
JEE Main 2024 Apr 4 Shift 1 Question Paper with Solution pdf is available for download here. Students found Mathematics easy and Physics hard. Chemistry carried the highest weightage and overall difficulty level was moderate.
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Let f : R → R be a function given by:
1 − cos(2x)/x², for x < 0
β√(1 − cos(x))/x, for x > 0.
If f is continuous at x = 0, then α² + β² is equal to:
To ensure continuity at x = 0, we require the left-hand and right-hand limits of f(x) to be equal:
For x < 0: limx→0⁻ f(x) = limx→0 (1 − cos(2x))/x² = 2. This result uses the small-angle approximation of cos(2x).
For x > 0: limx→0⁺ f(x) = β limx→0 (√(1 − cos(x))/x) = β√1/2 = β/√2.
Equating the two limits, we have β/√2 = 2, so β = 2√2. Finally, α = 2 and α² + β² = 4 + 8 = 12.
Three urns A, B, and C contain 7 red, 5 black; 5 red, 7 black; and 6 red, 6 black balls, respectively. One of the urns is selected at random, and a ball is drawn. If the ball drawn is black, then the probability that it is drawn from urn A is:
We use Bayes' theorem to find P(A|Black).
The prior probabilities are P(A) = P(B) = P(C) = 1/3. The likelihoods of drawing a black ball from each urn are:
The total probability of drawing a black ball is: P(Black) = P(A)P(Black|A) + P(B)P(Black|B) + P(C)P(Black|C) = 1/3 × (5/12 + 7/12 + 6/12) = 6/12.
Finally, using Bayes' theorem: P(A|Black) = (P(A) × P(Black|A)) / P(Black) = (1/3 × 5/12) / (6/12) = 5/18.
The vertices of a triangle are A(−1, 3), B(−2, 2), and C(3,−1). A new triangle is formed by shifting the sides of the triangle one unit inwards. Then the equation of the side of the new triangle nearest to the origin is:
The equation of line AC is x + y = 2. To find a parallel line shifted inward by 1 unit, we calculate:
The perpendicular distance between two parallel lines is given by |c₁ − c₂| / √(a² + b²). For the original line x + y = 2, the shifted line becomes x + y = 2 − √2. This ensures the new line is 1 unit closer to the origin.
If the solution y = y(x) of the differential equation (x⁴ + 2x³ + 3x² + 2x + 2) dy = (2x² + 2x + 3) dx satisfies y(−1) = −π/4, then y(0) is equal to:
Separating variables and integrating:
The differential equation becomes (dy/dx) = (2x² + 2x + 3) / (x⁴ + 2x³ + 3x² + 2x + 2). Integrating using partial fractions and applying the initial condition y(−1) = −π/4, we find y(0) = π/4.
Let the sum of the maximum and minimum values of the function f(x) = (2x² − 3x + 8)/(2x² + 3x + 8) be m/n, where gcd(m,n) = 1. Then m + n is equal to:
The function is a ratio of two quadratic polynomials.
The range is determined by setting the derivative of f(x) equal to zero to find critical points. Solving yields the minimum value as 5/11 and the maximum value as 1. Their sum is 16/11, and with m/n = 16/11, we have m + n = 16 + 11 = 201.
One of the points of intersection of the curves y = 1 + 3x − 2x² and y = 1/x is (1/2, 2). Let the area of the region enclosed by these curves be 1/24 (ℓ√5 + m) − n ln(1 + √5), where ℓ, m, n ∈ N. Then ℓ + m + n is equal to:
The area is calculated using the integral of (1 + 3x − 2x² − 1/x).
Simplifying the integral and solving, we find that the values of ℓ, m, and n satisfy the given area formula. The final value of ℓ + m + n = 30.
If the system of equations x + (√2 sinα) y + (√2 cosα) z = 0, x + (cosα) y + (sinα) z = 0, x + (sinα) y − (cosα) z = 0 has a non-trivial solution, then α ∈ (0, π/2) is equal to:
The determinant of the coefficient matrix must be zero for the system to have a non-trivial solution.
Solving the determinant condition, we find that α = 5π/24 satisfies the equation.
There are 5 points P₁, P₂, P₃, P₄, P₅ on the side AB, excluding A and B, of a triangle ABC. Similarly, there are 6 points P₆, P₇, ..., P₁₁ on the side BC and 7 points P₁₂, P₁₃, ..., P₁₈ on the side CA of the triangle. The number of triangles that can be formed using the points P₁, P₂, ..., P₁₈ as vertices is:
The total number of ways to select 3 points from 18 is 816. Subtracting the collinear cases:
The collinear points on AB, BC, and CA are excluded, which amounts to 65 cases. Thus, the total number of triangles is 816 − 65 = 751.
Let f(x) =
{
x - 2, -2 < x < 0
1 - 2x, 0 ≤ x ≤ 2
}
and h(x) = f(x) + f(x). Then ∫h(x) dx is equal to:
The function h(x) = 0 on both intervals, so the integral ∫₋₂² h(x) dx = 2.
The sum of all rational terms in the expansion of (1/2⁵ + 1/5³)¹⁵ is equal to:
The binomial expansion results in a mix of rational and irrational terms.
The rational terms are calculated based on integer powers of both terms in the binomial. Their sum is found to be 3133.
Let a unit vector which makes an angle of 60° with 2i + 2j − k and an angle of 45° with i − k be C. Then C + (−1/2 i + (1/3√2 j − √2/3 k)) is:
By applying the angle conditions and solving, the resulting vector is obtained.
The dot product conditions are used to find the components of vector C. Adding the given vector results in the final solution as √2/3 i − 1/2 k.
Let the first three terms 2, p, and q, with q ≠ 2, of a G.P. be respectively the 7th, 8th, and 13th terms of an A.P. If the 5th term of the G.P. is the nth term of the A.P., then n is equal to:
Using the relations for terms in the G.P. and A.P., we calculate n:
The G.P. has terms in the ratio p/2. For the A.P., the common difference is derived from the relationship between terms. Substituting and solving yields n = 163.
Let a, b ∈ R. Let the mean and the variance of 6 observations −3, 4, 7, −6, a, b be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:
Using the given mean and variance conditions, the values of a and b are calculated:
Solving for the mean, we find a + b = 10. Using the variance formula, we calculate the individual deviations and compute the mean deviation as 13/3.
If 2 and 6 are the roots of the equation ax² + bx + 1 = 0, then the quadratic equation, whose roots are 1/(2a+b) and 1/(6a+b), is:
Using the sum and product of the roots of the quadratic equations:
The roots of the new quadratic are derived from the relationships 1/(2a+b) and 1/(6a+b). Substituting these into the quadratic formula gives x² + 8x + 12 = 0.
Let α and β be the sum and the product of all the non-zero solutions of the equation z² + |z| = 0, z ∈ C. Then 4(α² + β²) is equal to:
The non-zero solutions are z = i and z = −i.
For these solutions, α = 0 and β = −1. Substituting into the formula 4(α² + β²), we find the result as 4.
Let the point, on the line passing through the points P(1,−2, 3) and Q(5,−4, 7), farther from the origin and at a distance of 9 units from the point P, be (α, β, γ). Then α² + β² + γ² is equal to:
Using the parametric equation of the line and the distance formula:
The parametric equation of the line is derived as (x, y, z) = (1 + 4t, −2 − 2t, 3 + 4t). Using the condition that the distance from P is 9, solving for t gives the point (7, −5, 9). Thus, α² + β² + γ² = 7² + (−5)² + 9² = 155.
A square is inscribed in the circle x² + y² − 10x − 6y + 30 = 0. One side of this square is parallel to y = x + 3. If (xi, yi) are the vertices of the square, then Σ(x²i + y²i) is equal to:
The center of the circle and its radius are calculated as follows:
The center of the circle is (5, 3), and the radius is 2. Using the properties of the inscribed square and its orientation, the vertices are calculated. The sum of the squares of their coordinates is found to be 152.
If the domain of the function sin⁻¹((3x − 22)/(2x − 19)) + ln((3x² − 8x + 5)/(x² − 3x − 10)) is (α, β), then 3α + 10β is equal to:
The domain is derived from the conditions of the functions:
Solving the inequalities for the arguments of sin⁻¹ and ln to be valid, we find α = 5 and β = 41/5. Substituting into the formula, 3α + 10β = 97.
Let f(x) = x⁵ + 2e^(x/4) for all x ∈ R. Consider a function g(x) such that (g ∘ f)(x) = x for all x ∈ R. Then the value of 8g′(2) is:
Using the chain rule, we calculate g′(f(x)) ⋅ f′(x) = 1:
At x = 0, f(0) = 0 and f′(0) = 5(0)⁴ + 1/2e^(0) = 1/2. Thus, g′(0) = 1/f′(0) = 2. At x = 2, g′(2) = g′(f(2)) ⋅ f′(2). Substituting values, we find 8g′(2) = 16.
Let α ∈ (0,∞) and A =
[1 2 α] [1 0 1] [0 1 2]
If det(adj(2A − Aᵀ) ⋅ adj(A − 2Aᵀ)) = 28, then (det(A))² is equal to:
Using determinant properties:
det(adj(B)) = (det(B))^(n−1). Substituting the given matrices and solving for α, we find det(A) = 4, so (det(A))² = 16.
If limx→1 ((5x + 1)^(1/3) − (x + 5)^(1/3)) / ((2x + 3)^(1/2) − (x + 4)^(1/2)) = m√5 / n(2n)^(2/3), where gcd(m, n) = 1, then 8m + 12n is equal to:
Using Taylor expansions and simplifications:
The numerator and denominator are expanded using first-order approximations. Simplifying gives m = 8 and n = 3. Substituting into the formula, 8m + 12n = 97.
In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics; and 10 studied none of these subjects. Let m and n respectively be the least and the most number of students who studied all three subjects. Then m + n is equal to:
Using inclusion-exclusion and solving for the range of students who studied all three subjects:
The total number of students studying at least one subject is calculated as 220 − 10 = 210. Using inclusion-exclusion for the three sets and calculating the intersections, the least number of students studying all three subjects is m = 10, and the maximum is n = 15. Hence, m + n = 25.
Let the solution y = y(x) of the differential equation (dy/dx) − y = 1 + 4sin(x) satisfy y(π) = 1. Then y(π/2) + 10 is equal to:
Solving the first-order linear differential equation using integrating factors:
The integrating factor is e^(−x). Multiplying through and solving with the initial condition y(π) = 1, we find y(π/2) = −3. Adding 10 gives y(π/2) + 10 = 7.
The shortest distance between the lines (x+2)/2 = (y+3)/3 = (z−5)/4 and (x−3)/1 = (y−2)/−3 = (z+4)/2 is 38/√5 k and ∫0k [x²] dx = α−√α, where [x] denotes the greatest integer function, then 6α³ is equal to:
Using vector cross-product formula for the shortest distance between skew lines and evaluating the integral:
The distance between the lines is calculated using the direction vectors and position vectors. For the integral, solving for α gives α = 2, and substituting into 6α³ yields 15.
Let A be a 3×3 matrix of non-negative real elements such that A [1 1 1] = 3 [1 1 1]. Then the maximum value of det(A) is:
Using matrix operations and maximizing determinant subject to the given constraints:
The eigenvalues of A are calculated to satisfy the condition A[1 1 1] = 3[1 1 1]. Maximizing the determinant subject to these constraints gives det(A) = 6.
Let a = 1 + (2C2)/3! + (3C2)/4! + (4C2)/5! + ... and b = 1 + (1C0)/1! + (1C1)/1! + (2C0)/2! + (2C1)/2! + ... . Then 2b/a² is equal to:
Evaluating the series for a and b:
The series are evaluated term by term using the definitions of binomial coefficients and factorials. Simplifying the ratio 2b/a² gives the result as 10.
Let A be a 3×3 matrix of non-negative real elements such that A [1 1 1] = 3 [1 1 1]. Then the maximum value of det(A) is:
From matrix determinant properties and the given conditions:
Analyzing the eigenvalues and properties of A gives the maximum determinant of det(A) = 20.
Let the length of the focal chord PQ of the parabola y² = 12x be 15 units. If the distance of PQ from the origin is p, then 10p² is equal to:
For the parabola y² = 12x, the length of a focal chord is 4a csc²(θ):
With the given length of the focal chord, solving for the parameter θ and the distance p gives 10p² = 72.
Let ABC be a triangle of area 15√2 and the vectors AB = i + 2j − 7k, BC = ai + bj + ck, and AC = 6i + dj − 2k, d > 0. Then the square of the length of the largest side of the triangle ABC is:
Using vector cross-product to find the area and then solving for the lengths of the sides:
The sides of the triangle are calculated using vector norms, and the largest side is determined to have a squared length of 54.
If ∫₀^(π/4) (sin²x / (1 + sinx cosx)) dx = (1 / a) ln(a / 3) + (π / b√3), then a + b is equal to:
By solving the integral:
The substitution and simplification give a = 2 and b = 6, hence a + b = 8.
The electron will continue to move with uniform velocity along the axis of the solenoid when:
Since the magnetic field is parallel to the motion of the electron:
The force on the electron due to the magnetic field is zero as F = q(v × B) and the velocity is parallel to the magnetic field. Therefore, the electron moves with uniform velocity along the axis of the solenoid.
The magnetic field induction of an electromagnetic wave is:
The magnetic field in an electromagnetic wave is perpendicular to both the electric field and the direction of propagation:
The electric field is aligned along î, the wave propagates in the z-direction, so the magnetic field must be in the ĵ direction to satisfy the cross-product relation E × B = c. The correct expression is B = ĵ 40 c cos(ω(t − z/c)).
Which of the following nuclear fragments corresponding to nuclear fission between neutron (¹⁰n) and uranium isotope (²³⁵₉₂U) is correct?
In nuclear fission of ²³⁵₉₂U:
When a neutron collides with a uranium-235 nucleus, it splits into ¹⁴⁴₅₆Ba and ⁸⁹₃₆Kr, releasing 3 neutrons. This is based on conservation of mass and atomic number in nuclear reactions.
In an experiment to measure focal length (f) of a convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are ∆u and ∆v, respectively. The error in the measurement of the focal length of the convex lens will be:
Using the lens formula 1/f = 1/v − 1/u, the errors are propagated:
The differential of the lens formula gives the error in f as proportional to ∆u/u² and ∆v/v². Combining these terms results in f² [ ∆u/u² + ∆v/v² ] as the total error.
Given below are two statements:
Statement I: When speed of liquid is zero everywhere, pressure difference at any two points depends on equation P₁ − P₂ = ρg(h₂ − h₁).
Statement II: In a venturi tube shown, 2gh = v₂² − v₁².
In the light of the above statements, choose the most appropriate answer from the options given below:
Analyzing both statements based on principles of fluid mechanics:
Statement I is correct as it applies to hydrostatic conditions. Statement II is incorrect because it does not accurately represent Bernoulli’s equation, and 2gh ≠ v₂² − v₁² for the venturi tube scenario.
The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are 8Ω and 10Ω respectively. After inserting it in a hot bath of temperature 400°C, the resistance of platinum wire is:
Using the temperature-resistance relationship:
The resistance of the platinum wire at a given temperature is proportional to its temperature change. Using the formula \( R_t = R_0 (1 + \alpha t) \), the resistance at 400°C is calculated to be 16Ω.
A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the center of the arc. The gravitational force on the particle by the wire is:
Using the formula for gravitational force from a uniform mass distribution along a semicircular arc:
For the wire bent into a semicircle, the gravitational force at the center is the result of integration over the mass elements along the arc. The net force is 2GMmπ/L².
On Celsius scale, the temperature of a body increases by 40°C. The increase in temperature on Fahrenheit scale is:
The relationship between temperature changes in Celsius and Fahrenheit scales is given by Δ°F = Δ°C × 1.8:
For a 40°C increase, the corresponding change on the Fahrenheit scale is 40 × 1.8 = 72°F.
An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25 D. Focal length of each of the convex lenses is:
For lenses in contact, the total power is the sum of the individual powers:
The total power is 25 D for 5 lenses, so each lens has a power of 5 D. The focal length of each lens is \( f = 100/P \), which gives 20 cm.
Which figure shows the correct variation of applied potential difference V with photoelectric current I at two different intensities of light (I₁ < I₂) of same wavelength:
In the photoelectric effect, the stopping potential is constant for the same wavelength, while the saturation current increases with intensity:
The correct graph is Figure 3, where the stopping potential remains the same, and the saturation current increases with light intensity.
A wooden block, initially at rest on the ground, is pushed by a force which increases linearly with time t. Which of the following curves best describes the acceleration of the block with time:
Since the force increases linearly with time, the acceleration will also increase linearly, as given by Newton's second law (F = ma):
The acceleration of the block is directly proportional to the force, and since force increases linearly with time, the acceleration also increases linearly. This is shown in Figure 2.
A rubber ball falls from a height h and rebounds up to the height of h/2, the percentage loss of total energy of the initial system as well as the velocity of the ball before it strikes the ground, respectively, are:
The ball loses 50% of its total energy during the bounce:
The potential energy at height h is proportional to h. Since the rebound height is h/2, the energy loss is 50%. The velocity before impact is calculated using energy conservation: \( v = √(2gh) \).
The equation of a stationary wave is: y = 2a sin(2πnt / λ) cos(2πx / λ). Which of the following is NOT correct?
From the stationary wave equation, we know the correct dimensions for each term:
The term n/λ represents a frequency, which has dimensions [T⁻¹]. Therefore, the dimensions of n/λ as [T] are incorrect.
A body travels 102.5 m in the nth second and 115.0 m in the (n + 2)th second. The acceleration is:
Using the formula for distance traveled in the nth second:
The distance in the nth second is \( s_n = u + \frac{a}{2}(2n - 1) \). Setting up equations for the nth and (n+2)th seconds and solving gives an acceleration of 6.25 m/s².
To measure the internal resistance of a battery, a potentiometer is used. For R = 10Ω, the balance point is observed at ℓ = 500 cm and for R = 1Ω the balance point is observed at ℓ = 400 cm. The internal resistance of the battery is approximately:
Using the balance point data, apply the formula for internal resistance in a potentiometer setup:
The formula \( r_s = R (\ell_1 - \ell_2) / \ell_2 \) is used. Substituting the values, \( r_s = (10)(500 - 400) / 400 = 0.3Ω \).
An infinitely long positively charged straight thread has a linear charge density λ Cm⁻¹. An electron revolves along a circular path having its axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of the electron as a function of the radius of the circular path from the wire is:
The kinetic energy of the electron depends on the electric field from the charged wire:
The electric field due to the wire decreases with distance from the wire, leading to a decrease in the centripetal force and, consequently, a decrease in kinetic energy. The correct graph is Figure 2.
The value of net resistance of the network as shown in the given figure is:
The resistances in the given network are simplified using series and parallel combinations:
Analyzing the network step by step, the equivalent resistance is found to be 6Ω using the rules for series and parallel resistance combinations.
P-T diagram of an ideal gas having three different densities ρ₁, ρ₂, ρ₃ (in three different cases) is shown in the figure. Which of the following is correct:
The relationship between pressure, temperature, and density is given by the ideal gas law:
At a constant temperature, the density is inversely proportional to the volume. Based on the P-T graph, the density ρ₁ is greater than ρ₂ due to the inverse relationship.
The coordinates of a particle moving in the x-y plane are given by:
x = 2 + 4t, y = 3t + 8t².
The motion of the particle is:
The y-component of the motion shows a quadratic dependence on time, while the x-component is linear:
The equations describe a parabolic path because the y-component is proportional to t², while the x-component is linear. The particle's motion is uniformly accelerated along this parabolic trajectory.
In an AC circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to:
In an LC circuit, the current leads or lags the voltage by 90°:
The phase difference between current and voltage in an LC circuit results in the current being zero when the voltage reaches its maximum value. This behavior occurs due to the resonant properties of the inductor-capacitor combination.
An infinite plane sheet of charge having uniform surface charge density +σ C/m² is placed on the x-y plane. Another infinitely long line charge having uniform linear charge density +λe C/m is placed at z = 4m plane and parallel to the y-axis. If the magnitude values |σ| = 2|λe| then at point (0, 0, 2), the ratio of magnitudes of electric field values due to sheet charge to that of line charge is π√n : 1. The value of n is:
The electric field due to an infinite sheet is constant, and the electric field due to a line charge decreases with distance:
By applying the electric field formulas for an infinite sheet and a line charge, the ratio of magnitudes is calculated as π√16:1. Hence, n = 16.
A hydrogen atom changes its state from n = 3 to n = 2. Due to recoil, the percentage change in the wavelength of emitted light is approximately 1×10⁻ⁿ. The value of n is:
The percentage change in the wavelength due to recoil is approximately 10⁻⁷:
Using energy conservation and the Rydberg formula, the recoil effect introduces a small wavelength change. Calculations confirm that n = 7.
The magnetic field existing in a region is given by B = 0.2(1 + 2x)k̂ T. A square loop of edge 50 cm carrying 0.5A current is placed in the x-y plane with its edges parallel to the x- and y-axes. The magnitude of the net magnetic force experienced by the loop is:
The net magnetic force is calculated by integrating the force on each segment of the loop due to the varying magnetic field:
The magnetic force on the square loop arises from the variation in the field along the x-direction. Summing the forces on all four sides, the net force is found to be 50 mN.
An alternating current at any instant is given by i = (6 + √56 sin(100πt + π/3)) A. The rms value of the current is:
The rms value of a current with both DC and sinusoidal components is:
\( I_{\text{rms}} = \sqrt{I_0^2 + I_1^2/2} \), where \( I_0 = 6 \) and \( I_1 = √56 \). Substituting these values, \( I_{\text{rms}} = 8 \) A.
Twelve wires each having resistance 2Ω are joined to form a cube. A battery of 6V emf is joined across point a and c. The voltage difference between e and f is V:
The network of resistors forms a cube, and by applying symmetry:
Using the concept of equivalent resistance and symmetry, the potential difference between points e and f is calculated to be 1V.
A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If the surface tension of the soap solution is 40 dyne/cm, the new radius is:
The work done in expanding a soap bubble is given by:
Using the formula \( W = 8πr²T \), where \( r \) is the radius and \( T \) is the surface tension, the new radius is found to be 7 cm after solving for \( r \).
Two wavelengths λ₁ and λ₂ are used in Young’s double slit experiment. λ₁ = 450 nm and λ₂ = 650 nm. The minimum order of fringe produced by λ₂ which overlaps with the fringe produced by λ₁ is n. The value of n is:
In Young's experiment, overlapping fringes occur when \( n_2λ_2 = n_1λ_1 \):
Solving the equation \( 450n_1 = 650n_2 \) for integer values of \( n_1 \) and \( n_2 \), the minimum value of \( n \) is determined to be 9.
An elastic spring under tension of 3N has a length a. Its length under tension 2N is b. For its length (3a − 2b), the value of tension will be:
Using Hooke's law and proportionality of tension to length in the spring:
The tension for length (3a − 2b) is calculated as \( T = k \cdot \Delta L \). Solving for \( T \), the value is determined to be 5 N.
Two forces F₁ and F₂ are acting on a body. One force has magnitude thrice that of the other force, and the resultant of the two forces is equal to the force of larger magnitude. The angle between F₁ and F₂ is cos⁻¹(1/n). The value of |n| is:
Using the equation for the resultant force magnitude and the given condition:
The angle between the forces is calculated using vector addition and trigonometry. The value of \( n = 6 \) is determined by solving \( R = √(F₁² + F₂² + 2F₁F₂ \cosθ) \).
A solid sphere and a hollow cylinder roll up without slipping on the same inclined plane with the same initial speed v. The sphere and the cylinder reach up to maximum heights h₁ and h₂, respectively, above the initial level. The ratio h₁ : h₂ is n/10. The value of n is:
Using conservation of energy and rotational kinetic energy expressions:
The height ratio is determined as \( h₁/h₂ = (1 + k₁²/r₁²)/(1 + k₂²/r₂²) \), where \( k \) is the radius of gyration. Solving gives \( n = 7 \).
What pressure (bar) of H₂ would be required to make the emf of the hydrogen electrode zero in pure water at 25°C?
Using the Nernst equation for the hydrogen electrode:
The emf becomes zero when \( E = 0 \). Using \( E = E° − \frac{RT}{nF} \ln(P_{H₂}[H⁺]²) \), the pressure required is \( 10⁻¹⁴ \) bar, considering the neutral pH of water (\([H⁺] = 10⁻⁷\)).
The correct sequence of ligands in the order of decreasing field strength is:
According to the Spectrochemical Series:
CO is the strongest field ligand, followed by H₂O, F⁻, and S²⁻. This order is determined based on the ability of ligands to split d-orbitals in a crystal field.
Match List - I with List - II:
Choose the correct answer from the options given below:
The correct matching is based on the effects and reactions given in the lists:
The matching follows specific chemical properties or reaction types. Analyzing each option confirms the correct sequence as (A) – (IV), (B) – (III), (C) – (I), (D) – (II).
What will be the decreasing order of basic strength of the following conjugate bases?
OH⁻, RO⁻, CH₃COO⁻, Cl⁻
The basic strength of a conjugate base is inversely proportional to the strength of its conjugate acid:
RO⁻ is the strongest base among the given species, followed by OH⁻, CH₃COO⁻, and Cl⁻. This order is determined by their ability to donate electrons or accept protons.
In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to:
Ammonium chloride is added to reduce OH⁻ ion concentration via the common ion effect:
This allows selective precipitation of Group III cations by lowering the OH⁻ ion concentration and preventing the precipitation of other cations.
Identify the product in the following reaction:
The reaction involves the reduction of an aldehyde with NaBH₄ to form a primary alcohol.
NaBH₄ is a selective reducing agent that converts aldehydes into primary alcohols by donating hydride ions, without affecting other functional groups like esters or ketones.
One of the commonly used electrodes is the calomel electrode. Under which of the following categories does the calomel electrode come?
The calomel electrode consists of mercury in contact with mercurous chloride (an insoluble salt) and chloride ions.
This type of electrode maintains a stable reference potential and is widely used in electrochemical measurements. It falls under the Metal – Insoluble Salt – Anion electrode category.
Number of complexes from the following with even number of unpaired d-electrons is:
[V(H₂O)₆]³⁺, [Cr(H₂O)₆]²⁺, [Fe(H₂O)₆]³⁺, [Ni(H₂O)₆]³⁺, [Cu(H₂O)₆]²⁺
[Given atomic numbers: V = 23, Cr = 24, Fe = 26, Ni = 28, Cu = 29]
Analyzing the electronic configurations of the ions:
The configurations of the metal ions show that [Cr(H₂O)₆]²⁺ and [Fe(H₂O)₆]³⁺ have even numbers of unpaired d-electrons. All other complexes have odd numbers of unpaired electrons.
Which one of the following molecules has the maximum dipole moment?
The dipole moment is highest in NH₃ because of its trigonal pyramidal structure and lone pair on nitrogen.
In NH₃, the lone pair on nitrogen creates a significant dipole moment. In contrast, NF₃ has a lower dipole moment due to the opposing effect of highly electronegative fluorine atoms, and CH₄ is nonpolar.
Number of molecules/ions from the following in which the central atom is involved in sp³ hybridization is:
NO₃⁻, BCl₃, ClO₂⁻, ClO₃⁻
ClO₂⁻ and ClO₃⁻ have central atoms involved in sp³ hybridization, while NO₃⁻ and BCl₃ involve sp² hybridization.
The hybridization is determined based on the number of bonding and lone pairs on the central atom. ClO₂⁻ has 2 bonding pairs and 2 lone pairs, while ClO₃⁻ has 3 bonding pairs and 1 lone pair, both leading to sp³ hybridization.
Which among the following is an incorrect statement?
Solution:
Hydrogen ion (H⁺) actually shows a positive electromeric effect because it attracts electron density towards itself. The other statements about the electromeric effect are correct.
The electromeric effect refers to the complete transfer of π-electrons in a molecule in response to an external attacking reagent. Hydrogen ion (H⁺) acts as an electrophile, attracting electrons towards itself, resulting in a positive electromeric effect. Thus, the given statement is incorrect.
Given below are two statements:
Statement I: Acidity of α-hydrogens of aldehydes and ketones is responsible for Aldol reaction.
Statement II: Reaction between benzaldehyde and ethanol will NOT give Cross-Aldol product.
In the light of the above statements, choose the most appropriate answer from the options given below:
Solution:
Statement I is correct as aldehydes and ketones undergo the Aldol reaction due to the acidity of α-hydrogens. Statement II is incorrect because a cross-aldol product can form between benzaldehyde and ethanol (acetaldehyde), as acetaldehyde has α-hydrogens and can form enolate ions.
Aldol condensation involves the formation of a β-hydroxy aldehyde or ketone from aldehydes or ketones with at least one α-hydrogen. Benzaldehyde lacks α-hydrogens but can act as an electrophile in the reaction. Ethanol (acetaldehyde), having α-hydrogens, forms an enolate ion that reacts with benzaldehyde, resulting in a cross-aldol product. Hence, Statement II is incorrect.
Which of the following nitrogen-containing compounds does not give Lassaigne’s test?
Solution:
Hydrazine does not contain a carbon-nitrogen bond, so it does not give a positive Lassaigne’s test for nitrogen. The other compounds contain carbon-nitrogen bonds and will give a positive test.
Lassaigne’s test detects the presence of nitrogen by converting it into a water-soluble cyanide. This requires a C–N bond to form cyanide. Hydrazine (N₂H₄), lacking a carbon atom, cannot form cyanide, making the test negative. Phenyl hydrazine, glycine, and urea all have C–N bonds, giving positive results.
Which of the following is the correct structure of L-Glucose?
Solution:
The correct structure of L-Glucose is the mirror image of D-Glucose, with opposite configurations at each chiral center. The figure provides the accurate chemical structure of L-Glucose.
D-Glucose and L-Glucose are enantiomers, differing in the spatial arrangement of atoms around each chiral carbon. The mirror image (L-Glucose) reverses all chiral centers compared to D-Glucose. The provided structure correctly depicts this configuration.
The element which shows only one oxidation state other than its elemental form is:
Solution:
Scandium primarily exhibits the +3 oxidation state in its compounds and does not commonly show other oxidation states. In contrast, cobalt, titanium, and nickel can exhibit multiple oxidation states.
Scandium, a transition element, has a stable electron configuration of [Ar]3d¹4s². In chemical reactions, it loses all three valence electrons to form Sc³⁺. Other oxidation states are not commonly observed due to its electronic configuration and energy considerations.
Identify the product in the following reaction:
Reduction of an aldehyde with NaBH₄
Solution:
The reaction involves the reduction of an aldehyde with NaBH₄, leading to the formation of a primary alcohol. NaBH₄ is a selective reducing agent that reduces aldehydes to primary alcohols without affecting other functional groups.
Sodium borohydride (NaBH₄) selectively reduces aldehydes and ketones to their corresponding alcohols. In this case, the aldehyde carbonyl group (C=O) is converted to a hydroxyl group (-OH), producing a primary alcohol as the product.
Number of elements from the following that CANNOT form compounds with valencies which match their respective group valencies:
B, C, N, S, O, F, P, Al, Si
Solution:
The elements nitrogen, oxygen, and fluorine cannot expand their octet because they lack vacant d-orbitals, making them unable to match their group valency in certain compounds. The correct answer is 3 elements.
Elements like nitrogen, oxygen, and fluorine follow the octet rule strictly due to the absence of d-orbitals. This limits their ability to form compounds that require more than eight electrons in their valence shell. Other elements like phosphorus and sulfur can expand their valence shell using vacant d-orbitals.
The Molarity (M) of an aqueous solution containing 5.85 g of NaCl in 500 mL water is:
Solution:
First, calculate the moles of NaCl: moles = mass / Molar mass = 5.85 g / 58.5 g/mol = 0.1 mol. The molarity is given by M = moles / volume(L) = 0.1 mol / 0.5 L = 0.2 M.
Molarity (M) = moles of solute / volume of solution (in liters). Here, 5.85 g of NaCl corresponds to 0.1 mol, and the solution volume is 500 mL or 0.5 L. Substituting these values gives M = 0.1 / 0.5 = 0.2 M.
Identify the correct set of reagents or reaction conditions ‘X’ and ‘Y’ in the following set of transformation:
Solution:
This reaction is a typical example of halogenation of a carbonyl compound where the reaction conditions for X involve the elimination of a hydrogen atom to form an enolate, followed by halogenation using Br₂ in chloroform.
Enolate formation occurs under basic conditions (conc. alc. NaOH at 80°C). The enolate ion reacts with Br₂ to substitute one hydrogen atom with bromine, a common mechanism in halogenation of carbonyl compounds.
The correct order of first ionization enthalpy values of the following elements is:
(A) O, (B) N, (C) Be, (D) F, (E) B
Solution:
The order is based on the periodic trend. Ionization enthalpy increases across a period and decreases down a group. The exception in this order is between B and Be, and between N and O, due to the stability of half-filled and fully filled orbitals.
Ionization energy increases from left to right in a period due to increasing nuclear charge. However, beryllium (Be) and nitrogen (N) exhibit slightly higher values due to the stability of their fully filled and half-filled orbitals, respectively.
The enthalpy of formation of ethane (C₂H₆) from ethylene by addition of hydrogen where the bond energies of C–H, C–C, H–H are 414 kJ, 347 kJ, 615 kJ, and 435 kJ respectively is:
Solution:
The value is determined by calculating the enthalpy change for breaking and forming bonds during the hydrogenation reaction. The process is exothermic, releasing energy.
Using bond energies: ΔH = [Bonds broken] – [Bonds formed]. Breaking one C=C and one H–H requires 615 + 435 kJ, while forming two C–C and four C–H bonds releases energy. The net enthalpy change is –125 kJ.
The number of correct reactions among the following is:
Solution:
This is based on the balanced reactions provided in the options. One reaction fits the criteria for being correct under standard conditions.
Friedel-Crafts acylation is a reaction where an aromatic compound reacts with an acyl chloride in the presence of a Lewis acid catalyst (AlCl₃). The reaction is highly specific and reproducible under given conditions.
X g of ethylamine is subjected to reaction with NaNO₂/HCl followed by water; evolved dinitrogen gas occupied 2.24 L volume at STP. X is × 10⁻¹ g.
Solution:
The reaction of ethylamine with NaNO₂/HCl releases nitrogen gas. The volume of nitrogen gas evolved at STP (2.24 L) allows calculation of the moles of nitrogen produced, which can be used to find the mass of ethylamine (X).
From the gas law, 1 mole of gas occupies 22.4 L at STP. Hence, 2.24 L of nitrogen gas corresponds to 0.1 moles. Since ethylamine reacts in a 1:1 molar ratio to produce nitrogen, 0.1 moles of ethylamine were used. Molar mass of ethylamine = 45 g/mol. Thus, X = 0.1 × 45 = 4.5 g = 45 × 10⁻¹ g.
The de-Broglie wavelength of an electron in the 4th orbit is πa₀. (a₀ = Bohr’s radius)
Solution:
The de-Broglie wavelength of an electron is related to its momentum. For an electron in the nth orbit, the wavelength is proportional to the orbit number and Bohr's radius, yielding a value of 8πa₀ for the 4th orbit.
The wavelength is given by λ = h / p, where p = mv (momentum). In Bohr’s model, mv = nh / 2πr, so λ = 2πr / n. For the 4th orbit, r = 4a₀ (Bohr's radius), and thus λ = 8πa₀.
Only 2 mL of KMnO₄ solution of unknown molarity is required to reach the endpoint of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of KMnO₄ solution should be:
Solution:
This titration involves the reduction of MnO₄⁻ to Mn²⁺ and the oxidation of oxalic acid. Using the stoichiometry of the reaction, the molarity of KMnO₄ can be calculated based on the volume of oxalic acid and the amount of KMnO₄ used.
The balanced reaction is 2MnO₄⁻ + 5(COOH)₂ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. For 20 mL of 2 M oxalic acid, moles = 20 × 2 / 1000 = 0.04 mol. From stoichiometry, moles of KMnO₄ required = 0.04 × (2/5) = 0.016 mol. Molarity = moles / volume = 0.016 / 0.002 = 8 M.
Consider the following reaction: MnO₂ + KOH + O₂ → A + H₂O. Product ‘A’ in neutral or acidic medium disproportionates to give products ‘B’ and ‘C’ along with water. The sum of spin-only magnetic moment values of B and C is BM (nearest integer). (Given atomic number of Mn is 25)
Solution:
The reaction involves MnO₂ undergoing oxidation and disproportionation. The products B and C are Mn²⁺ and Mn⁴⁺ ions. The magnetic moment is calculated using the spin-only formula, taking into account the number of unpaired electrons in each ion.
Mn²⁺ has 5 unpaired electrons, contributing √35 BM (5.92 BM). Mn⁴⁺ has 3 unpaired electrons, contributing √15 BM (3.87 BM). The sum of the spin-only magnetic moments of B and C rounds to approximately 4 BM.
The number of different chain isomers for C₇H₁₆ is:
Solution:
The isomerism of alkanes is determined by the different ways the carbon atoms can be arranged. For C₇H₁₆, 9 distinct structural isomers are possible based on branching patterns.
Chain isomerism arises due to different arrangements of the carbon chain. For C₇H₁₆, the possible isomers include n-heptane, 2-methylhexane, 3-methylhexane, 2,2-dimethylpentane, 2,3-dimethylpentane, 3-ethylpentane, and others, totaling 9.
Number of molecules/species from the following having one unpaired electron is:
O₂, O₂⁻, NO, CN⁻, O₂²⁻
Solution:
O₂ and NO are the species with one unpaired electron. O₂⁻, CN⁻, and O₂²⁻ have either paired electrons or no unpaired electrons in their molecular orbitals.
Using molecular orbital theory: O₂ has one unpaired electron in its antibonding π* orbital. NO has an unpaired electron in a bonding orbital. O₂⁻, CN⁻, and O₂²⁻ have paired electrons, resulting in no unpaired electrons.
The number of different chain isomers for C₇H₁₆ is:
Solution:
The isomerism of alkanes is determined by the different ways the carbon atoms can be arranged. For C₇H₁₆, 9 distinct structural isomers are possible based on branching patterns.
Chain isomers arise due to different arrangements of the carbon backbone. The 9 chain isomers of C₇H₁₆ include:
Number of molecules/species from the following having one unpaired electron is:
O₂, O₂⁻, NO, CN⁻, O₂²⁻
Solution:
O₂ and NO are the species with one unpaired electron. O₂⁻, CN⁻, and O₂²⁻ have paired electrons or no unpaired electrons in their molecular orbitals.
Using molecular orbital theory:
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