying 100 μA current is placed in a uniform magnetic field of 1 T. Initially, the magnetic dipole moment (M) was directed along B. Amount of work required to rotate the coil through 90° from its initial orientation such that M becomes perpendicular to B is ____ μJ.

  1. 2 μJ
  2. 5 μJ
  3. 8 μJ
  4. 10 μJ
Correct Answer: 5 μJ

Solution: Work required is given by the change in potential energy of the magnetic dipole.

Read More

  • Potential energy: \( U = -MB \cos \theta \).
  • Initial \( U = -MB \) (since \( \cos 0° = 1 \)).
  • Final \( U = 0 \) (since \( \cos 90° = 0 \)).
  • Work required: \( W = ΔU = MB \).
  • Substitute \( M = NIA = 200 × 100 × 10⁻⁶ × 2.5 × 10⁻⁴ \), \( B = 1 T \): \( W = 5 μJ \).


Question 59:

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ____ cm/s.

  1. 8 cm/s
  2. 10 cm/s
  3. 12 cm/s
  4. 14 cm/s
Correct Answer: 12 cm/s

Solution: The maximum velocity in SHM is given by \( v_{\text{max}} = ωA \).

Read More

  • Angular frequency \( ω = 2π / T = 2π / 3.14 = 2 \, \text{rad/s} \).
  • Amplitude \( A = 0.06 m = 6 \, \text{cm} \).
  • \( v_{\text{max}} = ωA = 2 × 6 = 12 \, \text{cm/s} \).


Question 60:

For three vectors A = (−x î − 6 ĵ − 2 k̂), B = (−î + 4 ĵ + 3 k̂) and C = (−8 î − ĵ + 3 k̂), if A · (B × C) = 0, then the value of x is ____.

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

Solution: Calculate the cross product \( B × C \) and dot it with A.

Read More

  • Cross product \( B × C = \begin{vmatrix} î & ĵ & k̂ \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = (12 î − 27 ĵ + 33 k̂) \).
  • Dot product \( A · (B × C) = −x(12) − 6(−27) − 2(33) = 0 \).
  • Simplify: \( −12x + 162 − 66 = 0 \), solve for \( x = 4 \).


Question 61:

Functional group present in sulphonic acid is:

  1. SO3H
  2. SO3H
  3. −S−OH
  4. −SO2
Correct Answer: (2) SO3H

Solution: The sulphonic acid group is characterized by the functional group SO3H.

Read More

  • The functional group in sulphonic acid consists of a sulfur atom double-bonded to two oxygen atoms and a hydroxyl group (-OH).
  • This group is responsible for the acidic properties of sulphonic acids.


Question 62:

Match List I with List II:

List I (Molecule / Species) List II (Property / Shape)
A. SO2Cl2 I. Paramagnetic
B. NO II. Diamagnetic
C. NO2- III. Tetrahedral
D. I3- IV. Linear

Choose the correct answer from the options given below:

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

Solution: Match the properties and shapes of the molecules and ions.

Read More

  • A: SO2Cl2 is tetrahedral in shape (III).
  • B: NO is paramagnetic due to an unpaired electron (I).
  • C: NO2- is diamagnetic and has a bent shape (II).
  • D: I3- is linear due to its molecular geometry (IV).


Question 63:

Given below are two statements:

  • Statement I: Picric acid is 2,4,6-trinitrotoluene.
  • Statement II: Phenol-2,4-disulphuric acid is treated with conc. HNO3 to get picric acid.

Choose the most appropriate answer from the options given below:

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

Solution: Picric acid is not trinitrotoluene; it is 2,4,6-trinitrophenol.

Read More

  • Statement I is incorrect as picric acid is 2,4,6-trinitrophenol, not trinitrotoluene.
  • Statement II is correct because phenol-2,4-disulphuric acid reacts with concentrated HNO3 to form picric acid.


Question 64:

Which of the following is a metamer of the given compound (X)?

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

Solution: Metamers have the same molecular formula and functional group but differ in the alkyl groups on either side of the functional group.

Read More

  • Analyze the alkyl groups around the functional group of the given compound.
  • The correct metamer has a different arrangement of alkyl groups but retains the same molecular formula and functional group.


Question 65:

DNA molecule contains 4 bases whose structures are shown below. One of the structures is not correct. Identify the incorrect base structure.

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

Solution: DNA contains adenine, guanine, cytosine, and thymine. The incorrect structure does not match these bases.

Read More

  • Examine the structure of the four bases.
  • Adenine and guanine are purines, while cytosine and thymine are pyrimidines.
  • The incorrect structure is one that does not correspond to any of these bases.


Question 66:

Match List I with List II:

List I (Hybridization) List II (Orientation in Space)
A. sp3 III. Tetrahedral
B. dsp2 IV. Square planar
C. sp3d I. Trigonal bipyramidal
D. sp3d2 II. Octahedral

Choose the correct answer from the options below:

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

Solution: Match the hybridization with its spatial orientation.

Read More
  • A (sp3): Tetrahedral structure with bond angles of approximately 109.5°.
  • B (dsp2): Square planar geometry, seen in coordination complexes like [Ni(CN)4]2-.
  • C (sp3d): Trigonal bipyramidal geometry with axial and equatorial positions.
  • D (sp3d2): Octahedral geometry, typical for complexes like [SF6].

Question 67:

Given below are two statements:

  • Statement I: Gallium is used in the manufacturing of thermometers.
  • Statement II: A thermometer containing gallium is useful for measuring the freezing point (256 K) of brine solution.

Choose the correct answer from the options below:

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

Solution: Gallium thermometers are designed for high-temperature measurements and are unsuitable for freezing points like 256 K.

Read More
  • Gallium's melting point is approximately 303 K, making it liquid at room temperature, ideal for high-temperature applications.
  • However, it cannot be used to measure low temperatures like the freezing point of brine solution (256 K).

Question 68:

Which of the following statements are correct?

  • A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.
  • B. Aniline can be purified by steam distillation because it decomposes at its normal boiling point.
  • C. Ethanol can be separated from ethanol-water mixture by azeotropic distillation because it forms azeotrope.
  • D. An organic compound is pure if mixed melting point remains unchanged.

Choose the correct answer from the options below:

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

Solution: Statements A, C, and D are correct as per the properties of the compounds and methods of purification.

Read More
  • A: Glycerol decomposes at its normal boiling point, hence vacuum distillation is used.
  • C: Ethanol and water form an azeotrope, separated by azeotropic distillation.
  • D: A pure organic compound retains its mixed melting point.
  • B is incorrect as aniline is immiscible in water but does not decompose at its boiling point.

Question 69:

Match List I with List II:

List I (Compound/Species) List II (Shape/Geometry)
A. SF4 III. See-saw
B. BrF3 IV. Bent T-shape
C. BrO3- II. Pyramidal
D. NH4+ I. Tetrahedral

Choose the correct answer from the options below:

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

Solution: Match the compounds with their shapes based on VSEPR theory.

Read More
  • A: SF4: See-saw due to one lone pair.
  • B: BrF3: T-shape (Bent) due to two lone pairs.
  • C: BrO3-: Pyramidal due to one lone pair.
  • D: NH4+: Tetrahedral as all valence electrons are bonded.

Question 70:

In Reimer-Tiemann reaction, phenol is converted into salicylaldehyde through an intermediate. The structure of the intermediate is:

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

Solution: The intermediate is the sodium salt of dichlorophenol, formed via the attack of dichlorocarbene (:CCl2) on phenol.

Read More
  • The Reimer-Tiemann reaction introduces a -CHO group ortho to the phenolic group.
  • The dichlorocarbene reacts with phenol to form the intermediate, which upon hydrolysis gives salicylaldehyde.

Question 71:

Which of the following material is not a semiconductor?

  1. Germanium
  2. Graphite
  3. Silicon
  4. Copper oxide
Correct Answer: (2) Graphite

Solution: Graphite is a good conductor of electricity, not a semiconductor.

Read More
  • Semiconductors like silicon and germanium have a specific band gap that allows them to conduct electricity under certain conditions.
  • Graphite, on the other hand, conducts electricity due to its delocalized π-electrons and does not have a band gap like semiconductors.

Question 72:

Consider the following complexes:

  • (A) [CoCl(NH3)5]2+
  • (B) [Co(CN)6]3-
  • (C) [Co(NH3)5(H2O)]3+
  • (D) [Cu(H2O)4]2+

The correct order of A, B, C, and D in terms of wavenumber of light absorbed is:

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

Solution: The ligand field strength determines the wavenumber of absorption. Strong field ligands like CN- cause higher energy transitions.

Read More
  • [CoCl(NH3)5]2+ has weaker field ligands compared to [Co(CN)6]3-, leading to a lower wavenumber absorption.
  • [Cu(H2O)4]2+ has the weakest field ligands, hence the lowest wavenumber.

Question 73:

Match List I with List II:

List I (Precipitating reagent and conditions) List II (Cation)
A. NH4Cl + NH4OH III. Al3+
B. NH4OH + Na2CO3 IV. Sr2+
C. NH4OH + NH4Cl + H2S gas II. Pb2+
D. Dilute HCl I. Mn2+

Choose the correct answer from the options below:

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

Solution: Match the reagents with the cations based on their precipitation reactions.

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  • A: NH4Cl + NH4OH precipitates Al(OH)3 (III).
  • B: NH4OH + Na2CO3 precipitates SrCO3 (IV).
  • C: NH4OH + NH4Cl + H2S precipitates PbS (II).
  • D: Dilute HCl precipitates MnCl2 (I).

Question 74:

The electron affinity values are negative for:

  • A. Be → Be-
  • B. N → N-
  • C. O → O-
  • D. Na → Na-
  • E. Al → Al-

Choose the most appropriate answer from the options below:

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

Solution: Electron affinity is negative for elements that do not readily accept an electron to form a stable anion.

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  • D: Sodium (Na) does not form a stable anion, leading to negative electron affinity.
  • E: Aluminum (Al) has a similar property, resulting in a negative electron affinity.

Question 75:

The number of elements from the following that do not belong to lanthanoids is:

Eu, Cm, Er, Tb, Yb, and Lu

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

Solution: Cm (Curium) is an actinide, while the rest belong to the lanthanoid series.

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  • Lanthanoids include Eu, Er, Tb, Yb, and Lu.
  • Cm (Curium) is an actinide and does not belong to the lanthanoid series.

Question 76:

The density of 'x' M solution ('x' molar) of NaOH is 1.12 g/mL, while in molality, the concentration of the solution is 3m (3 molal). Then x is:

  1. 3.5
  2. 3.0
  3. 3.8
  4. 2.8
Correct Answer: (2) 3.0

Solution: The relationship between molarity and molality is used with the given density, resulting in x = 3.0 M.

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Molarity (M) = (Molality × Density × 1000) / (1000 + (Molality × Molar Mass of Solute)). Substituting values, we find x = 3.0.


Question 77:

Which among the following aldehydes is most reactive towards nucleophilic addition reactions?

  1. HCHO
  2. C2H5CHO
  3. CH3CHO
  4. C3H7CHO
Correct Answer: (1) HCHO

Solution: Formaldehyde (HCHO) is the most reactive due to the absence of alkyl groups, leading to reduced steric hindrance and a higher partial positive charge on the carbonyl carbon.

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As alkyl groups increase, steric hindrance and electron donation to the carbonyl carbon reduce reactivity towards nucleophiles.


Question 78:

At −20°C and 1 atm pressure, a cylinder is filled with an equal number of H2, I2, and HI molecules for the reaction:

H2(g) + I2(g) ⇌ 2HI(g)

Kp for the process is x × 10−1. x =?

  1. 2
  2. 1
  3. 10
  4. 0.01
Correct Answer: (3) 10

Solution: Using the equilibrium constant expression, Kp = (PHI)² / (PH2 × PI2), and substituting mole fractions, x = 10.

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The partial pressures of the gases and the stoichiometry of the reaction allow for the calculation of Kp.


Question 79:

Match List I with List II:

List I (Compound) List II (Uses)
A. Iodoform III. Antiseptic
B. Carbon tetrachloride I. Fire extinguisher
C. CFC IV. Refrigerants
D. DDT II. Insecticide

Choose the correct answer from the options below:

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

Solution: A-III: Iodoform is an antiseptic; B-I: Carbon tetrachloride is a fire extinguisher; C-IV: CFCs are refrigerants; D-II: DDT is an insecticide.

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  • Iodoform has antiseptic properties due to its iodine content.
  • Carbon tetrachloride is a non-flammable liquid used in fire extinguishers.
  • CFCs (chlorofluorocarbons) are widely used in refrigeration systems.
  • DDT is an effective pesticide but is banned in many countries due to its environmental impact.

Question 80:

A conductivity cell with two electrodes (dark side) is half filled with an infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will:

  1. Increase sharply
  2. Remain same or cannot be measured accurately
  3. Decrease sharply
  4. Depend upon type of electrolyte
Correct Answer: (2) Remain same or cannot be measured accurately

Solution: For infinitely dilute solutions, molar conductivity becomes constant, and further dilution does not change it.

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At infinite dilution, ion interactions become negligible, and the limiting molar conductivity remains unaffected by additional water.


Question 81:

Consider the dissociation of the weak acid HX as given below:

HX(aq) ⇌ H+(aq) + X(aq), Ka = 1.2×10−5

The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ___ ×10−2 bar (nearest integer).

Given: R = 0.083 L·bar·mol−1·K−1

Correct Answer: 76

Solution: The osmotic pressure is calculated using the formula:

Π = iCRT, where i is the van 't Hoff factor considering dissociation.

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Considering the degree of dissociation and using the formula for i, the total concentration is adjusted to account for dissociation, leading to an osmotic pressure of approximately 76 × 10−2 bar.


Question 82:

The difference in the ‘spin-only’ magnetic moment values of KMnO4 and the manganese product formed during titration of KMnO4 against oxalic acid in acidic medium is ___ BM (nearest integer).

Correct Answer: 6

Solution: In KMnO4, Mn is in the +7 oxidation state (no unpaired electrons, μ = 0 BM). In the reduced product (Mn2+), there are 5 unpaired electrons, giving a spin-only magnetic moment of 6 BM.

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The difference is due to the change in the oxidation state of Mn from +7 (KMnO4) to +2 (Mn2+), where the latter has unpaired electrons contributing to its magnetic moment.


Question 83:

Time required for 99.9% completion of a first-order reaction is ___ times the time required for completion of 90% reaction (nearest integer).

Correct Answer: 3

Solution: The time for a given percentage completion is proportional to ln(remaining concentration). For 99.9% completion:

t99.9% = (ln(1000) / ln(10)) × t90%, which simplifies to 3 times t90%.

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The relationship is derived from the first-order reaction equation: t = (1/k) ln([A]0 / [A]).


Question 84:

Number of molecules from the following which can exhibit hydrogen bonding is ___ (nearest integer):

CH3OH, H2O, C6H6, C6H5NO2, HF, NH3

Correct Answer: 5

Solution: Hydrogen bonding is exhibited by CH3OH, H2O, HF, NH3, and C6H5NO2, due to the presence of N-H, O-H, or F-H bonds.

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C6H6 (benzene) cannot form hydrogen bonds as it lacks the necessary functional groups for hydrogen bonding.


Question 85:

9.3 g of pure aniline upon diazotization followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _____ g (nearest integer).

Correct Answer: 20

Solution: The molecular weight of aniline is 93 g/mol. Given 9.3 g of aniline, this corresponds to 0.1 mol. Each mole of aniline produces 1 mole of dye, and the molecular weight of the dye is 200 g/mol, so:

Mass of dye = 0.1 × 200 = 20 g.

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The reaction is a 1:1 stoichiometric conversion, ensuring all aniline converts to dye in 100% yield.


Question 86:

The major product of the following reaction is P:

CH3C≡C-CH3
(i) Na/liq. NH3 → (ii) dil. KMnO4, 273 K → P
The number of oxygen atoms present in product P is _____ (nearest integer).

Correct Answer: 2

Solution: The reaction produces a vicinal diol as the final product, with two hydroxyl groups attached to the adjacent carbon atoms, contributing two oxygen atoms.

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Na in liquid NH3 reduces the alkyne to a trans-alkene, and KMnO4 hydroxylates the double bond, forming a vicinal diol.


Question 87:

The frequency of the de-Broglie wave of an electron in Bohr’s first orbit of the hydrogen atom is ___ ×1013 Hz (nearest integer).

Given: RH (Rydberg constant) = 2.18×10−18 J, h (Planck’s constant) = 6.6×10−34 J·s.

Correct Answer: 661

Solution: The frequency is calculated using the relation ν = E/h, where E is the kinetic energy of the electron in the first orbit.

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The energy of the first orbit is derived from E = RH, and substituting values gives ν ≈ 661 ×1013 Hz.


Question 88:

The major products from the following reaction sequence are product A and product B:

B (i) Br2 → (ii) alc. KOH (3 eq.) → A
B (i) Br2 → (ii) Na+/O (1.0 eq.) → B
The total sum of π electrons in product A and product B are ____ (nearest integer).

Correct Answer: 8

Solution: Product A is benzene (6 π electrons), and product B is an alkene (2 π electrons). The total sum is 8 π electrons.

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The reaction sequence involves dehydrohalogenation and elimination, leading to aromatic and alkenic structures as products.


Question 89:

Among CrO, Cr2O3, and CrO3, the sum of spin-only magnetic moment values of basic and amphoteric oxides is ___ ×10−2 BM (nearest integer).

Given: Atomic number of Cr is 24.

Correct Answer: 877

Solution: CrO (Cr2+) has 4 unpaired electrons, Cr2O3 (Cr3+) has 3 unpaired electrons, and CrO3 (Cr6+) has no unpaired electrons.

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The spin-only magnetic moment is μ = √n(n + 2), where n is the number of unpaired electrons. Adding the moments for CrO and Cr2O3 gives 8.77 BM (or 877 ×10−2 BM).


Question 90:

An ideal gas, CV = 5/2 R, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively, then the final temperature is ____ K (nearest integer).

Correct Answer: 274

Solution: Using the adiabatic relation PVγ = constant, where γ = CP/CV, the final temperature is calculated.

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For γ = 1.4 (diatomic gas), the relationship T1V1γ−1 = T2V2γ−1 is applied, leading to T2 ≈ 274 K.


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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