
JEE Main 2024 Apr 5 Shift 2 Question Paper with Solution pdf is available for download here. Students found Physics easy and Mathematics hard. Chemistry carried the highest weightage and overall difficulty level was moderate.
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Let f : [−1, 2] → R be given by f(x) = 2x² + x + ⌊x²⌋ − ⌊x⌋, where ⌊t⌋ denotes the greatest integer less than or equal to t. The number of points where f is not continuous is:
The function involves floor terms ⌊x²⌋ and ⌊x⌋, which are discontinuous at integer values of x. For x ∈ [−1, 2], these discontinuities occur at x = −1, 0, 1, and 2. Therefore, f(x) is discontinuous at 4 points.
The floor function ⌊t⌋ creates jumps at integer points. When combined with x² and x in the given range, the points where discontinuities occur are directly associated with −1, 0, 1, and 2. Evaluating these points confirms the behavior of f(x) at each discontinuity.
The differential equation of the family of circles passing the origin and having the center on the line y = x is:
The general equation for a circle with the center on y = x passing through the origin is:
(x − h)² + (y − h)² = r², where h is the parameter. Differentiating and simplifying yields:
(x² − y² − 2xy)dx = (x² − y² + 2xy)dy.
To derive the differential equation, substitute the center coordinates (h, h) into the circle equation. Expanding and differentiating with respect to x and y produces the desired relationship. Verification of terms ensures consistency with the differential equation provided.
Let S1 = {z ∈ ℂ : |z| ≤ 5},
S2 = {z ∈ ℂ : Im((z + 1)/(z − 1)) ≥ 0},
S3 = {z ∈ ℂ : Re(z) ≥ 0}.
The area common to S1, S2, and S3 is:
The area common to S1, S2, and S3 is the intersection of a semicircle (S1), a region in the complex plane defined by S2, and the right half-plane (S3). Solving the geometry yields the area as (125/12)π.
Each region defines a geometric constraint. S1 is a circle with radius 5, S2 describes the upper half-plane with a specific boundary condition, and S3 restricts the region to the right of the imaginary axis. Their intersection is calculated geometrically.
The area enclosed between the curves y = x|x| and y = x − |x| is:
The area is computed by integrating the difference between the given curves over their intersection interval. For y = x|x| and y = x − |x|, solving the integration gives an enclosed area of 4/3.
The two curves intersect at x = 0 and x = 1. Integrating the absolute differences of the two functions over this interval provides the enclosed area.
60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50th word is:
To find the 50th word in dictionary order, arrange the letters of "BHBJO" alphabetically: B, B, H, J, O. Count the permutations for each starting letter until reaching the 50th word. The 50th word is "OBBJH".
Breaking the arrangement into groups based on the first letter and computing the number of permutations for each case helps identify the 50th word.
Let a = 2î + 5ĵ − k̂, b = 2î − 2ĵ + 2k̂, and c be three vectors such that:
(c + î) × (a + b + î) = a × (c + î) and a · c = −29.
Then c · (−2î + ĵ + k̂) is equal to:
Given the condition (c + î) × (a + b + î) = a × (c + î), simplify the cross product by substituting a + b + î = 5î + 3ĵ + k̂.
Additionally, using a · c = −29, solve for components of c. Finally, calculate c · (−2î + ĵ + k̂), resulting in 5.
Substitute given vectors and solve for c using dot product and cross product identities. Verify the solution with the constraints.
Consider three vectors a, b, c. Let |a| = 2, |b| = 3, and a = b × c. If α ∈ (0, π/3) is the angle between the vectors b and c, then the minimum value of 27|c − a|² is equal to:
Using the relation ⃗a = ⃗b × ⃗c and |⃗a| = |⃗b||⃗c|sinα, compute the minimum value of |⃗c − ⃗a|² by substituting values and minimizing the expression based on given conditions. The result is 124.
The cross product determines a perpendicular vector, and the magnitude is given by |b||c|sinα. Minimizing the squared distance involves substituting |b| = 3, |a| = 2, and simplifying |c − a|².
Let A(−1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of ∆PAB is 10. If the locus of P is ax + by = 15, then 5a + 2b is:
The area of ∆PAB gives the distance of P from line AB. Solving for the locus of P and equating to the given area constraint results in the equation ax + by = 15, with 5a + 2b = −12/5.
The area of a triangle formula provides the perpendicular distance of P from AB. Using this distance and the equation of the line AB, solve for the coefficients a and b in the locus equation.
Let (α, β, γ) be the point (8, 5, 7) on the line (x−1)/2 = (y+1)/3 = (z−2)/5. Then α + β + γ is equal to:
Substituting the parametric equations of the line, solve for α, β, γ. Summing these coordinates gives α + β + γ = 14.
Using the parametric form of the line equations, substitute values to find the coordinates of the point (α, β, γ). Add these values to verify the result.
If the constant term in the expansion of √(3/5)x + 2x/√(3)5)^12, x ≠ 0, is αx² × √(3/5), then 25α is equal to:
Using the binomial expansion formula, identify the constant term by equating the powers of x. Compute α and then 25α, resulting in 693.
Expand the given expression using the binomial theorem. Match the term where the exponent of x becomes zero, calculate α, and multiply by 25 to find the final value.
Let f, g : R → R be defined as:
f(x) = |x − 1| and
g(x) = ex, x ≥ 0 x + 1, x ≤ 0
Then the function f(g(x)) is:
For f(g(x)), the function involves the composition of g(x) with f(x). Since g(x) is not one-one across its domain (as it includes two distinct pieces: exponential and linear), and f(x) applied to g(x) does not cover all possible outputs, the resulting function is neither one-one nor onto.
Analyzing g(x), it is clear that the exponential part and the linear part overlap at x = 0. Applying the absolute value function f(x) further reduces the range of the composition.
Let the circle C1: x2 + y2 − 2(x + y) + 1 = 0 and C2 be a circle having its centre at (−1, 0) and radius 2. If the line of the common chord of C1 and C2 intersects the y-axis at the point P, then the square of the distance of P from the centre of C1 is:
The equation of the common chord is obtained by subtracting the equations of C1 and C2. The intersection of this line with the y-axis gives point P. Calculating the square of the distance of P from the center of C1 (1, 1) results in 2.
Subtract the equations of the two circles to derive the common chord. Substitute x = 0 (y-axis condition) to find P, and then compute the distance from P to the center of C1.
Let the set S = {2, 4, 8, 16, . . . , 512} be partitioned into 3 sets A, B, C with an equal number of elements such that A∪B∪C = S and A∩B = B∩C = A∩C = ∅. The maximum number of such possible partitions of S is equal to:
The set S contains 9 elements, and each of the sets A, B, and C must contain 3 elements. The number of ways to partition these elements into three distinct sets with no intersection is calculated as the number of combinations, yielding 1680 possible partitions.
The total number of elements in S is divided equally among A, B, and C. The number of ways to arrange them is derived using combinatorial methods, ensuring all sets are distinct.
The values of m, n for which the system of equations
x + y + z = 4,
2x + 5y + 5z = 17,
x + 2y + mz = n
has infinitely many solutions, satisfy the equation:
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. Solving for m and n based on this condition results in the equation m2 + n2 − mn = 39.
Calculate the determinant of the coefficient matrix formed by the given equations. Set the determinant to zero and solve for m and n to derive the relationship.
The coefficients a, b, c in the quadratic equation ax2 + bx + c = 0 are from the set {1, 2, 3, 4, 5, 6}. If the probability of this equation having one real root bigger than the other is p, then 216p equals:
The equation will have one real root bigger than the other if the discriminant (Δ) is positive and the roots are real and distinct. The total possible combinations of a, b, and c are 6 × 6 × 6 = 216. After calculating the favorable outcomes, we find that 216p = 38.
Calculate the discriminant and evaluate for positive values to ensure real and distinct roots. Use the total and favorable cases to find the probability p.
Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies:
By using the geometry of the square and properties of tangents, we can derive the equation of the radius r of the circle passing through point F and touching the sides BC and CD. This leads to the equation r2 − 8r + 8 = 0.
The geometry of the square helps in positioning the circle. Applying tangent properties and the Pythagoras theorem leads to the derivation of the quadratic equation for r.
Let β(m, n) = ∫01 xm−1(1 − x)n−1 dx, m, n > 0. If
∫01 (1 − x10)20 dx = a × β(b, c),
then 100(a + b + c) equals:
The given integral can be expressed in terms of the Beta function. By comparing the given integral with the standard form of the Beta function, we find the values of a, b, and c. The value of 100(a + b + c) equals 2120.
Expand the integral using the binomial theorem and compare it with the Beta function definition. Solve for the parameters a, b, and c, and compute the final value.
Let αβ ≠ 0 and A =
[ β α 3 ] [ α α β ] [ −β α 2α ]
If B =
[ 3α −9 3α ] [ −α 7 −2α ] [ −2α 5 −2β ]
is the matrix of cofactors of the elements of A, then det(AB) is equal to:
Using the property det(AB) = det(A) × det(B), and knowing the matrix of cofactors B, we can compute det(AB). The value of det(AB) results in 216.
Calculate det(A) and det(B) separately using cofactor expansion and matrix properties. Multiply the two determinants to find det(AB).
If
y(θ) = (2 cos θ + cos 2θ − 1) / (4 cos³ θ + 8 cos² θ + 5 cos θ + 2),
then at θ = π/2, y'' + y' + y is equal to:
The given function is differentiated twice to compute y' and y''. Substituting θ = π/2 into the derivatives and the original function, we find that y'' + y' + y equals 2 at this value of θ.
Differentiate the function step-by-step and evaluate at θ = π/2. Combine the results for y'', y', and y to get the final value.
For x ≥ 0, the least value of K for which
41+x, 41−x, K2, 16x + 16−x
are three consecutive terms of an arithmetic progression (A.P.) is equal to:
For the terms to be in A.P., the middle term must be the average of the first and third terms. Solving for K from the arithmetic progression condition gives K = 10 as the least value.
Set up the A.P. condition for the given terms, substitute expressions, and solve for K using algebraic manipulation. Simplify to find the minimum value of K.
Let the mean and the standard deviation of the probability distribution given by:
X: 1, 0, −3
P(X): 1/3, K, 1/6, 1
be µ and σ, respectively. If σ − µ = 2, then σ + µ is equal to:
By using the properties of probability distributions and the given condition that σ − µ = 2, we can calculate the values of σ and µ. The sum σ + µ is found to be 5.
Compute the mean (µ) as the expected value and the standard deviation (σ) using the variance formula. Solve the system of equations σ − µ = 2 to find σ + µ.
Let y = y(x) be the solution of the differential equation:
(dy/dx) + (2x / (1 + x2)) y = (x * e) / (1 + x2), y(0) = 0.
Then the area enclosed by the curve f(x) = y(x)e1/(1+x²) and the line y − x = 4 is equal to:
The solution to the differential equation is obtained using integration methods, and the area enclosed by the curve f(x) and the line y − x = 4 is computed as 18.
Apply an integrating factor to solve the linear differential equation. Then use definite integration to calculate the enclosed area.
The number of solutions of:
sin2x + (2 + 2x − x2) sin x − 3(x − 1)2 = 0,
where −π ≤ x ≤ π, is:
Solving the trigonometric equation and analyzing the roots within the given range −π ≤ x ≤ π, we find that there are 2 solutions.
Break the equation into cases based on the trigonometric function values. Solve for x and verify each root lies within the given interval.
Let the point (−1, α, β) lie on the line of the shortest distance between the lines:
(x + 2) / −3 = (y − 2) / 4 = (z − 5) / 2
and
(x + 2) / −1 = (y + 6) / 2 = (z − 1) / 0
Then (α − β)2 is equal to:
Using the shortest distance formula between skew lines and solving for the coordinates (α, β), we find that (α − β)2 = 25.
Find the line of shortest distance using vector equations. Substitute to solve for α and β and calculate (α − β)2.
If
(1 + √3 − √2 / 2) (√3 + 5 − 2√6 / 18) + (9√3 − 11√2 / 36√3) + (49 − 20√6 / 180) + . . . up to ∞ = 2 q (b / a + 1) log(e)
Then 11a + 18b is equal to:
By simplifying the given infinite series, the values of a and b are determined to be 5 and 7 respectively, and 11a + 18b equals 76.
Use series expansion and factorization to simplify the terms. Solve for a and b and compute 11a + 18b.
Let a > 0 be a root of the equation 2x2 + x − 2 = 0.
If limx→1/a (16 (1 − cos(2 + x − 2x2)) / (1 − ax2)) = α + β √17, where α, β ∈ Z, then α + β is equal to:
The quadratic equation 2x2 + x − 2 = 0 gives roots a = (−1 + √17) / 4. Simplifying the given limit using trigonometric identities and substitution at x = 1/a, we find that α = 153 and β = 17. Therefore, α + β = 170.
Find the roots of the quadratic equation, then substitute x = 1/a into the limit. Use trigonometric approximations and simplifications to find α and β.
If f(t) = ∫0π (2x / (1 − cos2t sin2x)) dx, 0 < t < π, then the value of ∫0π/2 π2 dt / f(t) is equal to:
The integral f(t) simplifies using trigonometric identities and substitution. Solving the second integral with the evaluated form of f(t) yields the final value as 1.
Evaluate f(t) by using standard trigonometric integrals. Substitute this into the second integral and simplify to find the solution.
Let the maximum and minimum values of (√(8x − x2) − 12 − 4)2 + (x − 7)2, x ∈ R, be M and m, respectively. Then M2 − m2 is equal to:
By analyzing the quadratic expression and solving for the maximum and minimum values, we determine that M = 49 and m = 9. Thus, M2 − m2 = 1600.
Find the critical points of the expression and calculate M and m. Use the difference of squares formula to find M2 − m2.
Let a line perpendicular to the line 2x − y = 10 touch the parabola y2 = 4(x − 9) at the point P. The distance of the point P from the center of the circle x2 + y2 − 14x − 8y + 56 = 0 is equal to:
The point P is calculated by solving the tangent equation to the parabola and the circle’s center coordinates. Using the distance formula, the distance between P and the circle’s center is found to be 10.
Determine the equation of the line perpendicular to 2x − y = 10. Find the point of tangency on the parabola and calculate the distance to the circle's center.
The number of real solutions of the equation x|x + 5| + 2|x + 7| − 2 = 0 is:
Solving the absolute value equation by breaking it into cases based on the signs of the terms, we find three distinct solutions within the defined domain.
Split the equation into different cases for x based on the critical points −5 and −7. Solve each case separately to find all valid solutions.
Given below are two statements:
Statement I: When white light passes through a prism, red light bends less than yellow and violet.
Statement II: The refractive indices are different for different wavelengths in dispersive media.
In the light of the above statements, choose the correct answer:
White light dispersion through a prism occurs because refractive indices vary with wavelength. Red light, having the longest wavelength, bends the least, while violet bends the most. Both statements are correct and consistent with this phenomenon.
Dispersion happens because refractive indices depend on the wavelength of light, which explains the varying degrees of bending for different colors in a prism.
Which of the following statements is NOT true about the stopping potential (V0)?
The stopping potential depends on the frequency of light and the material’s work function but not on light intensity. Light intensity only affects the number of photoelectrons emitted, not their energy or the stopping potential.
According to the photoelectric equation, stopping potential is related to the frequency of the incident light and the work function, independent of light intensity.
The angular momentum of an electron in a hydrogen atom is proportional to (where r is the radius of the orbit):
According to Bohr’s model, angular momentum (L) is proportional to the quantum number n, and the orbit radius (r) is proportional to n². Hence, L is proportional to √r.
The relationship between radius and angular momentum comes from combining Bohr’s quantization condition and the centripetal force equation.
A galvanometer of resistance 100Ω is connected in series with a 400Ω resistor to measure up to 10V. The value of resistance required to convert the galvanometer into an ammeter to read up to 10A is x × 10-2Ω. The value of x is:
To convert the galvanometer into an ammeter, a shunt resistance is calculated as S = igRg / (I - ig). Substituting the given values, S is found to be 0.2Ω, or 20 × 10-2Ω, giving x = 20.
Use the shunt resistance formula with ig = 10mA and I = 10A to find the required value of S. Simplify to express it as x × 10-2Ω.
The vehicles carrying inflammable fluids usually have metallic chains touching the ground:
Metallic chains are used to discharge static electricity generated due to friction with air while moving, preventing sparking that could ignite inflammable fluids.
The chains provide a grounding mechanism to dissipate static charges safely, avoiding hazards related to inflammable materials.
If n is the number density and d is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e., mean free path) is represented by:
The mean free path (λ) of a gas molecule is derived from kinetic theory and is given by λ = 1/√(2πd2n), where n is the number density and d is the molecular diameter.
Mean free path is calculated considering the collision cross-section and the number of molecules in a given volume. The formula is derived from statistical mechanics.
A particle moves in the x-y plane under the influence of a force F such that its linear momentum is P⃗(t) = î cos(kt) − ĵ sin(kt). If k is constant, the angle between F⃗ and P⃗ will be:
Differentiating P⃗(t) to find the force F⃗ shows that F⃗ is perpendicular to P⃗ because their dot product is zero. Hence, the angle between F⃗ and P⃗ is π/2.
Force F⃗ is the derivative of momentum P⃗ with respect to time. Compute the dot product F⃗ · P⃗ and observe that it equals zero, indicating perpendicularity.
The electrostatic force (F1) and magnetic force (F2) acting on a charge q moving with velocity v can be written as:
The electrostatic force is F1 = qE, and the magnetic force is F2 = q(v × B), as per the Lorentz force law.
According to the Lorentz force equation, the total force on a charged particle is F⃗ = q(E⃗ + v⃗ × B⃗). Separate the electric and magnetic components to identify F1 and F2.
A man carrying a monkey on his shoulder cycles smoothly on a circular track of radius 9m and completes 120 revolutions in 3 minutes. The magnitude of the centripetal acceleration of the monkey is (in m/s2):
The angular velocity is ω = 4π/3 rad/s. Using ac = ω2R, the centripetal acceleration is ac = 16π2 m/s2.
Calculate the angular velocity ω from the number of revolutions and time. Substitute ω and R into the centripetal acceleration formula to find the result.
A series LCR circuit is subjected to an AC signal of 200V, 50Hz. If the voltage across the inductor (L = 10mH) is 31.4V, then the current in this circuit is:
The inductive reactance is XL = 2πfL = 3.14Ω. Using V = IXL, the current is I = 31.4/3.14 = 10 A.
Find the inductive reactance XL using the formula XL = 2πfL. Use Ohm's law to calculate the current: I = V/XL.
What is the dimensional formula of ab-1 in the equation:
(P + a/V2)(V − b) = RT,
where letters have their usual meaning?
Using the dimensional formula of pressure ([ML-1T-2]) and volume ([L3]), the formula for ab-1 simplifies to [ML2T-2].
Dimensional analysis of the equation relates pressure and volume terms. Substituting their dimensions and simplifying leads to the result.
The output (Y) of the logic circuit given below is 0 only when:
Analyzing the logic gates, the output Y is 0 only when both inputs A and B are 0, as the OR gate and AND gate conditions are not satisfied simultaneously.
Examine the truth table for the given logic circuit. The combination where both inputs are 0 ensures the output is 0.
A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to:
For a body moving under constant power, velocity is proportional to √t, and displacement (integral of velocity) is proportional to t3/2.
Constant power implies force varies inversely with velocity. Integrating velocity over time gives displacement proportional to t3/2.
Match List-I with List-II:
List-I (EM-Wave) List-II (Wavelength Range)
(A) Infra-red (I) < 10-3 nm
(B) Ultraviolet (II) 400 nm to 1 nm
(C) X-rays (III) 1 mm to 700 nm
(D) Gamma rays (IV) 1 nm to 10-3 nm
Choose the correct option:
Infra-red corresponds to 1 mm to 700 nm, ultraviolet corresponds to 400 nm to 1 nm, X-rays correspond to 1 nm to 10-3 nm, and gamma rays correspond to wavelengths less than 10-3 nm.
Electromagnetic waves are categorized by their wavelengths. Refer to their standard ranges to match them correctly.
During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of CP/CV for the gas is:
For an adiabatic process, P ∝ T3. Using the relation P ∝ V-γ, γ = 7/5 is obtained, where γ = CP/CV.
Apply the ideal gas law and the adiabatic condition to derive the proportionality and solve for γ = CP/CV.
Choose the answer from the options given below:
The matching between List-I and List-II is as follows:
(A) A force that restores an elastic body of unit area to its original state corresponds to Stress (III).
(B) Two equal and opposite forces parallel to opposite faces correspond to Shear modulus (IV).
(C) Forces perpendicular everywhere to the surface per unit area correspond to Bulk modulus (I).
(D) Two equal and opposite forces perpendicular to opposite faces correspond to Young’s modulus (II).
Analyze the definitions and roles of different mechanical properties to match them correctly with their descriptions.
A vernier calipers has 20 divisions on the vernier scale, which coincides with the 19th division on the main scale. The least count of the instrument is 0.1mm. One main scale division is equal to:
From the given data, 20 vernier scale divisions coincide with 19 main scale divisions. The least count is given as 1 main scale division minus 1 vernier scale division. Substituting the values, one main scale division is calculated to be 2mm.
The least count formula is LC = 1 MSD - 1 VSD. Solving for MSD using the relationship between MSD and VSD yields 2mm.
A heavy box of mass 50kg is moving on a horizontal surface. If the coefficient of kinetic friction between the box and the surface is 0.3, then the force of kinetic friction is:
The normal force acting on the box is equal to its weight, calculated as 50 × 9.8 = 490N. The force of kinetic friction is given by µk × N = 0.3 × 490 = 147N.
Apply the frictional force formula Fk = µk × N, where N is the normal force. Substituting the values gives Fk = 147N.
A satellite revolving around a planet in a stationary orbit has a time period of 6 hours. The mass of the planet is one-fourth the mass of Earth. The radius of the orbit of the planet is:
Using Kepler’s third law and the given ratio of the masses and time periods, the radius of the orbit is calculated as 1.05×10⁴ km, using the proportional relationship r³ ∝ T²/M.
Kepler’s third law is T² ∝ r³/M. Substituting the given values for T and M and solving for r gives the orbit radius as 1.05×10⁴ km.
The ratio of heat dissipated per second through the resistances 5Ω and 10Ω in the circuit given below is:
The 5Ω and 10Ω resistors are connected in parallel. The current through each resistor is inversely proportional to its resistance. The power dissipated in a resistor is proportional to the square of the current and the resistance. This gives a ratio of heat dissipation of 2:1.
Using P = I²R and the inverse proportionality of current to resistance in parallel circuits, calculate the power ratio between the two resistors.
A solenoid of length 0.5m has a radius of 1 cm and is made up of m number of turns. It carries a current of 5A. If the magnitude of the magnetic field inside the solenoid is 6.28×10⁻³T, then the value of m is:
The magnetic field inside a solenoid is given by B = µ₀ni, where n is the number of turns per unit length. Rearranging to find m, we use the given values of B, µ₀, and i to calculate m as 500.
Use the formula n = m/L and B = µ₀ni. Substituting the given parameters simplifies to m = 500.
The shortest wavelength of the spectral lines in the Lyman series of the hydrogen spectrum is 915 Å. The longest wavelength of spectral lines in the Balmer series will be:
The longest wavelength in the Balmer series corresponds to the transition from n=3 to n=2. Using the formula for energy levels and the given Lyman series data, the wavelength is calculated as 6588 Å.
Use the Rydberg formula for the wavelength of spectral lines and solve for the longest wavelength in the Balmer series.
In a single-slit experiment, a parallel beam of green light of wavelength 550 nm passes through a slit of width 0.20 mm. The transmitted light is collected on a screen 100 cm away. The distance of the first-order minima from the central maximum will be x×10⁻⁵ m. The value of x is:
The distance of the first-order minima is given by y = λD/d. Substituting the given values, the distance is calculated as 275×10⁻⁵ m, where x=275.
Calculate the position of the minima using y = λD/d with given parameters for wavelength (λ), distance (D), and slit width (d).
A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with a fundamental frequency of 600 Hz under the same tension is:
The length of the sonometer wire is inversely proportional to the frequency. Using the ratio L₁/L₂ = f₂/f₁, the resonating length for 600 Hz is calculated as 60 cm.
Apply the inverse relationship between length and frequency to find the new resonating length.
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is x/5. The value of x is:
The total kinetic energy is the sum of rotational and translational kinetic energy. For a hollow sphere, the ratio of rotational to total kinetic energy is 2/5, giving x=2.
Calculate the ratio using the moment of inertia of a hollow sphere and the relationship between rotational and translational kinetic energies.
A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of 10 N is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is:
Using Pascal's law, the pressures in both arms are equal. The force is proportional to the cross-sectional area of each arm. Given the ratio of diameters, the force on the thicker arm is calculated to be 1000 N.
Apply Pascal’s principle: Pressure = Force/Area. Use the area ratio from diameters to find the force.
The electric field at point P due to an electric dipole is E. The electric field at point R on the equatorial line will be E/x. The value of x is:
For an electric dipole, the field on the axial line is twice as strong as the field on the equatorial line at the same distance. The ratio E_axial/E_equatorial gives x = 16.
Electric field along the axial line is proportional to 2p/r³, while the equatorial line field is proportional to p/r³. Taking the ratio gives x = 16.
The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is:
The maximum height of a projectile is proportional to the square of the initial velocity. Halving the velocity reduces the height to one-fourth. Thus, the new height is 64/4 = 16 m.
Maximum height H = (v₀²sin²θ)/(2g). Reducing v₀ to v₀/2 decreases H to (1/4)H.
A wire of resistance 20 Ω is divided into 10 equal parts. A combination of two parts is connected in parallel, and so on. Now the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is:
Each part of the wire has a resistance of 2 Ω. When two parts are connected in parallel, the equivalent resistance is 1 Ω. Five such pairs connected in series result in a total resistance of 5 Ω.
Resistance of each part = 20/10 = 2 Ω. Parallel combination of two parts gives R_parallel = 1 Ω. Total resistance = 5 × 1 = 5 Ω.
The current in an inductor is given by I = (3t + 8), where t is in seconds. The magnitude of the induced emf produced in the inductor is 12 mV. The self-inductance of the inductor is:
The induced emf is given by |ε| = L (dI/dt). Differentiating I = 3t + 8 gives dI/dt = 3. Substituting |ε| = 12×10⁻³ V, L is calculated as 4 mH.
Using |ε| = L(dI/dt), calculate L = ε/dI/dt. Substituting values gives L = 4 mH.
Match List-I with List-II:
List-I:
(A) ICl
(B) ICl₃
(C) ClF₅
(D) IF₇
List-II:
(I) T-Shape
(II) Square pyramidal
(III) Pentagonal bipyramidal
(IV) Linear
The molecular geometry of the compounds is determined by VSEPR theory:
- (A) ICl: Linear (IV).
- (B) ICl₃: T-Shape (I).
- (C) ClF₅: Square pyramidal (II).
- (D) IF₇: Pentagonal bipyramidal (III).
Geometry is based on the steric number and lone pairs around the central atom.
While preparing crystals of Mohr’s salt, dilute H₂SO₄ is added to a mixture of ferrous sulfate and ammonium sulfate. Before dissolving this mixture in water, dilute H₂SO₄ is added here to:
Dilute H₂SO₄ is added to prevent the hydrolysis of ferrous sulfate. Without the acidic medium, ferrous sulfate would hydrolyze to form ferric hydroxide, which interferes with crystal formation.
The acidic medium ensures the stability of Fe²⁺ ions, preventing oxidation or hydrolysis.
Identify the major product in the following reaction:
The reaction involves elimination via the E2 mechanism, where the base removes a β-hydrogen and the leaving group departs, forming a double bond. The major product follows Zaitsev’s rule, favoring the more substituted alkene: cyclopentene.
Zaitsev’s rule predicts that elimination will form the more stable, highly substituted alkene.
The nomenclature for the following compound is:
The parent chain has 7 carbons with a double bond at position 6. Functional groups include a formyl group at position 2, a hydroxyl group at position 4, and a carboxylic acid at the terminal position. The correct name is 2-Formyl-4-hydroxyhept-6-enoic acid.
Follow IUPAC rules to assign positions and prioritize functional groups in naming.
Given below are two statements:
Assertion (A): NH₃ and NF₃ molecules have a pyramidal shape with a lone pair of electrons on the nitrogen atom. The resultant dipole moment of NH₃ is greater than that of NF₃.
Reason (R): In NH₃, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the N-H bonds. F is the most electronegative element.
NH₃ has a higher dipole moment because the lone pair dipole aligns with the N-H bond dipoles. In NF₃, the lone pair dipole opposes the N-F bond dipoles. Both statements are true, and the reason explains the assertion.
Analyze the direction and relative magnitude of bond and lone pair dipoles in NH₃ and NF₃.
Given below are two statements:
Statement I: On passing HCl(g) through a saturated solution of BaCl₂ at room temperature, white turbidity appears.
Statement II: When HCl(g) is passed through a saturated solution of NaCl, sodium chloride is precipitated due to the common ion effect.
HCl(g) increases Cl⁻ concentration, reducing the solubility of BaCl₂ and forming white turbidity due to the common ion effect. However, NaCl is highly soluble, and its solubility is not significantly affected by HCl. Hence, Statement I is correct, but Statement II is incorrect.
The solubility product (Ksp) of BaCl₂ decreases in the presence of additional Cl⁻ ions from HCl.
The metal atom present in the complex MABXL (where A, B, X, and L are unidentate ligands and M is a metal) involves sp³ hybridization. The number of geometrical isomers exhibited by the complex is:
In sp³ hybridization, the complex adopts a tetrahedral geometry. Geometrical isomerism is not possible in tetrahedral complexes because all positions are equivalent. Thus, the number of geometrical isomers is 0.
Tetrahedral complexes lack cis and trans arrangements, as all ligands are symmetrically distributed.
Match List-I with List-II:
List-I (Pair of Compounds):
(A) n-Propanol and isopropanol
(B) Methoxypropane and ethoxyethane
(C) Propanone and propanal
(D) Neopentane and isopentane
List-II (Type of Isomerism):
(I) Metamerism
(II) Chain Isomerism
(III) Position Isomerism
(IV) Functional Isomerism
- (A) n-Propanol and isopropanol exhibit functional isomerism (IV).
- (B) Methoxypropane and ethoxyethane exhibit metamerism (I).
- (C) Propanone and propanal exhibit position isomerism (III).
- (D) Neopentane and isopentane exhibit chain isomerism (II).
The type of isomerism is determined by differences in structure, functional groups, or connectivity of atoms.
The quantity of silver deposited when one coulomb of charge is passed through AgNO₃ solution is:
The amount of a substance deposited during electrolysis is proportional to its electrochemical equivalent. For one coulomb of charge, the quantity of silver deposited is equal to its electrochemical equivalent.
Use Faraday’s first law of electrolysis: m = ZQ, where Z is the electrochemical equivalent.
Which one of the following reactions is NOT possible?
Phenol does not react with HCl to form chlorobenzene because the −OH group is strongly bonded to the benzene ring. Thus, Reaction (2) is not possible, while the other reactions are feasible under appropriate conditions.
The −OH group in phenol is not replaced by Cl⁻ under normal acidic conditions due to resonance stabilization.
Given below are two statements:
Statement I: The metallic radius of Na is 1.86 Å, and the ionic radius of Na⁺ is lesser than 1.86 Å.
Statement II: Ions are always smaller in size than the corresponding elements.
The metallic radius of Na is 1.86 Å, and its ionic radius decreases due to the loss of an electron, which reduces electron-electron repulsion. Statement II is false because anions (e.g., Cl⁻) are larger than their corresponding neutral atoms due to increased electron repulsion.
For cations, the size decreases due to fewer electrons, while for anions, additional electrons increase repulsion and size.
Consider the above reaction sequence and identify the major product P:
The reaction sequence involves oxidation of ethanol to acetic acid, further oxidation to carbon dioxide, and subsequent decarboxylation with soda lime to yield methane as the final product. Thus, the major product is methane (CH₄).
Decarboxylation is a reaction that removes a carboxyl group, releasing carbon dioxide.
Consider the given chemical reaction. Product 'A' is:
Cyclohexane undergoes oxidation in the presence of KMnO₄ and H₂SO₄, resulting in the cleavage of CH₂ groups to carboxylic acid groups. This leads to the formation of adipic acid (HOOC-(CH₂)₄-COOH).
Adipic acid is a dicarboxylic acid commonly produced through the oxidation of cyclohexane derivatives.
For the electrochemical cell M—M²⁺||X—X²⁻, if E°(M²⁺/M) = 0.46V and E°(X/X²⁻) = 0.34V, which of the following is correct?
The standard cell potential E°cell is calculated as E°cathode - E°anode. Here, E°cell = 0.34V - 0.46V = -0.12V. Since E°cell is negative, the reverse reaction M²⁺ + X²⁻ → M + X is spontaneous.
A negative E°cell indicates that the reverse reaction is thermodynamically favored.
The number of moles of methane required to produce 11 g of CO₂ after complete combustion is:
From the combustion reaction CH₄ + 2O₂ → CO₂ + 2H₂O, one mole of methane produces one mole of CO₂. Given the molar mass of CO₂ is 44 g/mol, 11 g corresponds to 0.25 moles of CO₂. Therefore, 0.25 moles of methane are required.
The stoichiometric ratio between CH₄ and CO₂ in combustion is 1:1.
The number of complexes from the following with no electrons in the t2 orbital is:
TiCl4, [MnO4]-, [FeO4]2-, [FeCl4]-, [CoCl4]2-
The number of complexes with no electrons in the t2 orbital is 3: TiCl4, [MnO4]-, and [FeO4]2-. Each has a 3d0 electronic configuration.
Complexes with no electrons in the t2 orbital have their d-electrons fully paired or absent in the orbital.
The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is:
Ti2+, Cr2+, V2+
All three ions (Ti2+, Cr2+, and V2+) are strong reducing agents capable of reacting with H+ ions to liberate H2 gas. Hence, the count is 3.
Reducing agents donate electrons to H+, producing H2.
Identify A and B in the given chemical reaction sequence:
Friedel-Crafts acylation followed by Clemmensen reduction
Friedel-Crafts acylation forms acetophenone (A), which undergoes Clemmensen reduction to produce ethylbenzene (B).
Friedel-Crafts acylation introduces an acyl group, and Clemmensen reduction converts the carbonyl group to a CH2 group.
The correct decreasing order of atomic radii of Group 13 elements is:
The correct order is Tl > In > Ga > Al > B, due to increasing nuclear charge and varying shielding effects down the group.
The atomic radii generally increase down a group, but poor shielding by d and f electrons causes deviations in the trend.
The number of ways the set S = {2, 4, 8, ..., 512} can be partitioned into three subsets of equal size is:
The set has 9 elements. To partition it into three subsets of equal size, use the formula for combinations and calculate the number of ways as 1680.
The partitioning involves distributing the elements into subsets such that no two subsets overlap, and each subset has an equal number of elements.
Combustion of 1 mole of benzene is expressed as:
C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)
The standard enthalpy of combustion of 2 moles of benzene is -x kJ. Calculate the value of x given the following data:
- Standard enthalpy of formation of C₆H₆(l): 48.5 kJ/mol
- Standard enthalpy of formation of CO₂(g): -393.5 kJ/mol
- Standard enthalpy of formation of H₂O(l): -286 kJ/mol
Using Hess's Law:
ΔH = ΣΔHf(products) − ΣΔHf(reactants).
ΔH = [(6 × -393.5) + (3 × -286)] − [2 × 48.5].
For 2 moles of benzene, ΔH = -6535 kJ.
The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products A and B along with the evolution of CO₂. The sum of spin-only magnetic moment values of A and B is ___ B.M. (Nearest integer).
Given atomic numbers: C = 6, Na = 11, O = 8, Fe = 26, Cr = 24
Chromite ore reacts with sodium carbonate to form Na₂CrO₄ and Fe₂O₃:
- Spin-only magnetic moment for Na₂CrO₄ (Cr⁶⁺) = 0 B.M. (no unpaired electrons).
- For Fe₃⁺ in Fe₂O₃: Magnetic moment = √35 ≈ 5.92 B.M. (nearest integer is 6).
In an atom, the total number of electrons having quantum numbers n = 4, |ml| = 1, and ms = -1/2 is:
For n = 4, |ml| = 1 corresponds to 3 orbitals (p, d, or f). Each orbital can hold one electron with ms = -1/2. Thus, the total number of electrons is 6.
Using the given figure, the ratio of Rf values of sample A and sample C is x × 10⁻². Value of x is:
Figure: Solvent front = 12.5 cm, Sample A = 5.0 cm, Sample C = 10.0 cm.
The Rf value is the ratio of the distance traveled by the sample to the distance traveled by the solvent front:
- Rf(A) = 5.0 / 12.5 = 0.4.
- Rf(C) = 10.0 / 12.5 = 0.8.
Ratio Rf(A) / Rf(C) = 0.4 / 0.8 = 0.5 = 50 × 10⁻².
In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is ________ g (nearest integer).
In the Claisen-Schmidt reaction:
1 mole of acetone reacts with 2 moles of benzaldehyde. Molar masses: Acetone = 58 g/mol, Benzaldehyde = 106 g/mol.
87 g of acetone corresponds to 1.5 moles. Benzaldehyde required = 1.5 × 2 × 106 = 318 g.
In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is ________ g (nearest integer).
In the Claisen-Schmidt reaction, 1 mole of acetone reacts with 2 moles of benzaldehyde. Given:
87 g of acetone corresponds to 1.5 moles. The required amount of benzaldehyde is:
1.5 × 2 × 106 = 318 g.
Consider the following single-step reaction in the gas phase at constant temperature:
2A(g) + B(g) → C(g)
The initial rate of the reaction is r₁ when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate r₂ is recorded when the pressure of C becomes 0.5 atm. The ratio r₁ : r₂ is ________ × 10⁻¹ (nearest integer).
Using the rate law for the reaction:
Rate ∝ [A]²[B].
Initial pressures: [A] = 1.5 atm, [B] = 0.7 atm.
After C reaches 0.5 atm, pressures are updated:
Calculating the ratio:
r₁ : r₂ = (1.5² × 0.7) : (0.5² × 0.2) = 315 × 10⁻¹.
The product C in the following sequence of reactions has ________ π bonds:
Reaction sequence: KMnO₄–KOH, ∆ → A; H₃O⁺ → B; Br₂/FeBr₃ → C
The sequence involves oxidation of the alkyl chain in a benzene derivative:
- Product C is para-bromobenzoic acid.
- It has 3 π bonds in the benzene ring and 1 π bond in the carboxyl group.
Total = 4 π bonds.
Considering acetic acid dissociates in water, its dissociation constant is 6.25 × 10⁻⁵. If 5 mL of acetic acid is dissolved in 1 liter of water, the solution will freeze at −x × 10⁻² °C, provided pure water freezes at 0 °C. x = _________. (Nearest integer).
Given: Kf(water) = 1.86 K·kg·mol⁻¹, density of acetic acid = 1.2 g·mL⁻¹, molar mass of acetic acid = 60 g·mol⁻¹, density of water = 1 g·cm⁻³.
Molality (m) of acetic acid:
Mass of acetic acid = 5 × 1.2 = 6 g, moles = 6 / 60 = 0.1 mol.
Molality = 0.1 / 1 = 0.1 m.
Considering dissociation:
Effective molality = 0.1 × (1 + α), where α = degree of dissociation.
∆Tf = i × Kf × m, substituting values gives x = 19.
Number of compounds from the following with zero dipole moment is ___________. HF, H₂, H₂S, CO₂, NH₃, BF₃, CH₄, CHCl₃, SiF₄, H₂O, BeF₂
Compounds with symmetrical geometry and no net dipole moment are:
- H₂, CO₂, BF₃, CH₄, SiF₄, BeF₂.
Total = 6 compounds.
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