
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.
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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:
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.
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.
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:
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.
We compute:
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:
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.
Calculate:
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?
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.
For transitivity:
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:
Solution:
Solve the equations for arithmetic and geometric progressions. Calculate the cube of the geometric mean, resulting in 120.
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.
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:
Solution:
Compute adj(A) and its powers, summing the matrices to find B. The total sum of elements in B is -88.
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.
Evaluate 1×2² + 2×3² + ... + 100×(101)² / (12×22 + 22×32 + ... + 100²×101):
Solution:
Using summation formulas for numerator and denominator, compute the ratio, yielding 305.
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.
Let f(x) = ∫₀ˣ (t + sin(1 − eᵗ)) dt, x ∈ R. Find lim(x→0) f(x) / x³:
Solution:
Apply L’Hopital’s Rule and evaluate the limit step by step, yielding -1/6.
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.
The area of the region {(x, y): y² ≤ 2x, y ≥ 4x − 1} is:
Solution:
Compute the intersection points of the parabola and line, then integrate to find the area as 9/32.
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.
The area of the region S = {z ∈ C : |z − 1|² ≤ 4, z + z̅ ≥ 2, Im(z) ≥ 0} is:
Solution:
Solve geometrically to identify the intersection region as a semicircle and compute its area as 3π/2.
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.
If the value of the integral ∫-11 (cos(αx) / (1 + 3x)) dx = (2 / π), then a value of α is:
Solution:
Recognizing symmetry and applying definite integral properties, α = π / 2 satisfies the integral equation. Substituting confirms the given integral value of (2 / π).
The integral is simplified by considering the even nature of cos(αx). Substitute α = π / 2, and verify the integral evaluates to 2 / π.
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:
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.
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:
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:
Solution:
Using the condition 2(nC5) = (nC4) + (nC6) with binomial coefficients, solve for n. Simplifying gives the maximum value of n as 14.
The condition for arithmetic progression is:
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:
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.
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 |
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.
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?
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.
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.
Given cos⁻¹x − sin⁻¹y = α, −π/2 ≤ α ≤ π. The minimum value of x² + y² + 2xy sinα is:
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.
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.
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:
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.
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).
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:
Solution:
Calculate the cross product a × b and dot it with c to find (a × b) · c = 11.
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:
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.
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:
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:
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.
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:
Solution:
Use integration by parts and substitution to evaluate the integral, identifying α and β. The calculation shows that 8(α + β) = 3.
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:
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.
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:
Solution:
Analyzing the recursive composition pattern, α is determined as 2¹⁰ − 1. Substituting into √(3α + 1) yields 1024.
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:
Solution:
Solve for the elements of A using the conditions, calculate A⁻¹, and equate to αA + βI. Simplifying gives α + β = 5.
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:
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.
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:
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.
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:
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.
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:
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.
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:
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.
The translational degrees of freedom (ft) and rotational degrees of freedom (fr) of CH4 molecule are:
Solution:
CH4 is a polyatomic molecule with a non-linear structure. For such molecules:
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:
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:
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:
Solution:
The work done in rotating the magnet is the change in potential energy:
Which of the diode circuits shows correct biasing used for the measurement of dynamic resistance of a p-n junction diode:
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.
Arrange the following in ascending order of wavelength:
(A) Gamma rays (λ₁)
(B) X-rays (λ₂)
(C) Infrared waves (λ₃)
(D) Microwaves (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.
Identify the logic gate given in the circuit:
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.
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:
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.
Correct formula for the height of a satellite from Earth’s surface is:
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.
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
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).
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:
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.
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
Solution:
Using conservation of energy: PE at A = KE at B.
- 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.
An electric bulb rated 50 W – 200 V is connected across a 100 V supply. The power dissipation of the bulb is:
Solution:
Power dissipation is calculated using the formula P = V²/R.
- Resistance of the bulb is R = V²/P = 200²/50 = 800 Ω.
- Power dissipation at 100 V is P = 100²/800 = 12.5 W.
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:
Solution:
At the angle of repose, tanθ = μs.
- 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.
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:
Solution:
The total mechanical energy in SHM is given by E = (1/2)kA².
- 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.
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:
Solution:
The number of photons decreases with frequency for a constant intensity.
- 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.
According to Bohr’s theory, the moment of momentum of an electron revolving in the 4th orbit of a hydrogen atom is:
Solution:
Bohr's angular momentum is given by L = n(h/2π).
- For the 4th orbit, n = 4.
- Substituting into L = n(h/2π), we get L = 4(h/2π) = 2h/π.
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:
Solution:
Using the adiabatic relation TVγ−1 = constant, calculate work done.
- Use the formula for work done in an adiabatic process: W = (P1V1 − P2V2)/(γ − 1).
- After substituting values and simplifying, W = RT[2 − √2].
A charge q is placed at the center of one surface of a cube. The flux linked with the cube is:
Solution:
Apply Gauss’s law to find the flux through the cube.
- 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.
Applying the principle of homogeneity of dimensions, determine which one is correct:
Solution:
Dimensional analysis is used to verify correctness.
- 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.
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
Solution:
Gravitational force depends on the distance from the center of the Earth.
- Gravitational acceleration at a distance of 3R from the Earth's center is g/9.
- Force F = mg = 90 × (10/9) = 100 N.
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:
Solution:
Velocity is given by v = dx/dt.
- 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.
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:
Solution:
Use the kinematic equation to calculate distance.
- 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.
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:
Solution:
Calculate the change in energy using the initial and final capacitance.
- Initial energy Ui = (1/2)C0V² = (1/2)(12.5 × 10⁻¹²)(12²) = 9 × 10⁻¹² J.
- Final energy Uf = Ui/εr = 9 × 10⁻¹²/6 = 1.5 × 10⁻¹² J.
- Change in energy ΔU = Ui − Uf = 0.75 × 10⁻¹² J = 750 × 10⁻¹² J.
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:
Solution:
Use the center of mass condition to calculate the displacement.
- Center of mass condition: m1Δx1 = m2Δx2.
- Substituting m1 = 3, Δx1 = 2 cm, m2 = 2 gives Δx2 = (3 × 2)/2 = 3 cm.
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
Solution:
Calculate Q using the mass defect and energy equivalence.
- 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.
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:
Solution:
Calculate lateral displacement using trigonometric relations.
- 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.
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:
Solution:
The potential difference is calculated using the formula for motional EMF.
- 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.
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:
Solution:
Calculate resistance using the relation R ∝ 1/r² for the same mass.
- 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.
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:
Solution:
Using the Biot-Savart law, calculate the magnetic fields at A and C due to each wire.
- 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.
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
Solution:
Calculate the total pressure and then find the force using the area of the tube.
- 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.
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)
Solution:
Use the relationship between equilibrium constants for reversed and scaled reactions.
- 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.
Find out the major product formed from the following reaction:
Solution:
The reaction proceeds via an SN2 mechanism, replacing bromine atoms with Me2N groups.
- 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.
When MnO2 and H2SO4 are added to a salt (A), the greenish-yellow gas liberated indicates that salt (A) is:
Solution:
The reaction liberates Cl2 gas, which is greenish-yellow.
- Reaction: 2NH4Cl + MnO2 + 2H2SO4 → MnSO4 + (NH4)2SO4 + 2H2O + Cl2.
- The greenish-yellow gas is chlorine (Cl2).
The correct statements about hydrogen bonding are:
Solution:
Hydrogen bonding depends on electronegativity, physical state, and molecular structure.
- 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.
In the above chemical reaction sequence, "A" and "B" respectively are:
Solution:
The reaction involves ozonolysis followed by a haloform reaction.
- Step 1: Ozonolysis cleaves the double bond to form compound A (aldehydes).
- Step 2: Haloform reaction converts aldehydes into carboxylates and secondary alcohols.
Common name of Benzene-1,2-diol is:
Solution:
The structure of Benzene-1,2-diol matches Catechol.
- 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.
Consider the above reactions. Identify product B and product C:
Solution:
The reactions involve oxidation and reduction steps.
- B (2-Propanol) is formed as a secondary alcohol.
- C (1-Propanol) retains the primary alcohol structure after further reaction.
The adsorbent used in adsorption chromatography is/are:
Solution:
Silica gel and alumina are common adsorbents in chromatography.
- Silica gel: Acidic and polar.
- Alumina: Basic and polar.
- These adsorbents separate compounds based on polarity and adsorption affinity.
The reaction shown below leads to the formation of a major product "P":
Solution:
The reaction undergoes E2 elimination, forming the most substituted alkene.
- Alcoholic KOH favors elimination reactions.
- The most substituted (Zaitsev’s rule) product is the major product.
Correct order of stability of carbanions is shown below:
Solution:
Stability of carbanions depends on resonance and aromaticity.
- d: Aromatic, maximum stability.
- c: Resonance-stabilized.
- b: Hyperconjugation.
- a: Least stable, anti-aromatic.
The correct order of the first ionization enthalpy is:
Solution:
Ionization enthalpy increases due to lanthanide contraction and effective nuclear charge.
- 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.
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:
Solution:
Iron (III) is in a low-spin state due to the strong field ligand cyanide.
- 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.
A fuel cell, using hydrogen and oxygen as fuels:
Solution:
Fuel cells are eco-friendly and used in spaceships; they are a type of galvanic cell.
- 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.
Choose the incorrect statement about Dalton's Atomic Theory:
Solution:
Dalton's theory states that compounds form in fixed, whole-number ratios.
- 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.
Match List I with List II:
List I:
List II:
Solution:
The correct match is based on the relationship between molecules.
- 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.
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.
Solution:
Both statements are accurate based on periodic trends.
- 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.
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:
Solution:
The slope unit is derived from molar conductivity against √concentration.
- 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.
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:
Solution:
The spin-only magnetic moment corresponds to 3 unpaired electrons.
- 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.
The number of unpaired d-electrons in [Co(H2O)6]3+ is:
Solution:
Co3+ has all d-electrons paired in a strong ligand field.
- 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.
The number of species from the following that have pyramidal geometry around the central atom is:
Solution:
Only SO32− has pyramidal geometry due to a lone pair on the central atom.
- S2O32−, SO42−, and S2O72− have tetrahedral or planar structures.
- SO32− has one lone pair on sulfur, leading to a pyramidal shape.
The maximum number of orbitals which can be identified with n = 4 and ml = 0 is:
Solution:
There are 4 orbitals with ml = 0 for n = 4.
- 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.
The number of compounds/species from the following with non-zero dipole moment is:
Solution:
The compounds with non-zero dipole moments are identified.
- Polar molecules: NF3, H2O, H2S, HBr, HCl.
- Non-polar molecules: BeCl2, BCl3, XeF4, CCl4, CO2, H2.
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:
Solution:
The work done equals the heat exchange (Q = -W).
- 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.
From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ___ × 10-1 g.
Solution:
The stoichiometric relationship is used to find the yield of acetanilide.
- Reaction: Aniline (93 g) → Acetanilide (135 g).
- For 6.55 g aniline, yield = (135/93) × 6.55 = 9.5 g = 95 × 10-1 g.
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:
Solution:
The first-order reaction formula is used to calculate time.
- 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.
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:
Solution:
The reaction forms N-benzylphthalimide.
- π-bonds from aromatic rings: 4.
- π-bonds from C=O groups: 2.
- Total: 6 π-bonds.
The total number of σ and π bonds in 2-oxohex-4-ynoic acid is:
Solution:
The structure is analyzed to count bonds.
- 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.
A first-row transition metal with the highest enthalpy of atomization forms Cr2O3 with a spin-only magnetic moment of:
Solution:
Spin-only magnetic moment is calculated for Cr2O3.
- Chromium in Cr2O3 is in +3 oxidation state.
- Configuration: d3, n = 3 unpaired electrons.
- μ = √[n(n+2)] = √[3(3+2)] = 5.92 BM.
2.7 kg each of water and acetic acid are mixed. The freezing point of the solution will be -x °C.
Solution:
The freezing point depression is calculated.
- 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.
The vanillin compound obtained from vanilla beans has a total sum of oxygen atoms and π electrons of:
Solution:
The molecular structure of vanillin is analyzed.
- Oxygen atoms: 3.
- π-electrons: 13 (aromatic rings and C=O bonds).
- Total: 3 + 13 = 16.
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